Class 12 vector notes jee mains and advance

GauravArora393961 16 views 280 slides Oct 07, 2024
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About This Presentation

This contains all concpets required for jee mains and advance with practice question. One can clearly understand the concepts from this pdf


Slide Content

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??????
��
�
Ԧ�×�
Ԧ�
�

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Table of Contents
Session 01
Physical Quantities
Vector
Types of Vectors
03
04
07
08
Session 02
Section Formula
Position vector of a centroid
Position vector of a Incentre
47
48
54
55
Session 03
Tangent at a point:
Dot product or Scalar product
Algebraic Identities
79
72
80
91
Session 04
Linear combination of Vectors
Vector Product of Two Vectors
Lagrange’s Identity
Geometrical Interpretation of
Vector Product
107
108
109
118
120
Interpretation Of Vector Product
As Vector Area
126
Vector Triple Product 187
Session 05
Product of three or more vectors
Scalar Triple Product
Properties of Scalar Triple
Product
140
141
142
158
Session 06
Properties of Scalar Triple
product
Volume Of Tetrahedron
171
172
181
Session 07
Scalar Product of Four
Vectors
Vector Product of Four
Vectors
198
209
217
Angle between two vectors 33
Triangle Law of Vector Addition
Parallelogram Law of Vector
Addition
Properties of Scalar
Multiplication
56
59
71
Angle between two vectors
Triangle Inequality
92
96
Projection of �on Ԧ�97
Scalar product in terms of
components
103
Tetrahedron 134
Parallelopiped 138
Properties of Vector Triple
Product
190
Session 08
Theorem in Space (Fundamental
theorem in Space)
Linearly Dependent and
Independently Vectors
224
225
231
Reciprocal System of Vectors 239
Parametric Vector Equation
of a Straight Line
248
Session 09
Vector Equation of Angle Bisectors
between Two Straight Lines
Shortest distance between 2
skew lines
253
254
261

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Introduction to vectors and
its types
Session 01
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Physical Quantities
Only Magnitude, No Direction
E.g.: Speed, Mass and
Volume
Magnitude as well as Direction
Obeys vector laws of Algebra
E.g.: Velocity, Acceleration and
Displacement.
Scalar Quantity: Vector Quantity:

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Net force : �
1+�
2
�
Forces cannot be added directly, they obey vector law.
�
1
�
2


Is Force a Vector Quantity ?
So, it is a vector quantity. ●

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Is Angle a Vector Quantity ?
It has magnitude.
It has direction i.e.clockwise or anti–clockwise
It does not obeysvector law of algebra.
We can simply add two angles.
No, it is not a vector quantity.
??????
??????
+ve
−ve
Clockwise
Anti-clockwise




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Vector
Any directed line segment is called a vector.
Line of support
It is represented by ��
Its magnitude(modulus) is represented by ��or ��
The bigger the length of the directed line segment, the greater
the magnitude is.
�
(Initial point) (Terminal point)
�



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Types of Vectors

Zero Vector / Null vector 0:


It has same initial and terminal point.
It can have any direction of your choice.
Its magnitude is zero.

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Types of Vectors
Unit Vector:
Number of unit vectors perpendicular to a
given plane in space is 2.
Number of unit vectors perpendicular to a
given line in space is infinite.
Magnitude =1●

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Types of Vectors
Free Vectors:
A vector that maintains the same magnitude and direction regardless
of its position.
Velocity, DisplacementExample
Both vectors
are same
�
�

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Alocalized vectoris a vector where line of action and position are as
important as magnitude and direction. These vectors change with change
in position or direction.
Types of vectors:
Localized/Fixed Vectors

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Two or more vectors are collinear if their directed line segments are parallel
disregard with their magnitude.
Same direction
Like vectors
Ԧ�
Ԧ�
??????=0
°
Opposite direction
Unlike vectors
Ԧ�
Ԧ�
??????=180
°
Parallel Vectors
Types of vectors:
Collinear and Parallel Vectors

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Relationship between two parallel vectors:
If Ԧ�andԦ�are two parallel vectors, then there exists a non-zero scalar �
such that Ԧ�=�Ԧ�;�∈ℝ.
If Ԧ�andԦ�are two parallel vectors, then there exist two non-zero scalar
quantities λand �so that λԦ�+�Ԧ�=0.
Or

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Relationship between two parallel vectors:
● If Ԧ�andԦ�be two non-zero non-parallel vectorsthen
If λԦ�+�Ԧ�=
⇒λ=0;�=0
Ԧ�=0;Ԧ�=0
0
λ= 0;�=0
Ԧ�∥Ԧ�
or
or
λԦ�+ �Ԧ�=0

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Types of vectors:
●Obtained by joining the point to origin
with direction towards the point.
Any vector can be expressed as a linear
combination of 3orthonormal triad of unit vectors.
��=�ƶ�+�ƶ�+�ƶ�


Position Vector:

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Types of vectors:



Position Vector of �:
��
��= �
1
ƶ�+�
1
ƶ�+�
1
ƶ�
Position Vector of �:��=�
2
ƶ�+�
2
ƶ�+�
2
ƶ�
=(�
2−�
1)ƶ�+(�
2−�
1)ƶ�+(�
2−�
1)ƶ�=��−��
��
2,�
2,�
2
�
�
��
1,�
1,�
1
�
Example If point �(2,1,3)& point �(1,2,4)
=��−��
Then, ��=2ƶ�+1ƶ�+3ƶ�,��=1ƶ�+2ƶ�+4ƶ�
=��−��
��
��
=−Ƹ�+Ƹ�+෠�
=Ƹ�−Ƹ�−෠�

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4A
−6B
D
C −12
1
The value of �when Ԧ�=2ƶ�−3ƶ�+ƶ�and �=8ƶ�+�ƶ�+4ƶ�are parallel is

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Since Ԧ�||Ԧ�
2
8
=
−3
�
=
1
4
⇒Ԧ�=�Ԧ�
⇒�=−12
4A
−6B
D
C −12
1
The value of �when Ԧ�=2ƶ�−3ƶ�+ƶ�and �=8ƶ�+�ƶ�+4ƶ�are parallel is
Solution:

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−12
C
4A
−6B
D 1
The value of �when Ԧ�=2ƶ�−3ƶ�+ƶ�and �=8ƶ�+�ƶ�+4ƶ�are parallel is

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Uniquevalueforx,wherex∈0,
??????
6
A
Uniquevalueforx,wherex∈
??????
6
,
??????
3
B
D
C Novalueofx
Infinitevalueforx
The vectors Ԧ��=cos�ƶ�+sin�ƶ�and ��=�ƶ�+sin�ƶ�are collinear
for:

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Solution:Ԧ�=�Ԧ�;Clearly , �=1
cos�=�;
��=�−cos�Lets take
�0=−1;�
??????
6
<0;&#3627408467;
??????
3
>0
Using I.V.T., there exists at least one
solution in
??????
6
,
??????
3
Moreover,&#3627408467;

&#3627408485;=sin&#3627408485;+1
&#3627408467;is increasing function , only one solution in
??????
6
,
??????
3
0 ??????
6
??????
3
The vectors Ԧ&#3627408462;&#3627408485;=cos&#3627408485;ƶ&#3627408470;+sin&#3627408485;ƶ&#3627408471;and &#3627408463;&#3627408485;=&#3627408485;ƶ&#3627408470;+sin&#3627408485;ƶ&#3627408471;are collinear
for:

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Uniquevalueforx,wherex∈0,
??????
6
A
Uniquevalueforx,wherex∈
??????
6
,
??????
3
B
D
C Novalueofx
Infinitevalueforx
The vectors Ԧ&#3627408462;&#3627408485;=cos&#3627408485;ƶ&#3627408470;+sin&#3627408485;ƶ&#3627408471;and &#3627408463;&#3627408485;=&#3627408485;ƶ&#3627408470;+sin&#3627408485;ƶ&#3627408471;are collinear
for:

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If Ԧ&#3627408462;and &#3627408463;are non-collinear vector the value of &#3627408485;for which the
vectors Ԧ&#3627409148;=&#3627408485;−2Ԧ&#3627408462;+&#3627408463;and Ԧ&#3627409149;=3+2&#3627408485;Ԧ&#3627408462;−2&#3627408463;are collinear ?

1
4
A
1
4
B
D
C 1
−1

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Since the vectors, Ԧ&#3627409148;and Ԧ&#3627409149;are collinear
Therefore, the ratio of respective
components will be same

&#3627408485;−2
3+2&#3627408485;
=
1
−2
⇒4&#3627408485;−1=0

1
4
A
1
4
B
D
C 1
−1
&#3627408485;=
1
4
Solution:
If Ԧ&#3627408462;and &#3627408463;are non-collinear vector the value of &#3627408485;for which the
vectors Ԧ&#3627409148;=&#3627408485;−2Ԧ&#3627408462;+&#3627408463;and Ԧ&#3627409149;=3+2&#3627408485;Ԧ&#3627408462;−2&#3627408463;are collinear ?

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1
4
B

1
4
A
D
C 1
−1
If Ԧ&#3627408462;and &#3627408463;are non-collinear vector the value of &#3627408485;for which the
vectors Ԧ&#3627409148;=&#3627408485;−2Ԧ&#3627408462;+&#3627408463;and Ԧ&#3627409149;=3+2&#3627408485;Ԧ&#3627408462;−2&#3627408463;are collinear ?

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0,
5
2
,
5
4
A
5
2
,
5
2
,
5
4
B
D
C
5
2
,0,
5
4
5
2
,
5
4
,0
If the position vector of the points &#3627408436;,&#3627408437;,&#3627408438;,&#3627408439;are 0,2,1,3,1,1,−5,3,2
(2,4,1)respectively and if &#3627408451;&#3627408436;+&#3627408451;&#3627408437;+&#3627408451;&#3627408438;+&#3627408451;&#3627408439;=0,then the position
vector of &#3627408451;is

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0,
5
2
,
5
4
A
5
2
,
5
2
,
5
4
B
D
C
5
2
,0,
5
4
5
2
,
5
4
,0
&#3627408451;&#3627408436;+&#3627408451;&#3627408437;+&#3627408451;&#3627408438;+&#3627408451;&#3627408439;=0,
&#3627408450;&#3627408436;−&#3627408450;&#3627408451;+&#3627408450;&#3627408437;−&#3627408450;&#3627408451;+&#3627408450;&#3627408438;−&#3627408450;&#3627408451;
+&#3627408450;&#3627408439;−&#3627408450;&#3627408451;=0,
⇒4&#3627408450;&#3627408451;−(&#3627408450;&#3627408436;+&#3627408450;&#3627408437;+&#3627408450;&#3627408438;+&#3627408450;&#3627408439;)=0,
⇒&#3627408450;&#3627408451;=
??????&#3627408436;+??????&#3627408437;+??????&#3627408438;+??????&#3627408439;
4
⇒&#3627408450;&#3627408451;=0,
10
4
,
5
4
&#3627408470;.&#3627408466;.0,
5
2
,
5
4
Solution:
If the position vector of the points &#3627408436;,&#3627408437;,&#3627408438;,&#3627408439;are 0,2,1,3,1,1,−5,3,2
(2,4,1)respectively and if &#3627408451;&#3627408436;+&#3627408451;&#3627408437;+&#3627408451;&#3627408438;+&#3627408451;&#3627408439;=0,then the position
vector of &#3627408451;is

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A
5
2
,
5
2
,
5
4
B
D
C
5
2
,0,
5
4
5
2
,
5
4
,0
0,
5
2
,
5
4
If the position vector of the points &#3627408436;,&#3627408437;,&#3627408438;,&#3627408439;are 0,2,1,3,1,1,−5,3,2
(2,4,1)respectively and if &#3627408451;&#3627408436;+&#3627408451;&#3627408437;+&#3627408451;&#3627408438;+&#3627408451;&#3627408439;=0,then the position
vector of &#3627408451;is

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Types of vectors:
Coplanar Vector:
Coplanar vectors are the vectors which lie on the same plane, in a three-
dimensional space. These are vectors which are parallel to the same plane.
Ԧ&#3627408462;
Ԧ&#3627408463;
Ԧ&#3627408464;

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Types of vectors:
Coplanar Vector:
Two Vectors are always co-planar i.e. two vectors always lie in a plane.
Coplanar vectors are the vectors which lie on the same plane, in a three-
dimensional space. These are vectors which are parallel to the same plane.

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Types of vectors:
Coplanar Vector:
Coplanar vectors are the vectors which lie on the same plane, in a three-
dimensional space. These are vectors which are parallel to the same plane.
3or more vectors may or may not be co-planar.
&#3627408485;
&#3627408486;
&#3627408487;
ƶ&#3627408471;
ƶ&#3627408470;
ƶ&#3627408472;

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Types of vectors:
Coplanar Vector:
Two Vectors are always co-planar.
3or more vectors may or may not be co-planar.
In other words, two vectors always lie in a plane.
Non co-planar:
&#3627408485;
&#3627408486;
&#3627408487;
ƶ&#3627408471;
ƶ&#3627408470;ƶ&#3627408472;
Example

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Angle between two vectors:
Connect tail/terminal point of one vector with tail of another vector.
0≤??????≤??????
Tail of one vector should be connected withtail of another vector.
Ԧ&#3627408463;
Ԧ&#3627408462;
??????
Ԧ&#3627408462;
Ԧ&#3627408463;
Ԧ&#3627408462;
Ԧ&#3627408463;
??????
Ԧ&#3627408462;
Ԧ&#3627408463;
??????
Note
??????
Consider two vectors Ԧ&#3627408462;and Ԧ&#3627408463;.
This angle ??????,is the angle between two vectors Ԧ&#3627408462;and Ԧ&#3627408463;.



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Angle between two vectors:
Ԧ&#3627408462;
Ԧ&#3627408463;
??????
Ԧ&#3627408463;
150°
=30°
Ԧ&#3627408462;
Connect tail/terminal point of one vector with tail of another vector.
0≤??????≤??????
Tail of one vector should be connected with tail of another vector.Note
Consider two vectors Ԧ&#3627408462;and Ԧ&#3627408463;.
This angle ??????,is the angle between two vectors Ԧ&#3627408462;and Ԧ&#3627408463;.



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Magnitude of vectors(2D):
Position vector &#3627408450;&#3627408451;=&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;
Magnitude of &#3627408450;&#3627408451;=&#3627408450;&#3627408451;=Length of directed line segment
&#3627408450;&#3627408451;=&#3627408485;
2
+&#3627408486;
2
&#3627408459;
&#3627408460;
&#3627408450;
&#3627408486;
&#3627408485;
&#3627408451;&#3627408485;,&#3627408486;

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&#3627408450;&#3627408452;=&#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
Position vector &#3627408452;:&#3627408450;&#3627408452;=&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;+&#3627408487;ƶ&#3627408472;
Magnitude of &#3627408450;&#3627408452;=&#3627408450;&#3627408452;=Length of directed line segment
&#3627408459;
&#3627408460;
&#3627408452;&#3627408485;,&#3627408486;,&#3627408487;
&#3627408486;
&#3627408485;
&#3627408485;
2
+&#3627408487;
2
&#3627408487;
&#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
&#3627408450;


Magnitude of vectors(3D):

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Unit vector in the direction of Ԧ&#3627408462;:ො&#3627408462;=
Ԧ&#3627408462;
Ԧ&#3627408462;
Unit vector in the direction of Ԧ&#3627408462;

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Find the vector in the direction of Ԧ&#3627408462;=ƶ&#3627408470;−2ƶ&#3627408471;that has magnitude
7units.
Solution:

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Find the vector in the direction of Ԧ&#3627408462;=ƶ&#3627408470;−2ƶ&#3627408471;that has magnitude
7units.
Unit vector in the direction of Ԧ&#3627408462;∶ො&#3627408462;=
&#3627408462;
&#3627408462;
ො&#3627408462;=
ƶ&#3627408470;−2ƶ&#3627408471;
5
Required vector: 7ො&#3627408462;=7
ƶ&#3627408470;−2ƶ&#3627408471;
5
Solution:

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If one side of the square is represented by 3Ƹ&#3627408470;+4Ƹ&#3627408471;+5෠&#3627408472;, then
the area(in sq. units) of the square is ______
50A
25B
D
C 100
125

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If one side of the square is represented by 3Ƹ&#3627408470;+4Ƹ&#3627408471;+5෠&#3627408472;, then
the area(in sq. units) of the square is ______
50A
25B
D
C 100
125
Let, Ԧ&#3627408462;=3Ƹ&#3627408470;+4Ƹ&#3627408471;+5෠&#3627408472;
⇒|Ԧ&#3627408462;|=3
2
+4
2
+5
2
=52
Thus the length of side of square is 52
Hence, area of square is 52
2
=50sq. units
Solution:

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50A
If one side of the square is represented by 3Ƹ&#3627408470;+4Ƹ&#3627408471;+5෠&#3627408472;, then
the area(in sq. units) of the square is ______
25B
D
C 100
125

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Distance Formula:
| &#3627408436;&#3627408437;|=(&#3627408485;
2−&#3627408485;
1)
2
+&#3627408486;
2−&#3627408486;
1
2
+(&#3627408487;
2−&#3627408487;
1)
2
Position Vector of &#3627408436;: &#3627408450;&#3627408436;= &#3627408485;
1
ƶ&#3627408470;+&#3627408486;
1
ƶ&#3627408471;+&#3627408487;
1
ƶ&#3627408472;
Position Vector of &#3627408437;:&#3627408450;&#3627408437;=&#3627408485;
2
ƶ&#3627408470;+&#3627408486;
2
ƶ&#3627408471;+&#3627408487;
2
ƶ&#3627408472;
&#3627408436;&#3627408437;=&#3627408450;&#3627408437;−&#3627408450;&#3627408436;
=(&#3627408485;
2−&#3627408485;
1)ƶ&#3627408470;+(&#3627408486;
2−&#3627408486;
1)ƶ&#3627408471;+(&#3627408487;
2−&#3627408487;
1)ƶ&#3627408472;
Distance between &#3627408436;&#3627408437;=Magnitude of &#3627408436;&#3627408437;
&#3627408459;
&#3627408460;
&#3627408436;(&#3627408485;
1,&#3627408486;
1,&#3627408487;
1)
&#3627408437;(&#3627408485;
2,&#3627408486;
2,&#3627408487;
2)
&#3627408450;



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A
2&#3627408462;&#3627408481;
1−2&#3627408462;&#3627408481;
2+2&#3627408462;&#3627408481;
3+&#3627408462;&#3627408481;
1
2
+&#3627408462;&#3627408481;
2
2
+&#3627408462;&#3627408481;
2
B
D
C
2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
22
+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
22
2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
22
+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
22
+2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
22
2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
2
+(2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
2
)+(2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
2
)
Magnitude of the vector joining the points&#3627408451;(&#3627408462;&#3627408481;
1
2
,&#3627408462;&#3627408481;
2
2
,&#3627408462;&#3627408481;
3
2
)
and&#3627408452;(2&#3627408462;&#3627408481;
1,2&#3627408462;&#3627408481;
2,2&#3627408462;&#3627408481;
3)is

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The vector joining the points &#3627408451;(&#3627408462;&#3627408481;
1
2
,&#3627408462;&#3627408481;
2
2
,&#3627408462;&#3627408481;
3
2
)and
&#3627408452;(2&#3627408462;&#3627408481;
1,2&#3627408462;&#3627408481;
2,2&#3627408462;&#3627408481;
3)
can be obtained by
&#3627408451;&#3627408452;=Position vector of &#3627408452;−Position vector of &#3627408451;
&#3627408451;&#3627408452;=2&#3627408462;&#3627408481;
1Ƹ&#3627408470;+2&#3627408462;&#3627408481;
2Ƹ&#3627408471;+2&#3627408462;&#3627408481;
3
෠&#3627408472;−&#3627408462;&#3627408481;
1
2
Ƹ&#3627408470;+&#3627408462;&#3627408481;
2
2
Ƹ&#3627408471;+&#3627408462;&#3627408481;
3
2෠&#3627408472;
∴&#3627408451;&#3627408452;=2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
2
Ƹ&#3627408470;+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
2
Ƹ&#3627408471;+2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
2෠&#3627408472;
∴Scalar components of &#3627408451;&#3627408452;
are (2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
2
),(2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
2
),(2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
2
)
So magnitude of vector &#3627408451;&#3627408452;
=2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
22
+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
22
+2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
22
Solution:
Magnitude of the vector joining the points&#3627408451;(&#3627408462;&#3627408481;
1
2
,&#3627408462;&#3627408481;
2
2
,&#3627408462;&#3627408481;
3
2
)
and&#3627408452;(2&#3627408462;&#3627408481;
1,2&#3627408462;&#3627408481;
2,2&#3627408462;&#3627408481;
3)is

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A
2&#3627408462;&#3627408481;
1−2&#3627408462;&#3627408481;
2+2&#3627408462;&#3627408481;
3+&#3627408462;&#3627408481;
1
2
+&#3627408462;&#3627408481;
2
2
+&#3627408462;&#3627408481;
2
B
C
2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
22
+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
22
2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
2
+(2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
2
)+(2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
2
)
D 2&#3627408462;&#3627408481;
1−&#3627408462;&#3627408481;
1
22
+2&#3627408462;&#3627408481;
2−&#3627408462;&#3627408481;
2
22
+2&#3627408462;&#3627408481;
3−&#3627408462;&#3627408481;
3
22
Magnitude of the vector joining the points&#3627408451;(&#3627408462;&#3627408481;
1
2
,&#3627408462;&#3627408481;
2
2
,&#3627408462;&#3627408481;
3
2
)
and&#3627408452;(2&#3627408462;&#3627408481;
1,2&#3627408462;&#3627408481;
2,2&#3627408462;&#3627408481;
3)is

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Addition of vectors
Session 02
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Section Formula:
&#3627408436;&#3627408437;=&#3627408450;&#3627408437;−&#3627408450;&#3627408436;
=&#3627408485;
2−&#3627408485;
1
ƶ&#3627408470;+&#3627408486;
2−&#3627408486;
1
ƶ&#3627408471;+&#3627408487;
2−&#3627408487;
1
ƶ&#3627408472;
Distance between &#3627408436;&&#3627408437;=Magnitude of &#3627408436;&#3627408437;
&#3627408436;&#3627408437;=&#3627408485;
2−&#3627408485;
1
2
+&#3627408486;
2−&#3627408486;
1
2
+&#3627408487;
2−&#3627408487;
1
2
&#3627408459;
&#3627408460;
&#3627408436;&#3627408485;
1,&#3627408486;
1,&#3627408487;
1
&#3627408437;&#3627408485;
2,&#3627408486;
2,&#3627408487;
2
&#3627408450;

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Point &#3627408438;divides line segment &#3627408436;&#3627408437;in &#3627408474;:&#3627408475;ratio internally.
&#3627408451;.??????.of point &#3627408436;Ԧ&#3627408462;
&#3627408451;.??????.of point &#3627408437;Ԧ&#3627408463;
&#3627408451;.??????.of point &#3627408438;Ԧ&#3627408464;
Ԧ&#3627408464;=
&#3627408474;&#3627408463;+&#3627408475;&#3627408462;
&#3627408474;+&#3627408475;
&#3627408474; &#3627408475;
&#3627408438;Ԧ&#3627408464; &#3627408437;Ԧ&#3627408463;&#3627408436;Ԧ&#3627408462;
Section Formula:

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Point &#3627408438;divides line segment &#3627408436;&#3627408437;in &#3627408474;:&#3627408475;ratio externally.
&#3627408451;.??????.of point &#3627408436;Ԧ&#3627408462;
&#3627408451;.??????.of point &#3627408437;Ԧ&#3627408463;
&#3627408451;.??????.of point &#3627408438;Ԧ&#3627408464;
Ԧ&#3627408464;=
&#3627408474;&#3627408463;−&#3627408475;&#3627408462;
&#3627408474;−&#3627408475;
&#3627408475;
&#3627408438;Ԧ&#3627408464;&#3627408437;Ԧ&#3627408463;&#3627408436;Ԧ&#3627408462;
&#3627408474;
&#3627408436;&#3627408438;
&#3627408437;&#3627408438;
=
&#3627408474;
&#3627408475;
Section Formula:

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Point &#3627408453;divides the line segment &#3627408451;&#3627408452;internally in the ratio 2:3.Find the &#3627408451;.??????.
of point &#3627408453;if &#3627408451;.??????.of &#3627408451;is ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;and that of &#3627408452;is ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;.
A
B
D
C
5ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;
5
5ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
5

5ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;
5

5ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
5

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Point &#3627408453;divides the line segment &#3627408451;&#3627408452;internally in the ratio 2:3.Find the &#3627408451;.??????.
of point &#3627408453;if &#3627408451;.??????.of &#3627408451;is ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;and that of &#3627408452;is ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;.
A
B
D
C
5ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;
5
5ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
5

5ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;
5

5ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
5

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Point &#3627408453;divides the line segment &#3627408451;&#3627408452;internally in the ratio 2:3.Find the &#3627408451;.??????.
of point &#3627408453;if &#3627408451;.??????.of &#3627408451;is ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;and that of &#3627408452;is ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;.
2 3
&#3627408453;Ԧ&#3627408479;&#3627408452;ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;&#3627408451;ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;&#3627408451;.??????.of &#3627408453;Ԧ&#3627408479;=
2ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;+3ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
2+3
⇒&#3627408453;Ԧ&#3627408479;=
5ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
5
Solution:

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&#3627408442;(Centroid) is the intersection point of all
three medians of a triangle. &#3627408436;&#3627408442;:&#3627408442;&#3627408448;=2:1
Position vector of a centroid:
&#3627408448;
&#3627408463;+Ԧ&#3627408464;
2
&#3627408442;
1
&#3627408464;&#3627408463;
&#3627408438;Ԧ&#3627408464;
&#3627408437;Ԧ&#3627408463;
&#3627408436;Ԧ&#3627408462;
2
It divides the median &#3627408436;&#3627408448;in the ratio 2:1.
&#3627408451;.??????.of centroid Ԧ&#3627408442;=
2
&#3627408463;+&#3627408464;
2
+&#3627408462;
2+1
&#3627408451;.??????.of centroid Ԧ&#3627408442;=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
3

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Position vector of a Incentre:
&#3627408439;
&#3627408463;&#3627408478;+&#3627408464;Ԧ&#3627408479;
&#3627408463;+&#3627408464;
??????
&#3627408464;&#3627408463;
&#3627408438;&#3627408479;
&#3627408437;Ԧ&#3627408478;
&#3627408436;Ԧ&#3627408477;
??????(Incentre) is the intersection point of three
bisectors of the interior angles of a triangle.
Incentre divides angle bisector &#3627408436;&#3627408439;in ratio &#3627408463;+&#3627408464;:&#3627408462;
&#3627408451;.??????.of Incentre=
&#3627408462;Ԧ&#3627408477;+&#3627408463;+&#3627408464;
&#3627408463;&#3627408478;+&#3627408464;&#3627408479;
&#3627408463;+&#3627408464;
&#3627408462;+&#3627408463;+&#3627408464;
&#3627408451;.??????.of Incentre =
&#3627408462;Ԧ&#3627408477;+&#3627408463;&#3627408478;+&#3627408464;Ԧ&#3627408479;
&#3627408462;+&#3627408463;+&#3627408464;

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Addition of Vectors:
Triangle Law of Vector Addition
If two vectors are represented by two consecutive sides of a triangle then their sum
is represented by the third side of the triangle but in the opposite direction.
Let &#3627408436;&#3627408437;=Ԧ&#3627408462;,&#3627408437;&#3627408438;=Ԧ&#3627408463;and &#3627408436;&#3627408438;=Ԧ&#3627408464;,
&#3627408436;
Ԧ&#3627408463;
Ԧ&#3627408462;
&#3627408437;
&#3627408437;
&#3627408438;
&#3627408436;
&#3627408438;
Ԧ&#3627408464;
Then, &#3627408436;&#3627408437;+&#3627408437;&#3627408438;=&#3627408436;&#3627408438;
⇒Ԧ&#3627408462;+Ԧ&#3627408463;=Ԧ&#3627408464;
Converse of triangle law is also true.

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Triangle Law of Vector Addition:
&#3627408436;
Ԧ&#3627408462;&#3627408437;
Ԧ&#3627408463;
&#3627408438;
Ԧ&#3627408464;
(Initial point)
(Final point)
Ԧ&#3627408462;+Ԧ&#3627408463;=Ԧ&#3627408464;
&#3627408436;
Ԧ&#3627408462;
&#3627408437;
Ԧ&#3627408463;
&#3627408438;
Ԧ&#3627408464;
(Initial point)
Ԧ&#3627408462;+Ԧ&#3627408464;=Ԧ&#3627408463;
(Final point)

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Ԧ&#3627408463;+Ԧ&#3627408464;=Ԧ&#3627408462; Ԧ&#3627408464;=Ԧ&#3627408463;−Ԧ&#3627408462;
(Final point)
&#3627408436;
Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;&#3627408438;
Ԧ&#3627408464;
(initial
point)
(Final
point)
&#3627408436;
Ԧ&#3627408462;
&#3627408437;
Ԧ&#3627408463;
Ԧ&#3627408464;
(Initial point)
Triangle Law of Vector Addition:

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Parallelogram Law of Vector Addition:
Iftwovectorsarerepresentedbytwoadjacentsidesofaparallelogram,then
theirsumisrepresentedbythediagonaloftheparallelogramwhoseinitialpoint
isthesameastheinitialpointofgivenvectors.
&#3627408436;
Ԧ&#3627408462;
&#3627408437;
&#3627408439;
Ԧ&#3627408463;
&#3627408438;
Ԧ&#3627408463;
Ԧ&#3627408462;
Ԧ&#3627408462;+Ԧ&#3627408463;
Ԧ&#3627408462;−Ԧ&#3627408463;

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&#3627408436;&#3627408437;&#3627408438;&#3627408439;is a parallelogram whose diagonals meet at &#3627408451;.If &#3627408450;is a fixed point,
then &#3627408450;&#3627408436;+&#3627408450;&#3627408437;+&#3627408450;&#3627408438;+&#3627408450;&#3627408439;equals?
A
B
C
D
&#3627408450;&#3627408451;
2&#3627408450;&#3627408451;
3&#3627408450;&#3627408451;
4&#3627408450;&#3627408451;
&#3627408436; &#3627408437;
&#3627408439; &#3627408438;
&#3627408451;
&#3627408450;

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=&#3627408450;&#3627408451;+&#3627408451;&#3627408436;+&#3627408450;&#3627408451;+&#3627408451;&#3627408437;+&#3627408450;&#3627408451;+&#3627408451;&#3627408438;+&#3627408450;&#3627408451;+&#3627408451;&#3627408439;
=4&#3627408450;&#3627408451;+&#3627408451;&#3627408436;+&#3627408451;&#3627408438;+&#3627408451;&#3627408437;+&#3627408451;&#3627408439;∵&#3627408451;bisect the two diagonals
=4&#3627408450;&#3627408451;+&#3627408451;&#3627408436;−&#3627408451;&#3627408436;+&#3627408451;&#3627408437;−&#3627408451;&#3627408437;
=4&#3627408450;&#3627408451;
&#3627408450;&#3627408436;+&#3627408450;&#3627408437;+&#3627408450;&#3627408438;+&#3627408450;&#3627408439;
Solution:
&#3627408436;&#3627408437;&#3627408438;&#3627408439;is a parallelogram whose diagonals meet at &#3627408451;.If &#3627408450;is a fixed point,
then &#3627408450;&#3627408436;+&#3627408450;&#3627408437;+&#3627408450;&#3627408438;+&#3627408450;&#3627408439;equals?

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If the midpoint of a consecutive sides of a quadrilateral are joined,
prove that the resulting quadrilateral is a parallelogram.
Solution:
&#3627408451;
&#3627408437;
&#3627408436;
&#3627408439;
&#3627408438;
&#3627408452;
&#3627408453;
&#3627408454;
In ∆&#3627408436;&#3627408439;&#3627408438;,(Mid point Theorem)
&#3627408453;&#3627408454;||&#3627408436;&#3627408438;and &#3627408453;&#3627408454;=
1
2
&#3627408436;&#3627408438;
Similarly, In ∆&#3627408436;&#3627408437;&#3627408438;,
&#3627408451;&#3627408452;||&#3627408436;&#3627408438;and &#3627408451;&#3627408452;=
1
2
&#3627408436;&#3627408438;
∴&#3627408451;&#3627408452;||&#3627408453;&#3627408454;and &#3627408451;&#3627408452;=&#3627408453;&#3627408454;
Similarly,
&#3627408451;&#3627408454;||&#3627408452;&#3627408453;and &#3627408451;&#3627408454;=&#3627408452;&#3627408454;
Hence it is a parallelogram

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If Ԧ&#3627408462;and Ԧ&#3627408463;represented by the sides &#3627408436;&#3627408437;and &#3627408437;&#3627408438;of a regular hexagon
&#3627408436;&#3627408437;&#3627408438;&#3627408439;&#3627408440;&#3627408441;,then vector represented by &#3627408441;&#3627408436;(in terms of Ԧ&#3627408462;and Ԧ&#3627408463;) will be ?
A
B
D
C
Ԧ&#3627408462;−Ԧ&#3627408463;
Ԧ&#3627408463;−Ԧ&#3627408462;
−Ԧ&#3627408462;−Ԧ&#3627408463;
Ԧ&#3627408462;+Ԧ&#3627408463;

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If Ԧ&#3627408462;and Ԧ&#3627408463;represented by the sides &#3627408436;&#3627408437;and &#3627408437;&#3627408438;of a regular hexagon
&#3627408436;&#3627408437;&#3627408438;&#3627408439;&#3627408440;&#3627408441;,then vector represented by &#3627408441;&#3627408436;(in terms of Ԧ&#3627408462;and Ԧ&#3627408463;) will be ?
A
B
D
C
Ԧ&#3627408462;−Ԧ&#3627408463;
Ԧ&#3627408463;−Ԧ&#3627408462;
−Ԧ&#3627408462;−Ԧ&#3627408463;
Ԧ&#3627408462;+Ԧ&#3627408463;

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If Ԧ&#3627408462;and Ԧ&#3627408463;represented by the sides &#3627408436;&#3627408437;and &#3627408437;&#3627408438;of a regular hexagon
&#3627408436;&#3627408437;&#3627408438;&#3627408439;&#3627408440;&#3627408441;,then vector represented by &#3627408441;&#3627408436;(in terms of Ԧ&#3627408462;and Ԧ&#3627408463;) will be ?
Solution:
&#3627408436;
Ԧ&#3627408462;&#3627408437;
&#3627408441;
&#3627408440;&#3627408439;
&#3627408438;
Ԧ&#3627408463;
&#3627408441;&#3627408438;=2Ԧ&#3627408462;
&#3627408436;&#3627408438;=Ԧ&#3627408462;+Ԧ&#3627408463;
(property of hexagon)
⇒&#3627408441;&#3627408436;+&#3627408436;&#3627408438;=&#3627408441;&#3627408438;
⇒&#3627408441;&#3627408436;+Ԧ&#3627408462;+Ԧ&#3627408463;=2Ԧ&#3627408462;
⇒&#3627408441;&#3627408436;=Ԧ&#3627408462;−Ԧ&#3627408463;
Using triangle law,

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Consider a parallelogram &#3627408436;&#3627408437;&#3627408438;&#3627408439;. &#3627408448;is mid point of &#3627408436;&#3627408439;.&#3627408437;&#3627408448;when extended
meet &#3627408438;&#3627408439;at &#3627408452;.&#3627408436;&#3627408438;when joined meet &#3627408437;&#3627408448;at &#3627408453;. Find &#3627408452;&#3627408453;:&#3627408453;&#3627408437;.
Solution:

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Consider a parallelogram &#3627408436;&#3627408437;&#3627408438;&#3627408439;. &#3627408448;is mid point of &#3627408436;&#3627408439;.&#3627408437;&#3627408448;when extended
meet &#3627408438;&#3627408439;at &#3627408452;.&#3627408436;&#3627408438;when joined meet &#3627408437;&#3627408448;at &#3627408453;. Find &#3627408452;&#3627408453;:&#3627408453;&#3627408437;.
&#3627408436;0
&#3627408439;Ԧ&#3627408465; &#3627408438;Ԧ&#3627408463;+Ԧ&#3627408465;
&#3627408448;
&#3627408453;
&#3627408437;Ԧ&#3627408463;
&#3627409158;
1
Position vector of &#3627408453;=&#3627408472;Ԧ&#3627408463;+Ԧ&#3627408465;
=
&#3627409158;&#3627408463;+
&#3627408465;
2
&#3627409158;+1
On comparing,
1
2&#3627409158;+1
=&#3627408472;
&#3627409158;
&#3627409158;+1
=&#3627408472;,
⇒&#3627409158;=
1
2
, &#3627408472;=
1
3
⇒&#3627408453;≡
1
3
Ԧ&#3627408463;+Ԧ&#3627408465;
&#3627408452;
Ԧ&#3627408465;
2
Solution:

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Consider a parallelogram &#3627408436;&#3627408437;&#3627408438;&#3627408439;. &#3627408448;is mid point of &#3627408436;&#3627408439;.&#3627408437;&#3627408448;when extended
meet &#3627408438;&#3627408439;at &#3627408452;.&#3627408436;&#3627408438;when joined meet &#3627408437;&#3627408448;at &#3627408453;. Find &#3627408452;&#3627408453;:&#3627408453;&#3627408437;.
&#3627408436;0
&#3627408439;Ԧ&#3627408465; &#3627408438;Ԧ&#3627408463;+Ԧ&#3627408465;
&#3627408448;
&#3627408453;
1
Ԧ&#3627408465;
2
&#3627409159;1
&#3627408480;
On comparing,1=
1
2
&#3627408480;+1,−&#3627409159;=−&#3627408480;
⇒&#3627409159;=&#3627408480;=1
Position vector of &#3627408452;≡
&#3627409159;+1Ԧ&#3627408465;−&#3627409159;&#3627408463;+Ԧ&#3627408465;
&#3627409159;+1−&#3627409159;
=Ԧ&#3627408465;−&#3627409159;Ԧ&#3627408463;
Position vector of &#3627408452;≡
&#3627408480;+1
&#3627408465;
2
−&#3627408480;&#3627408463;
&#3627408480;+1−&#3627408480;
=&#3627408480;+1
Ԧ&#3627408465;
2
−&#3627408480;Ԧ&#3627408463;
∴Position vector of &#3627408452;≡Ԧ&#3627408465;−Ԧ&#3627408463;
&#3627408452;&#3627408453;
&#3627408453;&#3627408437;
=
1
3
&#3627408463;+Ԧ&#3627408465;−Ԧ&#3627408465;−&#3627408463;
&#3627408463;−
1
3
&#3627408463;+Ԧ&#3627408465;
=
2
1
⇒&#3627408453;≡
1
3
Ԧ&#3627408463;+Ԧ&#3627408465; &#3627408452;
&#3627408437;Ԧ&#3627408463;
Solution:

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Linear combination of Vectors :
A vector Ԧ&#3627408479;is said to be a linear combination of the vectors
&#3627408462;
1,&#3627408462;
2,&#3627408462;
3,…,&#3627408462;
&#3627408475;,if there exist scalars &#3627408474;
1,&#3627408474;
2,…,&#3627408474;
&#3627408475;such that
Ԧ&#3627408479;=&#3627408474;
1&#3627408462;
1+&#3627408474;
2&#3627408462;
2+⋯+&#3627408474;
&#3627408475;&#3627408462;
&#3627408475;→linear combination of vectors
Fundamental Theorem in Plane :
If Ԧ&#3627408462;and Ԧ&#3627408463;are two non-zero non collinear vectorsthen
any vector Ԧ&#3627408479;coplanar with them can be expressed as a
linear combination Ԧ&#3627408479;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;.
Ԧ&#3627408462;
Ԧ&#3627408479;
&#3627408463;

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If Ԧ&#3627408462;=&#3627408485;Ƹ&#3627408470;+&#3627408486;Ƹ&#3627408471;+&#3627408487;෠&#3627408472;
then, &#3627408472;Ԧ&#3627408462;=&#3627408472;&#3627408485;Ƹ&#3627408470;+&#3627408472;&#3627408486;Ƹ&#3627408471;+&#3627408472;&#3627408487;෠&#3627408472;
Ԧ&#3627408462;
3&#3627408462;
Example:Ԧ&#3627408462;=2Ƹ&#3627408470;+Ƹ&#3627408471;+3෠&#3627408472;
then, 3Ԧ&#3627408462;=6Ƹ&#3627408470;+3Ƹ&#3627408471;+9෠&#3627408472;
3Ԧ&#3627408462;has same direction as vector &#3627408462;,with magnitude3times that of Ԧ&#3627408462;.
−2Ԧ&#3627408462;=−4Ƹ&#3627408470;+−2Ƹ&#3627408471;+−6෠&#3627408472;
−2Ԧ&#3627408462;hasoppositedirectionasvector
&#3627408462;,withmagnitude2timesthatofԦ&#3627408462;.
Ԧ&#3627408462;
2Ԧ&#3627408462;
Multiplication of a Vector by a Scalar:

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Properties of Scalar Multiplication:
For&#3627408472;,&#3627408473;,&#3627408474;∈ℝ
&#3627408472;Ԧ&#3627408462;+Ԧ&#3627408463;=&#3627408472;Ԧ&#3627408462;+&#3627408472;Ԧ&#3627408463;
&#3627408472;&#3627408474;Ԧ&#3627408462;=&#3627408472;&#3627408474;Ԧ&#3627408462;
&#3627408472;+&#3627408473;Ԧ&#3627408462;=&#3627408472;Ԧ&#3627408462;+&#3627408473;Ԧ&#3627408462;
Multiplication of a Vector by a Scalar:

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Collinearity of three points:
IfԦ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;andbepositionvectorofthreepoints&#3627408436;,&#3627408437;and&#3627408438;respectivelyand&#3627408485;,&#3627408486;,&#3627408487;be
threenon-zeroscalarsthenthenecessaryandsufficientconditionsforthree
pointstobecollinearis&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;=0;where&#3627408485;+&#3627408486;+&#3627408487;=0.
Three points &#3627408436;,&#3627408437;and &#3627408438;are collinear, if &#3627408436;&#3627408437;=&#3627409158;&#3627408437;&#3627408438;
Proof:&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;=0
⇒&#3627408487;Ԧ&#3627408464;=−&#3627408485;Ԧ&#3627408462;−&#3627408486;Ԧ&#3627408463;
⇒Ԧ&#3627408464;=−
&#3627408485;&#3627408462;+&#3627408486;&#3627408463;
&#3627408487;
∵&#3627408485;+&#3627408486;+&#3627408487;=0⇒&#3627408487;=−&#3627408485;+&#3627408486;
⇒Ԧ&#3627408464;=
&#3627408485;&#3627408462;+&#3627408486;&#3627408463;
&#3627408485;+&#3627408486;
⇒section formula

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If 2Ԧ&#3627408462;−3Ԧ&#3627408463;,Ԧ&#3627408463;and Ԧ&#3627408462;−Ԧ&#3627408463;are position vectors of three points &#3627408436;,&#3627408437;and &#3627408438;
then they are:
A
B
D
C
Collinear
Non-collinear
Cannot say
None of these

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If 2Ԧ&#3627408462;−3Ԧ&#3627408463;,Ԧ&#3627408463;and Ԧ&#3627408462;−Ԧ&#3627408463;are position vectors of three points &#3627408436;,&#3627408437;and &#3627408438;
then they are:
A
B
D
C
Collinear
Non-collinear
Cannot say
None of these

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If 2Ԧ&#3627408462;−3Ԧ&#3627408463;,Ԧ&#3627408463;and Ԧ&#3627408462;−Ԧ&#3627408463;are position vectors of three points &#3627408436;,&#3627408437;and &#3627408438;
then they are:
⇒12Ԧ&#3627408462;−3Ԧ&#3627408463;+1Ԧ&#3627408463;−2Ԧ&#3627408462;−Ԧ&#3627408463;=0
⇒1+1−2=0
∴Points are collinear.
Solution:
&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;=0
&#3627408485;+&#3627408486;+&#3627408487;=0

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&#3627408436;=2Ƹ&#3627408470;+3Ƹ&#3627408471;;&#3627408437;=&#3627408477;Ƹ&#3627408470;+9Ƹ&#3627408471;;&#3627408438;=Ƹ&#3627408470;−Ƹ&#3627408471;are collinear, then the value of &#3627408477;is?
A
B
D
C
1
2
3
2
7
2
5
2

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&#3627408436;=2Ƹ&#3627408470;+3Ƹ&#3627408471;;&#3627408437;=&#3627408477;Ƹ&#3627408470;+9Ƹ&#3627408471;;&#3627408438;=Ƹ&#3627408470;−Ƹ&#3627408471;are collinear, then the value of &#3627408477;is?
A
B
D
C
1
2
3
2
7
2
5
2

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&#3627408436;=2Ƹ&#3627408470;+3Ƹ&#3627408471;;&#3627408437;=&#3627408477;Ƹ&#3627408470;+9Ƹ&#3627408471;;&#3627408438;=Ƹ&#3627408470;−Ƹ&#3627408471;are collinear, then the value of &#3627408477;is?
Solution:
&#3627408436;&#3627408437;=&#3627408477;−2Ƹ&#3627408470;+6Ƹ&#3627408471;
&#3627408436;&#3627408438;=−Ƹ&#3627408470;−4Ƹ&#3627408471;
&#3627408436;&#3627408437;∥&#3627408436;&#3627408438;;

&#3627408477;−2
−1
=
6
−4
⇒&#3627408477;=
7
2

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Scalar Product of vectors
Session 03
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The dot product or scalar product of two vectors Ԧ&#3627408462;
and Ԧ&#3627408463;,denoted by Ԧ&#3627408462;⋅Ԧ&#3627408463;is equal to
Dot product or Scalar product:
0≤??????≤??????
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;cos??????
where ??????is the angle between their tails or heads
Ԧ&#3627408463;
Ԧ&#3627408462;
??????

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Properties of Dot Product
➢Ԧ&#3627408462;⋅Ԧ&#3627408463;∈ℝ
➢The dot product of a zero and non-zero vector is a
scalar zero i.e.0⋅Ԧ&#3627408462;=0
Ԧ&#3627408462;⋅Ԧ&#3627408463;>0
Ԧ&#3627408462;⋅Ԧ&#3627408463;<0
Ԧ&#3627408462;⋅Ԧ&#3627408463;=0
or at least one of Ԧ&#3627408462;or Ԧ&#3627408463;is a zero vector.
⇒??????∈0,
??????
2
⇒??????∈
??????
2
,??????
⇒??????=
??????
2
➢If angle between Ԧ&#3627408462;and Ԧ&#3627408463;is ??????,

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Properties of Dot Product
➢Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408462;
➢Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408464;=Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408462;⋅Ԧ&#3627408464;
➢If ??????=0
➢If ??????=??????
➢If ො&#3627408462;and ෠&#3627408463;are unit vectors,
➢Ԧ&#3627408462;⋅Ԧ&#3627408462;=Ԧ&#3627408462;
2
⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=|Ԧ&#3627408462;||Ԧ&#3627408463;|(Like vectors)
⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=−|Ԧ&#3627408462;||Ԧ&#3627408463;|(Unlike vectors)
⇒Ԧ&#3627408462;=Ԧ&#3627408462;⋅Ԧ&#3627408462;
then ො&#3627408462;⋅෠&#3627408463;=cos??????
➢If Ԧ&#3627408462;⊥Ԧ&#3627408463;⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=0
orԦ&#3627408463;=0or Ԧ&#3627408462;⊥Ԧ&#3627408463;but if Ԧ&#3627408462;⋅Ԧ&#3627408463;=0

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If angle between Ԧ&#3627408462;and Ԧ&#3627408463;is 120°and their magnitudes are respectively 2
and 3, then Ԧ&#3627408462;⋅Ԧ&#3627408463;equals
A
B
D
C
3
−3
3
−3

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Solution:
If angle between Ԧ&#3627408462;and Ԧ&#3627408463;is 120°and their magnitudes are respectively 2
and 3, then Ԧ&#3627408462;⋅Ԧ&#3627408463;equals
Ԧ&#3627408462;⋅Ԧ&#3627408463;
=23cos120°
=23−
1
2
=−3
=Ԧ&#3627408462;Ԧ&#3627408463;cos??????

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B
If angle between Ԧ&#3627408462;and Ԧ&#3627408463;is 120°and their magnitudes are respectively 2
and 3, then Ԧ&#3627408462;⋅Ԧ&#3627408463;equals
D
C
−3
3
−3
A 3

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➢If Ԧ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;and &#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;
then Ԧ&#3627408462;⋅Ԧ&#3627408463;=&#3627408462;
1&#3627408463;
1+&#3627408462;
2&#3627408463;
2+&#3627408462;
3&#3627408463;
3
Properties of Dot Product
➢If ƶ&#3627408470;,ƶ&#3627408471;and ෠&#3627408472;are unit vectors along the rectangular
coordinate axes &#3627408450;&#3627408459;,&#3627408450;&#3627408460;and &#3627408450;&#3627408461;then
ƶ&#3627408470;⋅ƶ&#3627408470;=ƶ&#3627408471;⋅ƶ&#3627408471;=෠&#3627408472;⋅෠&#3627408472;=1
ƶ&#3627408470;⋅ƶ&#3627408471;=ƶ&#3627408471;⋅෠&#3627408472;=෠&#3627408472;⋅ƶ&#3627408470;=0
&#3627408485;
&#3627408486;
&#3627408487;
ƶ&#3627408471;
ƶ&#3627408470;
ƶ&#3627408472;

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Properties of Dot Product
➢Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;⋅Ԧ&#3627408464;
➢(Ԧ&#3627408462;⋅Ԧ&#3627408463;)⋅Ԧ&#3627408464;is meaningless
Ԧ&#3627408462;=0
⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;−Ԧ&#3627408464;=0 Ԧ&#3627408463;=Ԧ&#3627408464;
Ԧ&#3627408462;⊥Ԧ&#3627408463;−Ԧ&#3627408464;

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If Ԧ&#3627408462;=3ƶ&#3627408470;+2ƶ&#3627408471;+ƶ&#3627408472;and Ԧ&#3627408463;=ƶ&#3627408470;−2ƶ&#3627408471;+5ƶ&#3627408472;then find Ԧ&#3627408462;⋅Ԧ&#3627408463;.
A
B
D
C
4
5
3
−3

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Ԧ&#3627408462;⋅Ԧ&#3627408463;
=3−4+5
=4
=31+2−2+15 Ԧ&#3627408462;⋅Ԧ&#3627408463;=&#3627408462;
1&#3627408463;
1+&#3627408462;
2&#3627408463;
2+&#3627408462;
3&#3627408463;
3
Solution:
If Ԧ&#3627408462;=3ƶ&#3627408470;+2ƶ&#3627408471;+ƶ&#3627408472;and Ԧ&#3627408463;=ƶ&#3627408470;−2ƶ&#3627408471;+5ƶ&#3627408472;then find Ԧ&#3627408462;⋅Ԧ&#3627408463;.

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If Ԧ&#3627408462;=3ƶ&#3627408470;+2ƶ&#3627408471;+ƶ&#3627408472;and Ԧ&#3627408463;=ƶ&#3627408470;−2ƶ&#3627408471;+5ƶ&#3627408472;then find Ԧ&#3627408462;⋅Ԧ&#3627408463;.
B
D
C
4
5
3
−3
A

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Algebraic Identities
➢(Ԧ&#3627408462;+Ԧ&#3627408463;)
2
=|Ԧ&#3627408462;|
2
+2Ԧ&#3627408462;⋅Ԧ&#3627408463;+|Ԧ&#3627408463;|
2
➢(Ԧ&#3627408462;−Ԧ&#3627408463;)
2
=|Ԧ&#3627408462;|
2
−2Ԧ&#3627408462;⋅Ԧ&#3627408463;+|Ԧ&#3627408463;|
2
➢Ԧ&#3627408462;+Ԧ&#3627408463;⋅Ԧ&#3627408462;−Ԧ&#3627408463;=|Ԧ&#3627408462;|
2
−|Ԧ&#3627408463;|
2
➢Ԧ&#3627408462;+Ԧ&#3627408463;=|Ԧ&#3627408462;|+|Ԧ&#3627408463;|
➢|Ԧ&#3627408462;+Ԧ&#3627408463;|
2
=|Ԧ&#3627408462;|
2
+|Ԧ&#3627408463;|
2
➢|Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;|
2
=Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;⋅Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;
=|Ԧ&#3627408462;|
2
+|Ԧ&#3627408463;|
2
+|Ԧ&#3627408464;|
2
+2Ԧ&#3627408462;⋅Ԧ&#3627408463;+2Ԧ&#3627408463;⋅Ԧ&#3627408464;+2Ԧ&#3627408464;⋅Ԧ&#3627408462;
⇒Ԧ&#3627408462;∥Ԧ&#3627408463;
⇒Ԧ&#3627408462;⊥Ԧ&#3627408463;⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=0⇒Ԧ&#3627408462;=0or Ԧ&#3627408463;=0

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Angle between two vectors
LetԦ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;and &#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;and
??????be the angle between them, then
⇒cos??????=
&#3627408462;
1&#3627408463;
1+&#3627408462;
2&#3627408463;
2+&#3627408462;
3&#3627408463;
3
&#3627408462;
1
2
+&#3627408462;
2
2
+&#3627408462;
3
2
∙&#3627408463;
1
2
+&#3627408463;
2
2
+&#3627408463;
3
2
⇒cos??????=
Ԧ&#3627408462;⋅Ԧ&#3627408463;
|Ԧ&#3627408462;|⋅|Ԧ&#3627408463;|
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;cos??????
Ԧ&#3627408462;
??????
Ԧ&#3627408463;
If Ԧ&#3627408462;and Ԧ&#3627408463;are perpendicular to each other then &#3627408462;
1&#3627408463;
1
+&#3627408462;
2&#3627408463;
2+&#3627408462;
3&#3627408463;
3=0
Note

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Angle between two vectors
LetԦ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;and &#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;and
??????be the angle between them, then
⇒cos??????=
Ԧ&#3627408462;⋅Ԧ&#3627408463;
|Ԧ&#3627408462;|⋅|Ԧ&#3627408463;|
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;cos??????
Ԧ&#3627408462;
??????
Ԧ&#3627408463;
Ԧ&#3627408462;⋅Ԧ&#3627408463;
&#3627408474;&#3627408462;&#3627408485;
=Ԧ&#3627408462;Ԧ&#3627408463;??????=0°
Ԧ&#3627408462;⋅Ԧ&#3627408463;
&#3627408474;&#3627408470;&#3627408475;
=−Ԧ&#3627408462;Ԧ&#3627408463;??????=??????
Cauchy Schwartz Inequality : −Ԧ&#3627408462;Ԧ&#3627408463;≤Ԧ&#3627408462;⋅Ԧ&#3627408463;≤Ԧ&#3627408462;Ԧ&#3627408463;

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Determine the values of &#3627408464;such that the vectors &#3627408464;&#3627408485;ƶ&#3627408470;−6ƶ&#3627408471;+3ƶ&#3627408472;and
&#3627408485;ƶ&#3627408470;+2ƶ&#3627408471;+2&#3627408464;&#3627408485;ƶ&#3627408472;∀&#3627408485;∈&#3627408453;make an obtuse angle with each other.
Let ??????be the angle between the given vectors
cos??????=
&#3627408464;&#3627408485;
2
−12+6&#3627408464;&#3627408485;
&#3627408464;
2
&#3627408485;
2
+6
2
+3
2
&#3627408485;
2
+2
2
+4&#3627408464;
2
&#3627408485;
2
If ??????is obtuse, then cos??????<0
⇒&#3627408464;<0and6&#3627408464;
2
−4(&#3627408464;)(−12)<0
⇒&#3627408464;&#3627408485;
2
+6&#3627408464;&#3627408485;−12<0∀&#3627408485;∈ℝ
cos??????=
&#3627408462;1&#3627408463;1+&#3627408462;2&#3627408463;2+&#3627408462;3&#3627408463;3
&#3627408462;
1
2
+&#3627408462;
2
2
+&#3627408462;
3
2
&#3627408463;
1
2
+&#3627408463;
2
2
+&#3627408463;
3
2
Ԧ&#3627408462;⋅Ԧ&#3627408463;<0
Ԧ&#3627408462;=&#3627408464;&#3627408485;Ƹ&#3627408470;−6Ƹ&#3627408471;+3෠&#3627408472;& Ԧ&#3627408463;=&#3627408485;Ƹ&#3627408470;+2Ƹ&#3627408471;+2&#3627408464;&#3627408485;෠&#3627408472;
cos??????=
&#3627408462;⋅&#3627408463;
&#3627408462;⋅&#3627408463;
<0
Solution:

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Determine the values of &#3627408464;such that the vectors &#3627408464;&#3627408485;ƶ&#3627408470;−6ƶ&#3627408471;+3ƶ&#3627408472;and
&#3627408485;ƶ&#3627408470;+2ƶ&#3627408471;+2&#3627408464;&#3627408485;ƶ&#3627408472;∀&#3627408485;∈&#3627408453;make an obtuse angle with each other.
⇒−
4
3
<&#3627408464;≤0
⇒&#3627408464;<0and6&#3627408464;
2
−4(&#3627408464;)(−12)<0
⇒&#3627408464;<0and&#3627408464;(3&#3627408464;+4)<0
for &#3627408464;=0∶
cos??????=
&#3627408462;1&#3627408463;1+&#3627408462;2&#3627408463;2+&#3627408462;3&#3627408463;3
&#3627408462;
1
2
+&#3627408462;
2
2
+&#3627408462;
3
2
&#3627408463;
1
2
+&#3627408463;
2
2
+&#3627408463;
3
2

4
3
0
&#3627408464;&#3627408485;
2
+6&#3627408464;&#3627408485;−12<0∀&#3627408485;∈ℝ
⇒&#3627408464;<0and3&#3627408464;
2
+4&#3627408464;<0
⇒&#3627408464;∈−
4
3
,0∪0
−12<0⇒cos??????<0→obtuse angle
&#3627408464;=0
Solution:
Let ??????be the angle between the given vectors

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Triangle Inequality
Ԧ&#3627408462;+Ԧ&#3627408463;≤Ԧ&#3627408462;+Ԧ&#3627408463;
Equality holds when ??????=0°
Ԧ&#3627408462;+Ԧ&#3627408463;=Ԧ&#3627408462;
2
+Ԧ&#3627408463;
2
+2Ԧ&#3627408462;Ԧ&#3627408463;cos??????
Ԧ&#3627408462;+Ԧ&#3627408463;≤Ԧ&#3627408462;
2
+Ԧ&#3627408463;
2
+2Ԧ&#3627408462;Ԧ&#3627408463;
1
2
Ԧ&#3627408462;
Ԧ&#3627408463;
??????
Ԧ&#3627408462;+Ԧ&#3627408463;
Ԧ&#3627408462;+Ԧ&#3627408463;=Ԧ&#3627408462;+Ԧ&#3627408463;⇒Ԧ&#3627408462;||Ԧ&#3627408463;
Ԧ&#3627408462;+Ԧ&#3627408463;
2
=Ԧ&#3627408462;
2
+2Ԧ&#3627408462;Ԧ&#3627408463;+Ԧ&#3627408463;
2
Maximise
cos??????
Ԧ&#3627408462;+Ԧ&#3627408463;=Ԧ&#3627408462;
2
+Ԧ&#3627408463;
2
+2Ԧ&#3627408462;Ԧ&#3627408463;

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Projection of &#3627408463;on Ԧ&#3627408462;
Geometrically, the scalar product of two vectors is equal to the product of the magnitude
of one and the projection of second in the direction of first vector.
Ԧ&#3627408462;⋅Ԧ&#3627408463;=|Ԧ&#3627408462;|(projection of Ԧ&#3627408463;in the direction of Ԧ&#3627408462;)
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;cos??????
Similarly, Ԧ&#3627408462;⋅Ԧ&#3627408463;=|Ԧ&#3627408463;|(projection of Ԧ&#3627408462;in the direction of Ԧ&#3627408463;)
Projection of Ԧ&#3627408463;on Ԧ&#3627408462;=Ԧ&#3627408463;cos??????=
&#3627408462;⋅&#3627408463;
|&#3627408462;|
=Ԧ&#3627408463;⋅ො&#3627408462;
Projection of Ԧ&#3627408462;on Ԧ&#3627408463;=Ԧ&#3627408462;cos??????=
&#3627408462;⋅&#3627408463;
|&#3627408463;|
=Ԧ&#3627408462;⋅෠&#3627408463;
=Ԧ&#3627408462;×&#3627408450;&#3627408448;⇒&#3627408450;&#3627408448;=Ԧ&#3627408463;cos??????
??????
Ԧ&#3627408462;
&#3627408450;
&#3627408448;
&#3627408436;
&#3627408437;
Ԧ&#3627408463;

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Projection of &#3627408463;on Ԧ&#3627408462;
Geometrically, the scalar product of two vectors is equal to the product of the magnitude
of one and the projection of second in the direction of first vector.
??????
Ԧ&#3627408462;
&#3627408450;
&#3627408448;
&#3627408436;
&#3627408437;
Ԧ&#3627408463;
Projection =
+&#3627408483;&#3627408466;;??????isacute,Ԧ&#3627408462;⋅Ԧ&#3627408463;>0
−&#3627408483;&#3627408466;;??????isobtuse,Ԧ&#3627408462;⋅Ԧ&#3627408463;<0
0;??????=
??????
2
,Ԧ&#3627408462;⋅Ԧ&#3627408463;=0
Projection of Ԧ&#3627408463;on Ԧ&#3627408462;=
&#3627408462;⋅&#3627408463;
|&#3627408462;|
Projection of Ԧ&#3627408462;on Ԧ&#3627408463;=
&#3627408462;⋅&#3627408463;
|&#3627408463;|

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The projection of vector ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;on the vector ƶ&#3627408470;−ƶ&#3627408471;+ƶ&#3627408472;is ____ .
A B
DC
3
1
3
2
3
23
Projection =
ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;ƶ&#3627408470;−ƶ&#3627408471;+ƶ&#3627408472;
|ƶ&#3627408470;−ƶ&#3627408471;+ƶ&#3627408472;|
=
1
3
=
1−1+1
1+1+1
Projection of Ԧ&#3627408463;on Ԧ&#3627408462;=
&#3627408462;⋅&#3627408463;
|&#3627408462;|
Solution:

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A
B
D
C
25
4
85
14
127
20
115
16
In a triangle &#3627408436;&#3627408437;&#3627408438;, if &#3627408437;&#3627408438;=8,&#3627408438;&#3627408436;=7and &#3627408436;&#3627408437;=10, then the projection of
the vector &#3627408436;&#3627408437;on &#3627408436;&#3627408438;is equal to:

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In a triangle &#3627408436;&#3627408437;&#3627408438;, if &#3627408437;&#3627408438;=8,&#3627408438;&#3627408436;=7and &#3627408436;&#3627408437;=10, then the projection of
the vector &#3627408436;&#3627408437;on &#3627408436;&#3627408438;is equal to:
Projection of &#3627408436;&#3627408437;on &#3627408436;&#3627408438;
|Ԧ&#3627408462;|=8
⇒Ԧ&#3627408463;⋅Ԧ&#3627408464;=
85
2
,|Ԧ&#3627408463;|=7,|Ԧ&#3627408464;|=10
Ԧ&#3627408462;Ԧ&#3627408464;
&#3627408463;
??????
&#3627408436; &#3627408438;
&#3627408437;
=Ԧ&#3627408464;⋅Ԧ&#3627408463;=
Ԧ&#3627408464;⋅&#3627408463;
&#3627408463;
=
&#3627408463;⋅Ԧ&#3627408464;
7
Ԧ&#3627408464;+Ԧ&#3627408462;=Ԧ&#3627408463;⇒Ԧ&#3627408462;=Ԧ&#3627408463;−Ԧ&#3627408464;
Ԧ&#3627408462;
2
=Ԧ&#3627408463;−Ԧ&#3627408464;
2
⇒64=Ԧ&#3627408463;
2
+Ԧ&#3627408464;
2
−2Ԧ&#3627408463;⋅Ԧ&#3627408464;
⇒2Ԧ&#3627408463;⋅Ԧ&#3627408464;=100+49−64
Since,Ԧ&#3627408464;⋅Ԧ&#3627408463;=
Ԧ&#3627408464;⋅&#3627408463;
&#3627408463;
=
&#3627408463;⋅Ԧ&#3627408464;
7
⇒Ԧ&#3627408464;⋅Ԧ&#3627408463;=
85
2
7
=
85
14
Solution:

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D
C
In a triangle &#3627408436;&#3627408437;&#3627408438;, if &#3627408437;&#3627408438;=8,&#3627408438;&#3627408436;=7and &#3627408436;&#3627408437;=10, then the projection of
the vector &#3627408436;&#3627408437;on &#3627408436;&#3627408438;is equal to:
25
4
85
14
127
20
115
16
A
B

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Scalar product in terms of components
➢For any vector Ԧ&#3627408462;,
➢Let Ԧ&#3627408462;and Ԧ&#3627408463;be two vectors such that
ThenԦ&#3627408462;⋅Ԧ&#3627408463;=&#3627408462;
1&#3627408463;
1+&#3627408462;
2&#3627408463;
2+&#3627408462;
3&#3627408463;
3
➢In particular,Ԧ&#3627408462;⋅Ԧ&#3627408462;=|Ԧ&#3627408462;|
2
Ԧ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;and Ԧ&#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;
Ԧ&#3627408462;=(Ԧ&#3627408462;⋅ƶ&#3627408470;)ƶ&#3627408470;+(Ԧ&#3627408462;⋅ƶ&#3627408471;)ƶ&#3627408471;+(Ԧ&#3627408462;⋅ƶ&#3627408472;)ƶ&#3627408472;

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Vector Components :
Ԧ&#3627408462;
??????
&#3627408450;
&#3627408449;
&#3627408436;
&#3627408437;Ԧ&#3627408463;&#3627408448;
Let Ԧ&#3627408462;and Ԧ&#3627408463;be two vectors represented by &#3627408450;&#3627408436;&
&#3627408450;&#3627408437;and ??????be the angle between them.
Component of Ԧ&#3627408462;along Ԧ&#3627408463;=&#3627408450;&#3627408448;=&#3627408450;&#3627408448;෠&#3627408463;
=
&#3627408462;⋅&#3627408463;
&#3627408463;
2
Ԧ&#3627408463;=Ԧ&#3627408462;⋅෠&#3627408463;෠&#3627408463;
Component of Ԧ&#3627408462;perpendicular to Ԧ&#3627408463;=&#3627408450;&#3627408449;
Ԧ&#3627408462;=&#3627408450;&#3627408448;+&#3627408448;&#3627408436;=&#3627408450;&#3627408448;+&#3627408450;&#3627408449;
&#3627408450;&#3627408449;=Ԧ&#3627408462;−&#3627408450;&#3627408448;
∴&#3627408450;&#3627408449;=Ԧ&#3627408462;−
&#3627408462;⋅&#3627408463;
&#3627408463;
2
Ԧ&#3627408463;

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Find the vector components of a vector 2ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;along and perpendicular
to non-zero vector 2ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;.
Solution:

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Let Ԧ&#3627408462;=2ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;and Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;
=
192ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;
9
Vector component of Ԧ&#3627408462;along Ԧ&#3627408463;
Ԧ&#3627408462;⋅Ԧ&#3627408463;=4+3+12=19
=
&#3627408462;⋅&#3627408463;
&#3627408463;
2
Ԧ&#3627408463;
Ԧ&#3627408463;=3
Vector component of Ԧ&#3627408462;perpendicular to Ԧ&#3627408463;
=2Ƹ&#3627408470;+3Ƹ&#3627408471;+6෠&#3627408472;−
19
9
2Ƹ&#3627408470;+Ƹ&#3627408471;+2෠&#3627408472;
=
18Ƹ&#3627408470;+27Ƹ&#3627408471;+54෠&#3627408472;−38Ƹ&#3627408470;−19Ƹ&#3627408471;−38෠&#3627408472;
9
=
1
9
−20Ƹ&#3627408470;+8Ƹ&#3627408471;+16෠&#3627408472;
Find the vector components of a vector 2ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;along and perpendicular
to non-zero vector 2ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;.
Solution:

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Vector Product of two
vectors
Session 04
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Linear combination of Vectors :
A vector Ԧ&#3627408479;is said to be a linear combination of the vectors
&#3627408462;
1,&#3627408462;
2,&#3627408462;
3,…,&#3627408462;
&#3627408475;,if there exist scalars &#3627408474;
1,&#3627408474;
2,…,&#3627408474;
&#3627408475;such that
Ԧ&#3627408479;−&#3627408474;
1&#3627408462;
1+&#3627408474;
2&#3627408462;
2+⋯+&#3627408474;
&#3627408475;&#3627408462;
&#3627408475;→linear combination of vectors
Fundamental Theorem in Plane :
If Ԧ&#3627408462;and Ԧ&#3627408463;are two non-zero non collinear vectorsthen
any vector Ԧ&#3627408479;coplanar with them can be expressed as a
linear combination Ԧ&#3627408479;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;.
Ԧ&#3627408462;
Ԧ&#3627408479;
&#3627408463;

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Vector Product of Two Vectors :
where ƶ&#3627408475;is a direction of vector Ԧ&#3627408462;×Ԧ&#3627408463;which isperpendicular
to the plane of Ԧ&#3627408462;and Ԧ&#3627408463;
Definition :
Ԧ&#3627408463;
Ԧ&#3627408462;
??????
ො&#3627408475;
Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;sin??????ො&#3627408475;
&#3627408514;
&#3627408515;
&#3627408514;×&#3627408515;

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Properties of Vector Product :
Ԧ&#3627408463;
Ԧ&#3627408462;
??????
ො&#3627408475;
Ԧ&#3627408462;×Ԧ&#3627408463;=−Ԧ&#3627408463;×Ԧ&#3627408462;
Ԧ&#3627408462;×Ԧ&#3627408462;=0
Ԧ&#3627408462;×(Ԧ&#3627408463;±Ԧ&#3627408464;)=Ԧ&#3627408462;×Ԧ&#3627408463;±Ԧ&#3627408462;×Ԧ&#3627408464;
For a scalar &#3627408474;,&#3627408474;Ԧ&#3627408462;×Ԧ&#3627408463;=&#3627408474;Ԧ&#3627408462;×Ԧ&#3627408463;
If Ԧ&#3627408462;||Ԧ&#3627408463;then ??????=0or ??????
IfԦ&#3627408462;⊥Ԧ&#3627408463;thenԦ&#3627408462;×Ԧ&#3627408463;=|Ԧ&#3627408462;||Ԧ&#3627408463;|ƶ&#3627408475;
=Ԧ&#3627408462;×&#3627408474;Ԧ&#3627408463;
⇒Ԧ&#3627408462;×Ԧ&#3627408463;=0
(or|Ԧ&#3627408462;×Ԧ&#3627408463;|=|Ԧ&#3627408462;||Ԧ&#3627408463;|)
Ԧ&#3627408462;=0
Ԧ&#3627408462;×Ԧ&#3627408463;=0 Ԧ&#3627408463;=0
Ԧ&#3627408462;∥Ԧ&#3627408463;

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ƶ&#3627408470;×ƶ&#3627408470;=ƶ&#3627408471;×ƶ&#3627408471;=ƶ&#3627408472;×ƶ&#3627408472;=0
ƶ&#3627408470;×ƶ&#3627408471;=ƶ&#3627408472;,ƶ&#3627408471;×ƶ&#3627408472;=ƶ&#3627408470;and ƶ&#3627408472;×ƶ&#3627408470;=ƶ&#3627408471;
If??????isanglebetweenԦ&#3627408462;andԦ&#3627408463;thensinθ=
|&#3627408462;×&#3627408463;|
|&#3627408462;∥&#3627408463;|
If Ԧ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;and Ԧ&#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;then
Ԧ&#3627408462;×Ԧ&#3627408463;=
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
=&#3627408462;
2&#3627408463;
3−&#3627408462;
3&#3627408463;
2
ƶ&#3627408470;+&#3627408462;
3&#3627408463;
1−&#3627408462;
1&#3627408463;
3
ƶ&#3627408471;+&#3627408462;
1&#3627408463;
2−&#3627408462;
2&#3627408463;
1
ƶ&#3627408472;
Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;sin??????ො&#3627408475;
Ԧ&#3627408462;
Ԧ&#3627408463;
Properties of Vector Product :

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If Ԧ&#3627408462;=2ƶ&#3627408470;+2ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408463;=6ƶ&#3627408470;−3ƶ&#3627408471;+2ƶ&#3627408472;then Ԧ&#3627408462;×Ԧ&#3627408463;equals:
A
B
D
C
2ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;
6ƶ&#3627408470;−3ƶ&#3627408471;+2ƶ&#3627408472;
ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
ƶ&#3627408470;−10ƶ&#3627408471;−18ƶ&#3627408472;

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Solution:
If Ԧ&#3627408462;=2ƶ&#3627408470;+2ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408463;=6ƶ&#3627408470;−3ƶ&#3627408471;+2ƶ&#3627408472;then Ԧ&#3627408462;×Ԧ&#3627408463;equals:
Ԧ&#3627408462;×Ԧ&#3627408463;=
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;
22−1
6−32
=ƶ&#3627408470;4−3−ƶ&#3627408471;4+6+ƶ&#3627408472;−6−12
=ƶ&#3627408470;−10ƶ&#3627408471;−18ƶ&#3627408472;

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If Ԧ&#3627408462;=2ƶ&#3627408470;+2ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408463;=6ƶ&#3627408470;−3ƶ&#3627408471;+2ƶ&#3627408472;then Ԧ&#3627408462;×Ԧ&#3627408463;equals:
A
B
D
C
2ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;
6ƶ&#3627408470;−3ƶ&#3627408471;+2ƶ&#3627408472;
ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
ƶ&#3627408470;−10ƶ&#3627408471;−18ƶ&#3627408472;

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If angle between Ԧ&#3627408462;=ƶ&#3627408470;−2ƶ&#3627408471;+3ƶ&#3627408472;and Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;is ??????then sin??????equals:
A
B
D
C
5
7
3
14
5
27
5
21

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If angle between Ԧ&#3627408462;=ƶ&#3627408470;−2ƶ&#3627408471;+3ƶ&#3627408472;and Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;is ??????then sin??????equals:
Solution:
We know that sin??????=
|&#3627408462;×&#3627408463;|
|&#3627408462;||&#3627408463;|
=
53
146
=
5
28
=
5
27
cos??????=
ො&#3627408462;⋅෠&#3627408463;
&#3627408462;⋅&#3627408463;
=
2−2+3
14×16
=
3
27
sin??????=1−cos
2
??????⇒sin??????=1−
3
28
=
25
28
=
5
27

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If angle between Ԧ&#3627408462;=ƶ&#3627408470;−2ƶ&#3627408471;+3ƶ&#3627408472;and Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;is ??????then sin??????equals:
A
B
D
C
5
7
3
14
5
27
5
21

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Lagrange’s Identity :
Ԧ&#3627408462;
2Ԧ&#3627408463;
2
sin
2
??????=Ԧ&#3627408462;
2Ԧ&#3627408463;
2
−Ԧ&#3627408462;
2Ԧ&#3627408463;
2
cos
2
??????
⇒Ԧ&#3627408462;×Ԧ&#3627408463;
2
=Ԧ&#3627408462;
2Ԧ&#3627408463;
2
−Ԧ&#3627408462;⋅Ԧ&#3627408463;
2

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Solution:
For any vector Ԧ&#3627408462;,provethat|Ԧ&#3627408462;×ƶ&#3627408470;|
2
+|Ԧ&#3627408462;×ƶ&#3627408471;|
2
+|Ԧ&#3627408462;×ƶ&#3627408472;|
2
=2|&#3627408462;|
2
=2&#3627408462;
2
Let Ԧ&#3627408462;=&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;+&#3627408487;ƶ&#3627408472;
Ԧ&#3627408462;×Ԧ&#3627408463;
2
=Ԧ&#3627408462;
2Ԧ&#3627408463;
2
−Ԧ&#3627408462;Ԧ&#3627408463;
2
Ԧ&#3627408462;×Ƹ&#3627408470;
2
=Ԧ&#3627408462;
2
Ƹ&#3627408470;
2
−Ԧ&#3627408462;Ƹ&#3627408470;
2
Taking L.H.S
|Ԧ&#3627408462;×ƶ&#3627408470;|
2
+|Ԧ&#3627408462;×ƶ&#3627408471;|
2
+|Ԧ&#3627408462;×ƶ&#3627408472;|
2
=Ԧ&#3627408462;
2
−&#3627408485;
2
+Ԧ&#3627408462;
2
−&#3627408486;
2
+Ԧ&#3627408462;
2
−&#3627408487;
=3Ԧ&#3627408462;
2
−&#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
=3Ԧ&#3627408462;
2
−Ԧ&#3627408462;
2

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Geometrical Interpretation of Vector Product:
The vector product of two vectors Ԧ&#3627408462;and Ԧ&#3627408463;we represent a vector whose
modulus is equal to area of the parallelogram with adjacent side Ԧ&#3627408462;and Ԧ&#3627408463;.
Ԧ&#3627408463;
Ԧ&#3627408462;
&#3627408463;sin??????
??????
Area of parallelogram =Base ×Height
=&#3627408462;&#3627408463;sin??????=Ԧ&#3627408462;×Ԧ&#3627408463;

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Find the area of a parallelogram whose two adjacent sides are
represented by Ԧ&#3627408462;=3Ƹ&#3627408470;+Ƹ&#3627408471;+2෠&#3627408472;and Ԧ&#3627408463;=2Ƹ&#3627408470;−2Ƹ&#3627408471;+4෠&#3627408472;.
Ԧ&#3627408463;
Ԧ&#3627408462;
Solution:
Area of parallelogram =Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408462;×Ԧ&#3627408463;=
Ƹ&#3627408470;Ƹ&#3627408471;෠&#3627408472;
312
2−24
=8Ƹ&#3627408470;−8Ƹ&#3627408471;−8෠&#3627408472;
∴Area =8Ƹ&#3627408470;−8Ƹ&#3627408471;−8෠&#3627408472;
⇒Area =83Sq. unit

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If Ԧ&#3627408462;=Ƹ&#3627408470;+&#3627409148;Ƹ&#3627408471;+3෠&#3627408472;and Ԧ&#3627408463;=3Ƹ&#3627408470;−&#3627409148;Ƹ&#3627408471;+෠&#3627408472;,the area of parallelogram whose
adjacent sides are represented by vectors Ԧ&#3627408462;and Ԧ&#3627408463;is 83square
units, then Ԧ&#3627408462;⋅Ԧ&#3627408463;is equal to:
JEE MAIN FEB 2021
Solution:

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If Ԧ&#3627408462;=Ƹ&#3627408470;+&#3627409148;Ƹ&#3627408471;+3෠&#3627408472;and Ԧ&#3627408463;=3Ƹ&#3627408470;−&#3627409148;Ƹ&#3627408471;+෠&#3627408472;,the area of parallelogram whose
adjacent sides are represented by vectors Ԧ&#3627408462;and Ԧ&#3627408463;is 83square
units, then Ԧ&#3627408462;⋅Ԧ&#3627408463;is equal to:
JEE MAIN FEB 2021
Solution:
Area of parallelogram =Ԧ&#3627408462;×Ԧ&#3627408463;
=Ƹ&#3627408470;+&#3627409148;Ƹ&#3627408471;+3෠&#3627408472;×3Ƹ&#3627408470;−&#3627409148;Ƹ&#3627408471;+෠&#3627408472;
⇒83=4&#3627409148;Ƹ&#3627408470;+8Ƹ&#3627408471;−4&#3627409148;෠&#3627408472;
⇒643=16&#3627409148;
2
+64+16&#3627409148;
2
⇒12=2&#3627409148;
2
+4
⇒&#3627409148;
2
=4

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If Ԧ&#3627408462;=Ƹ&#3627408470;+&#3627409148;Ƹ&#3627408471;+3෠&#3627408472;and Ԧ&#3627408463;=3Ƹ&#3627408470;−&#3627409148;Ƹ&#3627408471;+෠&#3627408472;,the area of parallelogram whose
adjacent sides are represented by vectors Ԧ&#3627408462;and Ԧ&#3627408463;is 83square
units, then Ԧ&#3627408462;⋅Ԧ&#3627408463;is equal to:
JEE MAIN FEB 2021
Solution:
⇒&#3627409148;
2
=4
Ԧ&#3627408462;=Ƹ&#3627408470;+&#3627409148;Ƹ&#3627408471;+3෠&#3627408472;,Ԧ&#3627408463;=3Ƹ&#3627408470;−&#3627409148;Ƹ&#3627408471;+෠&#3627408472;
Ԧ&#3627408462;⋅Ԧ&#3627408463;=3−&#3627409148;
2
+3
=6−&#3627409148;
2
=6−4
⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=2

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Geometrical interpretation of Vector product:
Area of a triangle:
Area of triangle &#3627408436;&#3627408437;&#3627408438;
&#3627408436;Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;
&#3627408438;Ԧ&#3627408464;
=
1
2
Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408464;×Ԧ&#3627408462;

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Interpretation Of Vector Product As Vector Area:
Vector area of plane figures:
Witheverycloseboundsurfacewhichhasbeen
describedinacertainspecificmannerandwhose
boundariesdonotcross,itispossibletoassociate
adirectedlinesegmentԦ&#3627408464;suchthat:
Ԧ&#3627408464;=Area enclosed by the plane figure.
The support of Ԧ&#3627408464;is perpendicular to the area.
The sense of description of the boundaries and the direction of
Ԧ&#3627408464;is in accordance with the Right Hand screw rule.
Ԧ&#3627408464;
Ԧ&#3627408464;
Into the
plane
Outward
Not
possible

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Interpretation Of Vector Product As Vector Area:
Note:
If 3points with position vectors Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;are collinear
⇒Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;=0

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Unit vector perpendicular to the plane of the Δ&#3627408436;&#3627408437;&#3627408438;
ො&#3627408475;=±
&#3627408462;×&#3627408463;+&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×&#3627408462;

where Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are the position vector of
the angular points of the triangle &#3627408436;&#3627408437;&#3627408438;.
&#3627408436;Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;&#3627408438;Ԧ&#3627408464;
ො&#3627408475;
Interpretation Of Vector Product As Vector Area:
Note:

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Vector Area of a Quadrilateral &#3627408436;&#3627408437;&#3627408438;&#3627408439;
&#3627408436; &#3627408437;
&#3627408438;&#3627408439;
=Vector area of Δ&#3627408436;&#3627408437;&#3627408438;+Vector area of Δ&#3627408436;&#3627408438;&#3627408439;
=
1
2
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;+
1
2
&#3627408436;&#3627408438;×&#3627408436;&#3627408439;
=
1
2
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;−&#3627408436;&#3627408439;×&#3627408436;&#3627408438;
=
1
2
&#3627408436;&#3627408437;−&#3627408436;&#3627408439;×&#3627408436;&#3627408438;
=
1
2
&#3627408436;&#3627408437;+&#3627408439;&#3627408436;×&#3627408436;&#3627408438;
=
1
2
&#3627408439;&#3627408437;×&#3627408436;&#3627408438;
∴Area of &#3627408436;&#3627408437;&#3627408438;&#3627408439;=
1
2
&#3627408439;&#3627408437;×&#3627408436;&#3627408438;=
1
2
&#3627408436;&#3627408438;×&#3627408437;&#3627408439;
Interpretation Of Vector Product As Vector Area:

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If Ԧ&#3627408462;=Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;,Ԧ&#3627408463;=2Ƹ&#3627408470;−෠&#3627408472;,Ԧ&#3627408464;=2Ƹ&#3627408471;+෠&#3627408472;are the position vectors of the angular
points of the triangle &#3627408436;&#3627408437;&#3627408438;,find
I.A Vector of magnitude 6perpendicular to the plane &#3627408436;&#3627408437;&#3627408438;.
II.Area of triangle &#3627408436;&#3627408437;&#3627408438;.
III.Length of the altitude from &#3627408436;.
Solution: &#3627408436;Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;
&#3627408437;2Ƹ&#3627408470;−෠&#3627408472;&#3627408438;2Ƹ&#3627408471;+෠&#3627408472;

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If Ԧ&#3627408462;=Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;,Ԧ&#3627408463;=2Ƹ&#3627408470;−෠&#3627408472;,Ԧ&#3627408464;=2Ƹ&#3627408471;+෠&#3627408472;are the position vectors of the angular
points of the triangle &#3627408436;&#3627408437;&#3627408438;,find
I.A Vector of magnitude 6perpendicular to the plane &#3627408436;&#3627408437;&#3627408438;.
II.Area of triangle &#3627408436;&#3627408437;&#3627408438;.
III.Length of the altitude from &#3627408436;.
&#3627408436;Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;
&#3627408437;2Ƹ&#3627408470;−෠&#3627408472;&#3627408438;2Ƹ&#3627408471;+෠&#3627408472;
I.A Vector of magnitude 6perpendicular to the
plane &#3627408436;&#3627408437;&#3627408438;.
=6×unit vector in direction of area vector of Δ&#3627408436;&#3627408437;&#3627408438;
=±6
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;
&#3627408436;&#3627408437;=Ƹ&#3627408470;+Ƹ&#3627408471;−3෠&#3627408472;
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;=8Ƹ&#3627408470;+4Ƹ&#3627408471;+4෠&#3627408472;
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;=8Ƹ&#3627408470;+4Ƹ&#3627408471;+4෠&#3627408472;
&#3627408436;&#3627408438;=−Ƹ&#3627408470;+3Ƹ&#3627408471;−෠&#3627408472;
Solution:

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If Ԧ&#3627408462;=Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;,Ԧ&#3627408463;=2Ƹ&#3627408470;−෠&#3627408472;,Ԧ&#3627408464;=2Ƹ&#3627408471;+෠&#3627408472;are the position vectors of the angular
points of the triangle &#3627408436;&#3627408437;&#3627408438;,find
I.A Vector of magnitude 6perpendicular to the plane &#3627408436;&#3627408437;&#3627408438;.
II.Area of triangle &#3627408436;&#3627408437;&#3627408438;.
III.Length of the altitude from &#3627408436;.
&#3627408436;&#3627408437;=Ƹ&#3627408470;+Ƹ&#3627408471;−3෠&#3627408472;
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;=8Ƹ&#3627408470;+4Ƹ&#3627408471;+4෠&#3627408472;
&#3627408436;&#3627408438;=−Ƹ&#3627408470;+3Ƹ&#3627408471;−෠&#3627408472; &#3627408436;Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;
&#3627408437;2Ƹ&#3627408470;−෠&#3627408472;&#3627408438;2Ƹ&#3627408471;+෠&#3627408472;
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;=46
=±6×
8Ƹ&#3627408470;+4Ƹ&#3627408471;+4෠&#3627408472;
46
=±2Ƹ&#3627408470;+Ƹ&#3627408471;+෠&#3627408472;
II.Area of triangle &#3627408436;&#3627408437;&#3627408438;.
1
2
&#3627408436;&#3627408437;×&#3627408436;&#3627408438;=26sq. unit
Solution:

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If Ԧ&#3627408462;=Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;,Ԧ&#3627408463;=2Ƹ&#3627408470;−෠&#3627408472;,Ԧ&#3627408464;=2Ƹ&#3627408471;+෠&#3627408472;are the position vectors of the angular
points of the triangle &#3627408436;&#3627408437;&#3627408438;,find
I.A Vector of magnitude 6perpendicular to the plane &#3627408436;&#3627408437;&#3627408438;.
II.Area of triangle &#3627408436;&#3627408437;&#3627408438;.
III.Length of the altitude from &#3627408436;.
Solution:
III.Length of the altitude from &#3627408436;.
Area of Δ&#3627408436;&#3627408437;&#3627408438;=
1
2
×Length of &#3627408437;&#3627408438;×Altitude
from &#3627408436;
⇒26=
1
2
×23×Altitude from &#3627408436;
&#3627408436;Ƹ&#3627408470;−Ƹ&#3627408471;+2෠&#3627408472;
&#3627408437;2Ƹ&#3627408470;−෠&#3627408472;&#3627408438;2Ƹ&#3627408471;+෠&#3627408472;
⇒Length of Altitude from &#3627408436;
=22

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Tetrahedron :
&#3627408436;Ԧ&#3627408462;
&#3627408437;&#3627408463;
&#3627408438;Ԧ&#3627408464;
&#3627408439;Ԧ&#3627408465;
&#3627408447;
&#3627408447;

&#3627408462;
1
&#3627408462;
2
&#3627408462;
4
&#3627408462;
3
&#3627408462;
1
&#3627408463;+Ԧ&#3627408464;+Ԧ&#3627408465;
3
→Δ&#3627408437;&#3627408438;&#3627408439;
&#3627408462;
2
&#3627408462;+Ԧ&#3627408464;+Ԧ&#3627408465;
3
→Δ&#3627408436;&#3627408438;&#3627408439;
&#3627408462;
3
&#3627408462;+&#3627408463;+Ԧ&#3627408465;
3
→Δ&#3627408436;&#3627408437;&#3627408439;
&#3627408462;
4
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
3
→ΔA&#3627408437;&#3627408438;
&#3627408447;
&#3627408463;+Ԧ&#3627408465;
2
&#3627408447;

&#3627408462;+Ԧ&#3627408464;
2
&#3627408450;
&#3627408462;+&#3627408463;+Ԧ&#3627408464;+Ԧ&#3627408465;
4

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Tetrahedron :
&#3627408436;Ԧ&#3627408462;
&#3627408437;&#3627408463;
&#3627408438;Ԧ&#3627408464;
&#3627408439;Ԧ&#3627408465;
&#3627408447;
&#3627408447;

&#3627408462;
1
&#3627408462;
2
&#3627408462;
4
&#3627408462;
3
Point of concurrency of
&#3627408436;&#3627408462;
1,&#3627408437;&#3627408462;
2,&#3627408438;&#3627408462;
3,&#3627408439;&#3627408462;
4is &#3627408450;
→Centroid of Tetrahedron

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Centre of tetrahedron:
Line segment joining vertices of tetrahedron to
the centroid of its opposite faces are
concurrent.
This point of concurrency is same as that of
the join of midpoint of each pair of opposite
edges of tetrahedron.
This point of concurrency is called centroid of the
tetrahedron.

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Tetrahedron:
Let &#3627408450;
1be a point which divides &#3627408436;&#3627408462;
1in 3:1ratio
Position vector of &#3627408450;
1=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;+Ԧ&#3627408465;
4
Similarly, &#3627408450;
2,&#3627408450;
3,&#3627408450;
4be the points which divides
&#3627408437;&#3627408462;
2,&#3627408438;&#3627408462;
3,&#3627408439;&#3627408462;
4in 3:1respectively.
P.V of &#3627408450;
2=P.V of &#3627408450;
4=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;+Ԧ&#3627408465;
4
Mid Point of &#3627408447;and &#3627408447;

=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;+Ԧ&#3627408465;
4
Hence these 4lines are concurrent.

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Parallelopiped:
A polyhedron with six faces hexahedron, each of which is a parallelogram.

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Parallelopiped:
Centre of Parallelopiped &#3627408451;:
All four diagonals are concurrent and are
bisected at their point of concurrency.
This point of concurrency is called its
center.
Position vector of midpoint of &#3627408465;
1&&#3627408465;
4=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
2
Position vector of midpoint of &#3627408465;
2&&#3627408465;
3=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
2
Note:If &#3627408450;is a point from where 3adjacent edges are
originating and &#3627408451;is the centre of parallelopiped, then
&#3627408450;&#3627408451;=
??????&#3627408436;+??????&#3627408437;+??????&#3627408438;
2

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Scalar Triple Product
Session 05
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Product of three or more vectors :
Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408464;⇒(Scalar) multiply (Vector)
Ԧ&#3627408462;⋅Ԧ&#3627408463;⋅Ԧ&#3627408464;⇒(Scalar) dot (Vector)
Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;⇒(Scalar) cross(Vector)
Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408464;⇒(Vector) multiply (Vector)
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;⇒(Vector) dot (Vector)Scalar triple Product
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;⇒(Vector) cross (Vector) Vector triple Product ⇒Possible
⇒Possible
⇒Possible
⇒Not Possible
⇒Not Possible
⇒Not Possible

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Scalar Triple Product :
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are three vectors, then their scalar triple product (or box product) is defined
as the dot product of two vectors Ԧ&#3627408462;and (&#3627408463;×Ԧ&#3627408464;)i.e. Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;→Scalar Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;.
It is denoted by Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;→Box product
Similarly other scalar triple products can be defined as Ԧ&#3627408463;⋅(Ԧ&#3627408464;×Ԧ&#3627408462;)which is Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;
& Ԧ&#3627408464;⋅(Ԧ&#3627408462;×Ԧ&#3627408463;)
Note :
Scalar triple product always results in a scalar quantity.

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Absolute value of scalar triple product of three
vectors is equal to the volume of the
parallelepiped whose three coterminous edges
are represented by the given vectors.
Volume of parallelopiped =Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
&#3627408460;
&#3627408459;
&#3627408461;
Ԧ&#3627408463;
Ԧ&#3627408462;
Ԧ&#3627408464;
Geometrical Interpretation of Scalar Triple Product
Scalar Triple Product :

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Scalar Triple Product :
ො&#3627408475;is the unit vector in the directionof Ԧ&#3627408463;×Ԧ&#3627408464;
Geometrical Interpretation of Scalar Triple Product
ො&#3627408475;
Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408464;sin??????ො&#3627408475;
=Ԧ&#3627408463;Ԧ&#3627408464;sin??????Ԧ&#3627408462;⋅ො&#3627408475;
=Ԧ&#3627408463;Ԧ&#3627408464;sin??????Ԧ&#3627408462;cos&#3627409148;
=(Area of base) ×(Height)
=Volume of parallelepiped

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Scalar Triple Product :
If Ԧ&#3627408462;=&#3627408462;
1
ƶ&#3627408470;+&#3627408462;
2
ƶ&#3627408471;+&#3627408462;
3
ƶ&#3627408472;,Ԧ&#3627408463;=&#3627408463;
1
ƶ&#3627408470;+&#3627408463;
2
ƶ&#3627408471;+&#3627408463;
3
ƶ&#3627408472;and Ԧ&#3627408464;=&#3627408464;
1
ƶ&#3627408470;+&#3627408464;
2
ƶ&#3627408471;+&#3627408464;
3
ƶ&#3627408472;
Formula of Scalar Triple Product :
[Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;]=
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3

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If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;be three non-zero vectors and Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;,then
A
B
D
C
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408464;=0
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;=0
Ԧ&#3627408464;⋅Ԧ&#3627408462;=Ԧ&#3627408462;⋅Ԧ&#3627408463;=0
Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;=0

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Solution:
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;be three non-zero vectors and Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;,then
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
⇒Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408464;cos&#3627409148;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
(where ??????is the angle between (Ԧ&#3627408462;×Ԧ&#3627408463;)and Ԧ&#3627408464;)
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;sin??????cos&#3627409148;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
(where ??????is the angle between Ԧ&#3627408462;and Ԧ&#3627408463;)
⇒sin??????=1and cos&#3627409148;=1⇒??????=
??????
2
and &#3627409148;=0
⇒Ԧ&#3627408464;⋅Ԧ&#3627408462;=Ԧ&#3627408464;⋅Ԧ&#3627408463;=0
⇒sin??????⋅cos&#3627409148;=1
≤1≤1
Ԧ&#3627408462;⊥
&#3627408479;Ԧ&#3627408463;⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=0,Ԧ&#3627408462;×Ԧ&#3627408463;→vector ⊥
&#3627408479;
to Ԧ&#3627408462;&Ԧ&#3627408463;
Ԧ&#3627408464;is⊥
&#3627408479;
to Ԧ&#3627408462;&Ԧ&#3627408463; ⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;=0

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If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;be three non-zero vectors and Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;,then
A
B
D
C
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408464;=0
Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;=0
Ԧ&#3627408464;⋅Ԧ&#3627408462;=Ԧ&#3627408462;⋅Ԧ&#3627408463;=0
Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;=0

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If Ԧ&#3627408462;=2ƶ&#3627408470;−3ƶ&#3627408471;,Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=3ƶ&#3627408470;−ƶ&#3627408472;represent three coterminous
edges of a parallelopiped, then the volume of that parallelopiped is
A
B
D
C
2
10
6
4

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If Ԧ&#3627408462;=2ƶ&#3627408470;−3ƶ&#3627408471;,Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=3ƶ&#3627408470;−ƶ&#3627408472;represent three coterminous
edges of a parallelopiped, then the volume of that parallelopiped is
A
B
D
C
2
10
6
4

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If Ԧ&#3627408462;=2ƶ&#3627408470;−3ƶ&#3627408471;,Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=3ƶ&#3627408470;−ƶ&#3627408472;represent three coterminous
edges of a parallelopiped, then the volume of that parallelopiped is
Solution:
Volume =Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
=
2−30
11−1
30−1
=−2+3×2
=4

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Scalar Triple Product :
Result :
&#3627408460;
&#3627408459;
Ԧ&#3627408463;
Ԧ&#3627408462;
Ԧ&#3627408464;
??????
??????
&#3627408461;
If ??????=90°and ??????=0
°
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;sin??????cos??????
??????is the angle between Ԧ&#3627408462;&Ԧ&#3627408463;×Ԧ&#3627408464;
??????is the angle between Ԧ&#3627408463;&Ԧ&#3627408464;
then Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;are mutually
perpendicular vectors.

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Let the volume of a parallelepiped whose coterminous edges are given
by &#3627408482;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;ƶ&#3627408472;,Ԧ&#3627408483;=ƶ&#3627408470;+ƶ&#3627408471;+3ƶ&#3627408472;and &#3627408484;=2ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;be 1cubic unit.
If ??????be the angle between the edges &#3627408482;and &#3627408484;, then cos??????can be :
A B DC
7
310
7
63
5
33
5
7
Solution:
±
11&#3627409158;
113
211
=1
⇒−&#3627409158;+3=±1⇒&#3627409158;=2or 4
For &#3627409158;=4, =
ƶ&#3627408470;+ƶ&#3627408471;+4ƶ&#3627408472;⋅2ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
1
2
+1
2
+4
2
2
2
+1
2
+1
2
=
7
63
=
2+1+4
186
cos??????=
&#3627408482;⋅&#3627408484;
|&#3627408482;|⋅|&#3627408484;|
JEE MAIN JAN 2020
&#3627408482;Ԧ&#3627408483;&#3627408484;=1⇒&#3627408482;Ԧ&#3627408483;&#3627408484;=±1
&#3627408438;
1→&#3627408438;
1−&#3627408438;
2⇒
01&#3627409158;
013
111
=±1

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If the volume of a parallelopiped, whose coterminous edges are given by
the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627408475;ƶ&#3627408472;,Ԧ&#3627408463;=2ƶ&#3627408470;+4ƶ&#3627408471;−&#3627408475;ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;+&#3627408475;ƶ&#3627408471;+3ƶ&#3627408472;(&#3627408475;≥0), is
158cubicunits,then :
JEE MAIN SEPT 2020
A
B
D
C
&#3627408475;=7
Ԧ&#3627408462;⋅Ԧ&#3627408464;=17
&#3627408475;=9
Ԧ&#3627408463;⋅Ԧ&#3627408464;=10

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If the volume of a parallelopiped, whose coterminous edges are given by
the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627408475;ƶ&#3627408472;,Ԧ&#3627408463;=2ƶ&#3627408470;+4ƶ&#3627408471;−&#3627408475;ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;+&#3627408475;ƶ&#3627408471;+3ƶ&#3627408472;(&#3627408475;≥0), is
158cubicunits,then :
JEE MAINS SEPT 2020

11&#3627408475;
24−&#3627408475;
1&#3627408475;3
=±158
⇒12+&#3627408475;
2
−6+&#3627408475;+&#3627408475;2&#3627408475;−4=±158
3&#3627408475;
2
−5&#3627408475;−152=0
⇒3&#3627408475;
2
−24&#3627408475;+19&#3627408475;−152=0
⇒3&#3627408475;+19(&#3627408475;−8)=0
⇒&#3627408475;=8(∵&#3627408475;≥0)
3&#3627408475;
2
−5&#3627408475;+164=0
&#3627408439;<0
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=158
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=±158
A
B
D
C
&#3627408475;=7
Ԧ&#3627408462;⋅Ԧ&#3627408464;=17
&#3627408475;=9
Ԧ&#3627408463;⋅Ԧ&#3627408464;=10
Solution:

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If the volume of a parallelopiped, whose coterminous edges are given by
the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627408475;ƶ&#3627408472;,Ԧ&#3627408463;=2ƶ&#3627408470;+4ƶ&#3627408471;−&#3627408475;ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;+&#3627408475;ƶ&#3627408471;+3ƶ&#3627408472;(&#3627408475;≥0), is
158cubicunits,then :
JEE MAINS SEPT 2020
A
B
D
C
&#3627408475;=7
Ԧ&#3627408462;⋅Ԧ&#3627408464;=17
&#3627408475;=9
Ԧ&#3627408463;⋅Ԧ&#3627408464;=10
&#3627408475;=8
∴Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;+8ƶ&#3627408472;
Ԧ&#3627408462;⋅Ԧ&#3627408464;
,Ԧ&#3627408463;=2ƶ&#3627408470;+4ƶ&#3627408471;−8ƶ&#3627408472;
andԦ&#3627408464;=ƶ&#3627408470;+8ƶ&#3627408471;+3ƶ&#3627408472;
=1+8+24=33
Ԧ&#3627408463;⋅Ԧ&#3627408464;=2+32−24=10
Solution:

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If the volume of a parallelopiped, whose coterminous edges are given by
the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627408475;ƶ&#3627408472;,Ԧ&#3627408463;=2ƶ&#3627408470;+4ƶ&#3627408471;−&#3627408475;ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;+&#3627408475;ƶ&#3627408471;+3ƶ&#3627408472;(&#3627408475;≥0), is
158cubicunits,then :
JEE MAIN SEPT 2020
A
B
D
C
&#3627408475;=7
Ԧ&#3627408462;⋅Ԧ&#3627408464;=17
&#3627408475;=9
Ԧ&#3627408463;⋅Ԧ&#3627408464;=10

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Properties of Scalar Triple Product :
The position of (.)and ×can be interchanged i.e.Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;
Value of scalar triple product remains unchanged if the same cyclic order
of Ԧ&#3627408462;,band Ԧ&#3627408464;is followed i.e. Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;=Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;
Sign of the scalar triple product is reversed if the cyclic order of
vectors is changed.
Ԧ&#3627408462;⋅(Ԧ&#3627408463;×Ԧ&#3627408464;)=−Ԧ&#3627408462;⋅Ԧ&#3627408464;×Ԧ&#3627408463;or Ԧ&#3627408462;Ԧ&#3627408463;Ԧc=−Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408463;
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408464;⋅Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408462;
Ԧ&#3627408464;Ԧ&#3627408463;→
Ԧ&#3627408462;
Ԧ&#3627408463;
Ԧ&#3627408464;
=−
Ԧ&#3627408462;
Ԧ&#3627408464;
Ԧ&#3627408463;

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Properties of Scalar Triple Product :
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;=Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;
Proof:
From &#3627408470;and &#3627408470;&#3627408470;
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;=Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;=−Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408463;=−Ԧ&#3627408464;Ԧ&#3627408463;Ԧ&#3627408462;
Ԧ&#3627408462;⋅Ԧ&#3627408463;×Ԧ&#3627408464;=−Ԧ&#3627408462;⋅Ԧ&#3627408464;×Ԧ&#3627408463;or Ԧ&#3627408462;Ԧ&#3627408463;Ԧc=−Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408463;⋯&#3627408470;&#3627408470;
⋯&#3627408470;

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Properties of Scalar Triple Product :
The scalar triple product of three vectors when
two of them are equal or parallel, is zero.
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408463;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408462;=0
The scalar triple product of three mutually
perpendicular unit vectors is ±1.
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;=1,ƶ&#3627408470;ƶ&#3627408472;ƶ&#3627408471;=−1
Ƹ&#3627408470;⋅Ƹ&#3627408471;×෠&#3627408472;=Ƹ&#3627408470;×Ƹ&#3627408470;=1
Ƹ&#3627408470;Ƹ&#3627408471;෠&#3627408472;=Ƹ&#3627408471;෠&#3627408472;Ƹ&#3627408470;=෠&#3627408472;Ƹ&#3627408470;Ƹ&#3627408471;=1

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Properties of Scalar Triple Product :
Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;arethreecoplanarvectors,thenecessary
andsufficientconditionforthreenon-zero
non-collinearvectorstobecoplanaris
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=0
Ԧ&#3627408463;
Ԧ&#3627408463;×Ԧ&#3627408464;
Ԧ&#3627408462;

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Properties of Scalar Triple Product :
For any vectors Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;;Ԧ&#3627408462;+Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;+Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;
Ԧ&#3627408462;+Ԧ&#3627408463;Ԧ&#3627408463;+Ԧ&#3627408464;Ԧ&#3627408464;+Ԧ&#3627408462;=2Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
⇒If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are coplanar then,Ԧ&#3627408462;+Ԧ&#3627408463;,Ԧ&#3627408463;+Ԧ&#3627408464;,Ԧ&#3627408464;+Ԧ&#3627408462;are also coplanar.
Note :
Volumeofparallelopipedwithedgesasfacediagonalsofparallelopiped
withcoterminousedgesԦ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;istwiceitsvolume.
Ԧ&#3627408462;−Ԧ&#3627408463;Ԧ&#3627408463;−Ԧ&#3627408464;Ԧ&#3627408464;−Ԧ&#3627408462;is always zero.
⇒Ԧ&#3627408462;−Ԧ&#3627408463;,Ԧ&#3627408463;−Ԧ&#3627408464;andԦ&#3627408464;−Ԧ&#3627408462;are coplanar.

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A
B
D
C
1
0
−1
None of these
If vectors &#3627408462;ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;,ƶ&#3627408470;+&#3627408463;ƶ&#3627408471;+ƶ&#3627408472;and ƶ&#3627408470;+ƶ&#3627408471;+&#3627408464;ƶ&#3627408472;&#3627408462;≠&#3627408463;≠&#3627408464;≠1are coplanar,
then
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
equals ______

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Solution:
Since vectors are coplanar,

&#3627408462;11
1&#3627408463;1
11&#3627408464;
=0

&#3627408462;1−&#3627408462;1−&#3627408462;
1&#3627408463;−10
10&#3627408464;−1
=0
&#3627408438;
2→&#3627408438;
2−&#3627408438;
1,&#3627408438;
3→&#3627408438;
3−&#3627408438;
1
⇒&#3627408462;&#3627408463;−1−1−1−&#3627408462;&#3627408464;−1+1−&#3627408462;1−&#3627408463;=0
⇒&#3627408454;.&#3627408455;.&#3627408451;.=0
⇒&#3627408462;−1+1&#3627408463;−1&#3627408464;−1+1−&#3627408462;1−&#3627408464;1−&#3627408463;=0
⇒&#3627408462;−1&#3627408463;−1&#3627408464;−1+1−&#3627408463;1−&#3627408464;+1−&#3627408464;1−&#3627408462;+1−&#3627408462;1−&#3627408463;=0
If vectors &#3627408462;ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;,ƶ&#3627408470;+&#3627408463;ƶ&#3627408471;+ƶ&#3627408472;and ƶ&#3627408470;+ƶ&#3627408471;+&#3627408464;ƶ&#3627408472;&#3627408462;≠&#3627408463;≠&#3627408464;≠1are coplanar,
then
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
equals ______

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Solution:
⇒&#3627408462;−1&#3627408463;−1&#3627408464;−1+1−&#3627408463;1−&#3627408464;+1−&#3627408464;1−&#3627408462;+1−&#3627408462;1−&#3627408463;=0
Divide the equation by 1−&#3627408462;1−&#3627408463;1−&#3627408464;
−1+
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
=0
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
=1
If vectors &#3627408462;ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;,ƶ&#3627408470;+&#3627408463;ƶ&#3627408471;+ƶ&#3627408472;and ƶ&#3627408470;+ƶ&#3627408471;+&#3627408464;ƶ&#3627408472;&#3627408462;≠&#3627408463;≠&#3627408464;≠1are coplanar,
then
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
equals ______

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A
B
D
C
1
0
−1
None of these
If vectors &#3627408462;ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;,ƶ&#3627408470;+&#3627408463;ƶ&#3627408471;+ƶ&#3627408472;and ƶ&#3627408470;+ƶ&#3627408471;+&#3627408464;ƶ&#3627408472;&#3627408462;≠&#3627408463;≠&#3627408464;≠1are
coplanar, then
1
1−&#3627408462;
+
1
1−&#3627408463;
+
1
1−&#3627408464;
equals ______

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If Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
and Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectorsand 3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0then
the value of &#3627409158;is _____.
JEE MAIN JAN 2020
Solution:

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If Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
and Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectorsand 3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0then
the value of &#3627409158;is _____.
JEE MAIN JAN 2020
Solution:
&#3627409158;=?
Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectors
3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0
As Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectors, Ԧ&#3627408451;&#3627408452;&#3627408453;
&#3627408462;+1&#3627408462;&#3627408462;
&#3627408462;&#3627408462;+1&#3627408462;
&#3627408462;&#3627408462;&#3627408462;+1
=0
⇒&#3627408462;+1+&#3627408462;+&#3627408462;=0
⇒&#3627408462;=−
1
3
&#3627408453;
2→&#3627408453;
2−&#3627408453;
1,&#3627408453;
3→&#3627408453;
3−&#3627408453;
1

&#3627408462;+1&#3627408462;&#3627408462;
−110
−101
=0

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If Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
and Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectorsand 3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0then
the value of &#3627409158;is _____.
JEE MAIN JAN 2020
Solution:
&#3627409158;=?
Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectors
3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0
⇒&#3627408462;=−
1
3
Ԧ&#3627408451;=
2
3
ƶ&#3627408470;−
1
3
ƶ&#3627408471;−
1
3
ƶ&#3627408472;,&#3627408452;=−
1
3
ƶ&#3627408470;+
2
3
ƶ&#3627408471;−
1
3
ƶ&#3627408472;, &#3627408453;=−
1
3
ƶ&#3627408470;−
1
3
ƶ&#3627408471;+
2
3
ƶ&#3627408472;
Ԧ&#3627408451;⋅&#3627408452;=
1
9
(−2−2+1)=−
1
3
&#3627408453;×&#3627408452;=
1
9
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;
−12−1
−1−12
=
1
9
3Ƹ&#3627408470;+3Ƹ&#3627408471;+3෠&#3627408472;
=
1
9
3ƶ&#3627408470;+3ƶ&#3627408471;+3ƶ&#3627408472;=
ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;
3

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If Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;, &#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
and Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectorsand 3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0then
the value of &#3627409158;is _____.
JEE MAIN JAN 2020
Solution:
&#3627409158;=?
Ԧ&#3627408451;=(&#3627408462;+1)ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408452;=&#3627408462;ƶ&#3627408470;+(&#3627408462;+1)ƶ&#3627408471;+&#3627408462;ƶ&#3627408472;
&#3627408453;=&#3627408462;ƶ&#3627408470;+&#3627408462;ƶ&#3627408471;+(&#3627408462;+1)ƶ&#3627408472;
Ԧ&#3627408451;,&#3627408452;,&#3627408453;are coplanar vectors
3(Ԧ&#3627408451;⋅&#3627408452;)
2
−&#3627409158;|&#3627408453;×&#3627408452;|
2
=0
⇒&#3627408462;=−
1
3
Ԧ&#3627408451;=
2
3
ƶ&#3627408470;−
1
3
ƶ&#3627408471;−
1
3
ƶ&#3627408472;,&#3627408452;=−
1
3
ƶ&#3627408470;+
2
3
ƶ&#3627408471;−
1
3
ƶ&#3627408472;, &#3627408453;=−
1
3
ƶ&#3627408470;−
1
3
ƶ&#3627408471;+
2
3
ƶ&#3627408472;
Ԧ&#3627408451;⋅&#3627408452;=−
1
3
,&#3627408453;×&#3627408452;=
3
3
=
1
3
3Ԧ&#3627408451;&#3627408452;−&#3627409158;&#3627408453;×෠&#3627408452;
2
=0

1
9
−&#3627409158;×
1
3
=0⇒1−&#3627409158;=0⇒&#3627409158;=1

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Vector Triple Product
Session 06
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Properties of Scalar Triple product:
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Ԧ&#3627408463;⋅Ԧ&#3627408462;Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408464;
Ԧ&#3627408464;⋅Ԧ&#3627408462;Ԧ&#3627408464;⋅Ԧ&#3627408463;Ԧ&#3627408464;⋅Ԧ&#3627408464;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3
&#3627408462;
1&#3627408463;
1&#3627408464;
1
&#3627408462;
2&#3627408463;
2&#3627408464;
2
&#3627408462;
3&#3627408463;
3&#3627408464;
3

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Theedgesofaparallelopipedareofunitlengthandareparallelto
non-coplanarunitvectorsො&#3627408462;,෠&#3627408463;,Ƹ&#3627408464;,suchthatො&#3627408462;⋅෠&#3627408463;=෠&#3627408463;⋅Ƹ&#3627408464;=ƶ&#3627408464;⋅ො&#3627408462;=
1
2
.
Thenthevolumeofparallelopipedis_________.
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Ԧ&#3627408463;⋅Ԧ&#3627408462;Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408464;
Ԧ&#3627408464;⋅Ԧ&#3627408462;Ԧ&#3627408464;⋅Ԧ&#3627408463;Ԧ&#3627408464;⋅Ԧ&#3627408464;
=
1
1
2
1
2
1
2
1
1
2
1
2
1
2
1
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
ො&#3627408462;⋅ො&#3627408462;=1
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
3
4

1
2
1
4
+
1
2

1
4
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
1
2
Volume=
1
2
Solution:
IIT JEE 2008

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Properties of Scalar Triple product:
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408477;×Ԧ&#3627408478;=
Ԧ&#3627408477;⋅Ԧ&#3627408462;Ԧ&#3627408478;⋅Ԧ&#3627408462;Ԧ&#3627408462;
Ԧ&#3627408477;⋅Ԧ&#3627408463;Ԧ&#3627408478;⋅Ԧ&#3627408463;Ԧ&#3627408463;
Ԧ&#3627408477;⋅Ԧ&#3627408464;Ԧ&#3627408478;⋅Ԧ&#3627408464;Ԧ&#3627408464;
=
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3
×
Ƹ&#3627408470;&#3627408477;
1&#3627408478;
1
Ƹ&#3627408471;&#3627408477;
2&#3627408478;
2
෠&#3627408472;&#3627408477;
3&#3627408478;
3
Ƹ&#3627408471;
Ƹ&#3627408470;
෠&#3627408472;
Ԧ&#3627408462;=&#3627408462;
1
Ԧ&#3627408473;+&#3627408462;
2&#3627408474;+&#3627408462;
3&#3627408475; Ԧ&#3627408462;=&#3627408462;
1Ƹ&#3627408470;+&#3627408462;
2Ƹ&#3627408471;+&#3627408462;
3
෠&#3627408472;

Ԧ&#3627408473;
&#3627408474;
&#3627408475;

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If Ԧ&#3627408462;=&#3627408462;
1
መ&#3627408473;+&#3627408462;
2ෝ&#3627408474;+&#3627408462;
3ො&#3627408475;,&#3627408463;=&#3627408463;
1
መ&#3627408473;+&#3627408463;
2ෝ&#3627408474;+&#3627408463;
3ො&#3627408475;and Ԧ&#3627408464;=&#3627408464;
1
መ&#3627408473;+&#3627408464;
2ෝ&#3627408474;+&#3627408464;
3ො&#3627408475;,then
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=
&#3627408462;
1&#3627408462;
2&#3627408462;
3
&#3627408463;
1&#3627408463;
2&#3627408463;
3
&#3627408464;
1&#3627408464;
2&#3627408464;
3
መ&#3627408473;ෝ&#3627408474;ො&#3627408475;
መ&#3627408473;,ෝ&#3627408474;,ො&#3627408475;are three mutually perpendicular triad of axes.
Properties of Scalar Triple product:

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Let Ԧ&#3627408464;be a vector perpendicular to the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408463;=ƶ&#3627408470;+2ƶ&#3627408471;+ƶ&#3627408472;.
If Ԧ&#3627408464;⋅ƶ&#3627408470;+ƶ&#3627408471;+3ƶ&#3627408472;=8,then the value of Ԧ&#3627408464;⋅Ԧ&#3627408462;×Ԧ&#3627408463;is equal to _________.
Solution:

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Let Ԧ&#3627408464;be a vector perpendicular to the vectors Ԧ&#3627408462;=ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408463;=ƶ&#3627408470;+2ƶ&#3627408471;+ƶ&#3627408472;.
If Ԧ&#3627408464;⋅ƶ&#3627408470;+ƶ&#3627408471;+3ƶ&#3627408472;=8,then the value of Ԧ&#3627408464;⋅Ԧ&#3627408462;×Ԧ&#3627408463;is equal to _________.
Solution:
Ԧ&#3627408464;=&#3627409158;Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408462;×Ԧ&#3627408463;=
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;
11−1
121
=3ƶ&#3627408470;−2ƶ&#3627408471;+ƶ&#3627408472;
Ԧ&#3627408464;⋅ƶ&#3627408470;+ƶ&#3627408471;+3ƶ&#3627408472;=&#3627409158;3ƶ&#3627408470;−2ƶ&#3627408471;+ƶ&#3627408472;⋅ƶ&#3627408470;+ƶ&#3627408471;+3ƶ&#3627408472;
⇒&#3627409158;4=8
⇒&#3627409158;=2
⇒Ԧ&#3627408464;=2Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408464;⋅Ԧ&#3627408462;×Ԧ&#3627408463;=2Ԧ&#3627408462;×Ԧ&#3627408463;
2
=28

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If Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+&#3627409149;ƶ&#3627408471;+3ƶ&#3627408472;,Ԧ&#3627408463;=−&#3627409149;ƶ&#3627408470;−&#3627409148;ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;such that Ԧ&#3627408462;⋅Ԧ&#3627408463;=1and
Ԧ&#3627408463;⋅Ԧ&#3627408464;=−3,then
1
3
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;is equal to _________.
Solution:

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If Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+&#3627409149;ƶ&#3627408471;+3ƶ&#3627408472;,Ԧ&#3627408463;=−&#3627409149;ƶ&#3627408470;−&#3627409148;ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;such that Ԧ&#3627408462;⋅Ԧ&#3627408463;=1and
Ԧ&#3627408463;⋅Ԧ&#3627408464;=−3,then
1
3
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;is equal to _________.
Solution:
Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+&#3627409149;ƶ&#3627408471;+3ƶ&#3627408472;
Ԧ&#3627408463;=−&#3627409149;ƶ&#3627408470;−&#3627409148;ƶ&#3627408471;−ƶ&#3627408472;
Ԧ&#3627408464;=ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;
Ԧ&#3627408462;⋅Ԧ&#3627408463;=1
Ԧ&#3627408463;⋅Ԧ&#3627408464;=−3
1
3
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;=?
Ԧ&#3627408462;⋅Ԧ&#3627408463;=1
⇒−&#3627409148;&#3627409149;−&#3627409148;&#3627409149;−3=1
⇒&#3627409148;&#3627409149;=−2⋯&#3627408470;
Ԧ&#3627408463;⋅Ԧ&#3627408464;=−3
⇒−&#3627409149;+2&#3627409148;+1=−3⋯&#3627408470;&#3627408470;

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If Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+&#3627409149;ƶ&#3627408471;+3ƶ&#3627408472;,Ԧ&#3627408463;=−&#3627409149;ƶ&#3627408470;−&#3627409148;ƶ&#3627408471;−ƶ&#3627408472;and Ԧ&#3627408464;=ƶ&#3627408470;−2ƶ&#3627408471;−ƶ&#3627408472;such that Ԧ&#3627408462;⋅Ԧ&#3627408463;=1and
Ԧ&#3627408463;⋅Ԧ&#3627408464;=−3,then
1
3
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;is equal to _________.
Solution:
&#3627409148;&#3627409149;=−2⋯&#3627408470;−&#3627409149;+2&#3627409148;+1=−3⋯&#3627408470;&#3627408470;
Solving &#3627408470;&&#3627408470;&#3627408470;,
&#3627409148;,&#3627409149;=−1,2
1
3
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=
1
3
&#3627409148;&#3627409149;3
−&#3627409149;−&#3627409148;−1
1−2−1
=
1
3
−123
−21−1
1−2−1
1
3
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=
1
3
×6=2

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Volume Of Tetrahedron:
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are position vectors of vertices &#3627408436;,&#3627408437;and
&#3627408438;with respect to &#3627408450;,then volume of tetrahedron
&#3627408450;&#3627408436;&#3627408437;&#3627408438;represented by ??????is given by
&#3627408436;Ԧ&#3627408462;
&#3627408438;Ԧ&#3627408464;
&#3627408450;0
&#3627408437;Ԧ&#3627408463;
&#3627408475;
??????=
1
3
Base Area×Height
Base Area =
1
2
Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;=
&#3627408475;
2
Let Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;=&#3627408475;
Height =Projection of Ԧ&#3627408462;on &#3627408475;=Ԧ&#3627408462;⋅ො&#3627408475;
=
&#3627408462;⋅&#3627408462;×&#3627408463;+&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×&#3627408462;
|&#3627408475;|
=
&#3627408462;⋅&#3627408463;×Ԧ&#3627408464;
|&#3627408475;|
=
&#3627408462;&#3627408463;Ԧ&#3627408464;
&#3627408475;

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∴??????=
1
3

1
2
|&#3627408475;|
&#3627408462;&#3627408463;Ԧ&#3627408464;
|&#3627408475;|
⇒??????=
1
6
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
??????=
1
3
Base Area×Height
Base Area =
&#3627408475;
2
Hight =
&#3627408462;&#3627408463;Ԧ&#3627408464;
&#3627408475;
Volume Of Tetrahedron:
&#3627408436;Ԧ&#3627408462;
&#3627408438;Ԧ&#3627408464;
&#3627408450;0
&#3627408437;Ԧ&#3627408463;
&#3627408475;

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Note:
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are position vectors of vertices
&#3627408436;,&#3627408437;,&#3627408438;,&#3627408439;of a tetrahedron &#3627408436;&#3627408437;&#3627408438;&#3627408439;,then:
It’s volume =
1
6
&#3627408436;&#3627408437;&#3627408436;&#3627408438;&#3627408436;&#3627408439;
=
1
6
Ԧ&#3627408463;−Ԧ&#3627408462;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408465;−Ԧ&#3627408462;
&#3627408436;Ԧ&#3627408462;
&#3627408438;Ԧ&#3627408464;
&#3627408450;0
&#3627408437;Ԧ&#3627408463;
&#3627408475;

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If the vertices of any tetrahedron be Ԧ&#3627408462;=ƶ&#3627408471;+2ƶ&#3627408472;,Ԧ&#3627408463;=3ƶ&#3627408470;+ƶ&#3627408472;,Ԧ&#3627408464;=4ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;
and Ԧ&#3627408465;=2ƶ&#3627408470;+3ƶ&#3627408471;+2ƶ&#3627408472;then find its volume.
Solution:
&#3627408439;Ԧ&#3627408465;
&#3627408438;Ԧ&#3627408464;
&#3627408436;Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;

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If the vertices of any tetrahedron be Ԧ&#3627408462;=ƶ&#3627408471;+2ƶ&#3627408472;,Ԧ&#3627408463;=3ƶ&#3627408470;+ƶ&#3627408472;,Ԧ&#3627408464;=4ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;
and Ԧ&#3627408465;=2ƶ&#3627408470;+3ƶ&#3627408471;+2ƶ&#3627408472;then find its volume.
Solution:
&#3627408439;Ԧ&#3627408465;
&#3627408438;Ԧ&#3627408464;
&#3627408436;Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;
Let the position vectors of the vertices
&#3627408436;,&#3627408437;,&#3627408438;,&#3627408439;with respect to &#3627408450;are Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;
and Ԧ&#3627408465;respectively then
&#3627408436;&#3627408437;=3ƶ&#3627408470;−ƶ&#3627408471;−ƶ&#3627408472;
&#3627408436;&#3627408438;=4ƶ&#3627408470;+2ƶ&#3627408471;+4ƶ&#3627408472;&
&#3627408436;&#3627408439;=2ƶ&#3627408470;+2ƶ&#3627408471;
Volume of tetrahedron =
1
6
&#3627408436;&#3627408437;&#3627408436;&#3627408438;&#3627408436;&#3627408439;

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If the vertices of any tetrahedron be Ԧ&#3627408462;=ƶ&#3627408471;+2ƶ&#3627408472;,Ԧ&#3627408463;=3ƶ&#3627408470;+ƶ&#3627408472;,Ԧ&#3627408464;=4ƶ&#3627408470;+3ƶ&#3627408471;+6ƶ&#3627408472;
and Ԧ&#3627408465;=2ƶ&#3627408470;+3ƶ&#3627408471;+2ƶ&#3627408472;then find its volume.
Solution:
&#3627408439;Ԧ&#3627408465;
&#3627408438;Ԧ&#3627408464;
&#3627408436;Ԧ&#3627408462;
&#3627408437;Ԧ&#3627408463;
Volume of tetrahedron =
1
6
&#3627408436;&#3627408437;&#3627408436;&#3627408438;&#3627408436;&#3627408439;
=
1
6
3−1−1
424
220
=−6
∴Required volume =6units

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Vector Triple Product:
Ԧ&#3627408463;
Ԧ&#3627408464;
Ԧ&#3627408462;
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;
The vector triple product of three vectors Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;
is defined as the vector product of two vectors
Ԧ&#3627408462;and (Ԧ&#3627408463;×Ԧ&#3627408464;). It is denoted by Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;.
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;is a vector which is coplanar with Ԧ&#3627408462;
and Ԧ&#3627408463;and perpendicular to Ԧ&#3627408464;.
Hence Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;⋯&#3627408470;
(linear combination of Ԧ&#3627408462;and Ԧ&#3627408463;)

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Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;⋯&#3627408470;
Taking dot with Ԧ&#3627408464;
Ԧ&#3627408464;⋅Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=&#3627408485;Ԧ&#3627408462;⋅Ԧ&#3627408464;+&#3627408486;Ԧ&#3627408463;⋅Ԧ&#3627408464;
⇒0=&#3627408485;Ԧ&#3627408462;⋅Ԧ&#3627408464;+&#3627408486;(Ԧ&#3627408463;⋅Ԧ&#3627408464;)
⇒&#3627408485;Ԧ&#3627408462;⋅Ԧ&#3627408464;=−&#3627408486;(Ԧ&#3627408463;⋅Ԧ&#3627408464;)

&#3627408485;
&#3627408486;
=
−&#3627408463;⋅Ԧ&#3627408464;
&#3627408462;⋅Ԧ&#3627408464;
=&#3627409158;
∴&#3627408485;=&#3627409158;Ԧ&#3627408463;⋅Ԧ&#3627408464;and &#3627408486;=−&#3627409158;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Vector Triple Product:

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Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;⋯&#3627408470;
∴&#3627408485;=&#3627409158;Ԧ&#3627408463;⋅Ԧ&#3627408464;and &#3627408486;=−&#3627409158;Ԧ&#3627408462;⋅Ԧ&#3627408464;Substituting the values of &#3627408485;and &#3627408486;in &#3627408470;
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=&#3627409158;Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408462;−&#3627409158;Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;This is an identity
Put Ԧ&#3627408462;=Ԧ&#3627408464;=ƶ&#3627408470;and Ԧ&#3627408463;=ƶ&#3627408471;
ƶ&#3627408470;×ƶ&#3627408471;×ƶ&#3627408470;=&#3627409158;ƶ&#3627408471;⋅ƶ&#3627408470;ƶ&#3627408470;−&#3627409158;ƶ&#3627408470;⋅ƶ&#3627408470;ƶ&#3627408471;
ƶ&#3627408471;=−&#3627409158;ƶ&#3627408471;
⇒&#3627409158;=−1
Hence Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;−Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408462;
Vector Triple Product:

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Properties of Vector Triple Product:
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;−Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408464;
Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408462;=Ԧ&#3627408463;⋅Ԧ&#3627408462;Ԧ&#3627408464;−Ԧ&#3627408464;⋅Ԧ&#3627408462;Ԧ&#3627408463;
[Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;]=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Ԧ&#3627408463;⋅Ԧ&#3627408462;Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408464;
Ԧ&#3627408464;⋅Ԧ&#3627408462;Ԧ&#3627408464;⋅Ԧ&#3627408463;Ԧ&#3627408464;⋅Ԧ&#3627408464;
Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;=Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408463;×Ԧ&#3627408464;⋅Ԧ&#3627408464;×Ԧ&#3627408462;=&#3627408482;×Ԧ&#3627408463;×Ԧ&#3627408464;⋅Ԧ&#3627408464;×Ԧ&#3627408462;
&#3627408482;
0
=&#3627408482;⋅Ԧ&#3627408464;Ԧ&#3627408463;−Ԧ&#3627408464;&#3627408482;⋅Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408462;
=&#3627408482;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408462;

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Properties of Vector Triple Product:
Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;=&#3627408482;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408462;
=&#3627408482;⋅Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2

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Note:
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are non coplanar vectors then Ԧ&#3627408462;×Ԧ&#3627408463;,Ԧ&#3627408463;×Ԧ&#3627408464;and Ԧ&#3627408464;×Ԧ&#3627408462;
will also be non coplanar vectors.
Vector triple product is a vector quantity.
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;≠Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;(Does not follow commutative law)
Unit vector coplanar with Ԧ&#3627408462;&Ԧ&#3627408463;and perpendicular to Ԧ&#3627408464;is ±
&#3627408462;×&#3627408463;×Ԧ&#3627408464;
|&#3627408462;×&#3627408463;×Ԧ&#3627408464;|

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ƶ&#3627408470;×ƶ&#3627408471;×ƶ&#3627408472;+ƶ&#3627408471;×ƶ&#3627408472;×ƶ&#3627408470;+ƶ&#3627408472;×(ƶ&#3627408470;×ƶ&#3627408471;)equals ________.
A
B
D
C
0
ƶ&#3627408470;
ƶ&#3627408472;
ƶ&#3627408471;

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ƶ&#3627408470;×ƶ&#3627408471;×ƶ&#3627408472;+ƶ&#3627408471;×ƶ&#3627408472;×ƶ&#3627408470;+ƶ&#3627408472;×(ƶ&#3627408470;×ƶ&#3627408471;)equals ________.
A
B
D
C
0
ƶ&#3627408470;
ƶ&#3627408472;
ƶ&#3627408471;

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ƶ&#3627408470;×ƶ&#3627408471;×ƶ&#3627408472;+ƶ&#3627408471;×ƶ&#3627408472;×ƶ&#3627408470;+ƶ&#3627408472;×(ƶ&#3627408470;×ƶ&#3627408471;)equals ________.
Solution:
ƶ&#3627408470;×ƶ&#3627408471;×ƶ&#3627408472;+ƶ&#3627408471;×ƶ&#3627408472;×ƶ&#3627408470;+ƶ&#3627408472;×ƶ&#3627408470;×ƶ&#3627408471;
⇒ƶ&#3627408470;×ƶ&#3627408470;+ƶ&#3627408471;×ƶ&#3627408471;+ƶ&#3627408472;×ƶ&#3627408472;
=0+0+0
=0

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If Ԧ&#3627408462;=2ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;,then the value of ƶ&#3627408470;×Ԧ&#3627408462;×ƶ&#3627408470;
2
+ƶ&#3627408471;×Ԧ&#3627408462;×ƶ&#3627408471;
2
+ƶ&#3627408472;×Ԧ&#3627408462;×ƶ&#3627408472;
2
is equal to _________.
Solution:
JEE MAINS SEPT 2020

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If Ԧ&#3627408462;=2ƶ&#3627408470;+ƶ&#3627408471;+2ƶ&#3627408472;,then the value of ƶ&#3627408470;×Ԧ&#3627408462;×ƶ&#3627408470;
2
+ƶ&#3627408471;×Ԧ&#3627408462;×ƶ&#3627408471;
2
+ƶ&#3627408472;×Ԧ&#3627408462;×ƶ&#3627408472;
2
is equal to _________.
JEE MAINS SEPT 2020
Let Ԧ&#3627408462;=&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;+&#3627408487;ƶ&#3627408472;
Now ƶ&#3627408470;×Ԧ&#3627408462;×ƶ&#3627408470;=ƶ&#3627408470;⋅ƶ&#3627408470;Ԧ&#3627408462;−ƶ&#3627408470;⋅Ԧ&#3627408462;ƶ&#3627408470;=&#3627408486;ƶ&#3627408471;+&#3627408487;ƶ&#3627408472;
Similarly,
ƶ&#3627408471;×(Ԧ&#3627408462;×ƶ&#3627408471;)=&#3627408485;ƶ&#3627408470;+&#3627408487;ƶ&#3627408472;
ƶ&#3627408472;×Ԧ&#3627408462;×ƶ&#3627408472;=&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;
Now &#3627408486;ƶ&#3627408471;+&#3627408487;ƶ&#3627408472;
2
+&#3627408485;ƶ&#3627408470;+&#3627408487;ƶ&#3627408472;
2
+&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;
2
=2&#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
=24+1+4=18
Solution:

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Product of four vectors
Session 07
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Let three vectors Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;be such that Ԧ&#3627408464;is coplanar with
Ԧ&#3627408462;and Ԧ&#3627408463;,Ԧ&#3627408462;⋅Ԧ&#3627408464;=7and Ԧ&#3627408463;is perpendicular to Ԧ&#3627408464;, where Ԧ&#3627408462;=−ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;and
Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408472;, then the value of 2|Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;|
2
is
Solution:
JEE Main Feb 2021
Ԧ&#3627408464;∥Ԧ&#3627408463;×Ԧ&#3627408462;×Ԧ&#3627408463;
⇒Ԧ&#3627408464;=&#3627409158;Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408462;−Ԧ&#3627408463;⋅Ԧ&#3627408462;Ԧ&#3627408463;=&#3627409158;5Ԧ&#3627408462;+Ԧ&#3627408463;
Taking dot with Ԧ&#3627408462;
⇒Ԧ&#3627408464;=&#3627409158;Ԧ&#3627408463;×Ԧ&#3627408462;×Ԧ&#3627408463;
⇒Ԧ&#3627408464;=&#3627409158;−3ƶ&#3627408470;+5ƶ&#3627408471;+6ƶ&#3627408472;
Ԧ&#3627408464;⋅Ԧ&#3627408462;=7
⇒&#3627409158;3+5+6=7⇒&#3627409158;=
1
2
⇒Ԧ&#3627408464;=
1
2
−3ƶ&#3627408470;+5ƶ&#3627408471;+6ƶ&#3627408472;
∴2Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;
2
=2−
3
2
ƶ&#3627408470;+
5
2
ƶ&#3627408471;+3෠&#3627408472;+2ƶ&#3627408470;+෠&#3627408472;+−ƶ&#3627408470;+ƶ&#3627408471;+෠&#3627408472;
2

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Let three vectors Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;be such that Ԧ&#3627408464;is coplanar with
Ԧ&#3627408462;and Ԧ&#3627408463;,Ԧ&#3627408462;⋅Ԧ&#3627408464;=7and Ԧ&#3627408463;is perpendicular to Ԧ&#3627408464;, where Ԧ&#3627408462;=−ƶ&#3627408470;+ƶ&#3627408471;+ƶ&#3627408472;and
Ԧ&#3627408463;=2ƶ&#3627408470;+ƶ&#3627408472;, then the value of 2|Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;|
2
is
Solution:
∴2Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;
2
=2−
3
2
ƶ&#3627408470;+
5
2
ƶ&#3627408471;+3෠&#3627408472;+2ƶ&#3627408470;+෠&#3627408472;+−ƶ&#3627408470;+ƶ&#3627408471;+෠&#3627408472;
2
=2
1
4
+
49
4
+25
=25+50
=75
=2−
1
2
ƶ&#3627408470;+
7
2
ƶ&#3627408471;+5෠&#3627408472;
2
=2
25
2
+25
JEE Main Feb 2021

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If Ԧ&#3627408462;&Ԧ&#3627408463;are perpendicular vectors, then Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408463;is equal to:
A
B
D
C
1
2
Ԧ&#3627408462;
4Ԧ&#3627408463;
Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408462;
4Ԧ&#3627408463;
0
JEE Main Feb 2021

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If Ԧ&#3627408462;&Ԧ&#3627408463;are perpendicular vectors, then Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408463;is equal to:
Solution:
=Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;−Ԧ&#3627408462;
2Ԧ&#3627408463;
=−Ԧ&#3627408462;
2
Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;−Ԧ&#3627408462;
2Ԧ&#3627408463;
=−Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;Ԧ&#3627408462;
2
+Ԧ&#3627408462;
4Ԧ&#3627408463;
=Ԧ&#3627408462;
4Ԧ&#3627408463;
∵Ԧ&#3627408462;⋅Ԧ&#3627408463;=0=−Ԧ&#3627408462;
2
Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408463;
=Ԧ&#3627408462;×Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;×Ԧ&#3627408462;−Ԧ&#3627408462;⋅Ԧ&#3627408462;×Ԧ&#3627408463;
JEE Main Feb 2021

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If Ԧ&#3627408462;&Ԧ&#3627408463;are perpendicular vectors, then Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408462;×Ԧ&#3627408463;is equal to:
A
B
D
C Ԧ&#3627408462;
4Ԧ&#3627408463;
JEE Main Feb 2021

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Let Ԧ&#3627408462;=3,Ԧ&#3627408463;=5,Ԧ&#3627408463;⋅Ԧ&#3627408464;=10angle between Ԧ&#3627408463;&Ԧ&#3627408464;equal to
??????
3
. If Ԧ&#3627408462;is
perpendicular to Ԧ&#3627408463;×Ԧ&#3627408464;then the find the value of Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;.
JEE Main Jan2020
Solution:

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Let Ԧ&#3627408462;=3,Ԧ&#3627408463;=5,Ԧ&#3627408463;⋅Ԧ&#3627408464;=10angle between Ԧ&#3627408463;&Ԧ&#3627408464;equal to
??????
3
. If Ԧ&#3627408462;is
perpendicular to Ԧ&#3627408463;×Ԧ&#3627408464;then the find the value of Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;.
Ԧ&#3627408463;⋅Ԧ&#3627408464;=Ԧ&#3627408463;Ԧ&#3627408464;cos
??????
3
=10
⇒5⋅Ԧ&#3627408464;⋅
1
2
=10
⇒Ԧ&#3627408464;=4
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408463;×Ԧ&#3627408464;sin
??????
2
=3×5×4×
3
2
=30
=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;sin
??????
3
JEE Main Jan2020
Solution:

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Let ො&#3627408462;,෠&#3627408463;and Ƹ&#3627408464;be three unit vectors such that ො&#3627408462;×෠&#3627408463;×Ƹ&#3627408464;=
3
2
(෠&#3627408463;+Ƹ&#3627408464;).
If ෠&#3627408463;is not parallel to Ƹ&#3627408464;then the angle between ො&#3627408462;and ෠&#3627408463;is ___.
JEE Main 2016
A
B
D
C
5??????
6
3??????
4
??????
2
2??????
3

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Let ො&#3627408462;,෠&#3627408463;and Ƹ&#3627408464;be three unit vectors such that ො&#3627408462;×෠&#3627408463;×Ƹ&#3627408464;=
3
2
(෠&#3627408463;+Ƹ&#3627408464;).
If ෠&#3627408463;is not parallel to Ƹ&#3627408464;then the angle between ො&#3627408462;and ෠&#3627408463;is ___.
JEE Main 2016
ො&#3627408462;⋅Ƹ&#3627408464;−
3
2
෠&#3627408463;−ො&#3627408462;⋅෠&#3627408463;+
3
2
Ƹ&#3627408464;=0
⇒ො&#3627408462;⋅෠&#3627408463;=−
3
2
⇒??????=
5??????
6
⇒ො&#3627408462;⋅Ƹ&#3627408464;෠&#3627408463;−ො&#3627408462;⋅෠&#3627408463;Ƹ&#3627408464;=
3
2
(෠&#3627408463;+Ƹ&#3627408464;)
ො&#3627408462;×෠&#3627408463;×Ƹ&#3627408464;=
3
2
(෠&#3627408463;+Ƹ&#3627408464;)
⇒cos??????=−
3
2
Solution:

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Let ො&#3627408462;,෠&#3627408463;and Ƹ&#3627408464;be three unit vectors such that ො&#3627408462;×෠&#3627408463;×Ƹ&#3627408464;=
3
2
(෠&#3627408463;+Ƹ&#3627408464;).
If ෠&#3627408463;is not parallel to Ƹ&#3627408464;then the angle between ො&#3627408462;and ෠&#3627408463;is ___.
JEE Main 2016
A
B
D
C
5??????
6

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Scalar Product of Four Vectors:
Proof:
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;−Ԧ&#3627408462;⋅Ԧ&#3627408465;Ԧ&#3627408463;⋅Ԧ&#3627408464;=
Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408462;⋅Ԧ&#3627408465;
Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=&#3627408482;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=&#3627408482;×Ԧ&#3627408464;⋅Ԧ&#3627408465;=Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;⋅Ԧ&#3627408465;
&#3627408482;
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;⋅Ԧ&#3627408465;=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;−Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408462;⋅Ԧ&#3627408465;
=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;−Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408462;⋅Ԧ&#3627408465;
=
Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408462;⋅Ԧ&#3627408465;
Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;

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Scalar Product of Four Vectors:
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;−Ԧ&#3627408462;⋅Ԧ&#3627408465;Ԧ&#3627408463;⋅Ԧ&#3627408464;=
Ԧ&#3627408462;⋅Ԧ&#3627408464;Ԧ&#3627408462;⋅Ԧ&#3627408465;
Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408463;⋅Ԧ&#3627408465;
(Ԧ&#3627408462;×Ԧ&#3627408463;)⋅(Ԧ&#3627408462;×Ԧ&#3627408463;)=|Ԧ&#3627408462;×Ԧ&#3627408463;|
2
=
Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408462;⋅Ԧ&#3627408463;
Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408463;
=Ԧ&#3627408462;
2Ԧ&#3627408463;
2
−(Ԧ&#3627408462;⋅Ԧ&#3627408463;)
2
Lagrange’s identity
=Ԧ&#3627408462;
2Ԧ&#3627408463;
2
−Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408462;

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Prove that acute angle between the two plane faces of a regular
tetrahedron is cos
−1
1
3
.
Solution:
&#3627408450;
&#3627408438;
&#3627408437;
&#3627408436;
Ԧ&#3627408462;
Ԧ&#3627408464;
&#3627408475;
1
&#3627408475;
2
Ԧ&#3627408463;
Let edge length of regular tetrahedron =1
&#3627408475;
2=normal vector to plane &#3627408450;&#3627408437;&#3627408438;=Ԧ&#3627408463;×Ԧc
&#3627408475;
1=normal vector to plane &#3627408450;&#3627408436;&#3627408437;=Ԧ&#3627408462;×Ԧ&#3627408463;
∴Acuteanglebetweenplanefaces&#3627408450;&#3627408436;&#3627408437;&&#3627408450;&#3627408437;&#3627408438;isgivenas
=
Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408464;
Ԧ&#3627408462;×Ԧ&#3627408463;||Ԧ&#3627408463;×Ԧ&#3627408464;
cos??????=
&#3627408475;
1.&#3627408475;
2
|&#3627408475;
1||&#3627408475;
2|
=
(Ԧ&#3627408462;×Ԧ&#3627408463;).(Ԧ&#3627408463;×Ԧ&#3627408464;)
|Ԧ&#3627408462;×Ԧ&#3627408463;||Ԧ&#3627408463;×Ԧ&#3627408464;|

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Prove that acute angle between the two plane faces of a regular
tetrahedron is cos
−1
1
3
.
&#3627408450;
&#3627408438;
&#3627408437;
&#3627408436;
Ԧ&#3627408462;
Ԧ&#3627408464;
&#3627408475;
1
&#3627408475;
2
Ԧ&#3627408463;
=
cos60°cos60°
cos0°cos60°
sin
2
60°
=
1
4

1
2
3
4
=
1
3
⇒??????=cos
−1
1
3
cos??????=
Ԧ&#3627408462;⋅Ԧ&#3627408463;Ԧ&#3627408462;⋅Ԧ&#3627408464;
Ԧ&#3627408463;⋅Ԧ&#3627408463;Ԧ&#3627408463;⋅Ԧ&#3627408464;
Ԧ&#3627408462;×Ԧ&#3627408463;||Ԧ&#3627408463;×Ԧ&#3627408464;
Solution:

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A
B
D
C
Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are non-coplanar
Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are non-coplanar
Ԧ&#3627408463;,Ԧ&#3627408465;are non-parallel
Ԧ&#3627408462;,Ԧ&#3627408465;are parallel and Ԧ&#3627408463;,Ԧ&#3627408464;are parallel
IIT-JEE 2009
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408465;are unit vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=1and Ԧ&#3627408462;⋅Ԧ&#3627408464;=
1
2
, then

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IIT-JEE 2009
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408465;are unit vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=1and Ԧ&#3627408462;⋅Ԧ&#3627408464;=
1
2
, then
Let ??????
1be the angle between Ԧ&#3627408462;and Ԧ&#3627408463;⇒Ԧ&#3627408462;^Ԧ&#3627408463;=??????
1
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;=1⇒Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408464;×Ԧ&#3627408465;cos??????=1
⇒sin??????
1sin??????
2cos??????=1
⇒??????
1=
??????
2
,??????
2=
??????
2
,??????=0
As ??????=0so Ԧ&#3627408462;×Ԧ&#3627408463;and Ԧ&#3627408464;×Ԧ&#3627408465;are parallel
??????
2be the angle between Ԧ&#3627408464;and Ԧ&#3627408465;⇒Ԧ&#3627408464;^Ԧ&#3627408465;=??????
2
??????be angle between Ԧ&#3627408462;×Ԧ&#3627408463;and Ԧ&#3627408464;×Ԧ&#3627408465;
⇒Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are coplanar
Ԧ&#3627408464;
Ԧ&#3627408462;
Ԧ&#3627408465;
&#3627408463;
⇒Ԧ&#3627408462;Ԧ&#3627408463;⋅sin??????
1⋅Ԧ&#3627408464;Ԧ&#3627408465;sin??????
2⋅cos??????=1
Solution:

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IIT-JEE 2009
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408465;are unit vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=1and Ԧ&#3627408462;⋅Ԧ&#3627408464;=
1
2
, then
⇒Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are coplanar
Ԧ&#3627408464;
Ԧ&#3627408462;
Ԧ&#3627408465;
&#3627408463;
⇒??????
3=
??????
3
⇒angle between Ԧ&#3627408462;and Ԧ&#3627408464;is
??????
3
∴Ԧ&#3627408463;and Ԧ&#3627408465;are non-parallel
⇒angle between Ԧ&#3627408463;and Ԧ&#3627408465;is
??????
3
⇒cos??????
3=
1
2
(where ??????
3is the angle between Ԧ&#3627408462;and Ԧ&#3627408464;)
Ԧ&#3627408462;⋅Ԧ&#3627408464;=
1
2
Solution:

Return to Top
A
B
D
C Ԧ&#3627408463;,Ԧ&#3627408465;are non-parallel
IIT-JEE 2009
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408465;are unit vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=1and Ԧ&#3627408462;⋅Ԧ&#3627408464;=
1
2
, then

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Vector Product of Four Vectors:
??????=Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;
=&#3627408482;×Ԧ&#3627408464;×Ԧ&#3627408465;
=&#3627408482;⋅Ԧ&#3627408465;Ԧ&#3627408464;−&#3627408482;⋅Ԧ&#3627408464;Ԧ&#3627408465;
=Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;⋯&#3627408470;
=Ԧ&#3627408462;⋅Ԧ&#3627408483;Ԧ&#3627408463;−Ԧ&#3627408463;⋅Ԧ&#3627408483;Ԧ&#3627408462;
⇒??????=Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408483;
??????=Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;
=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408463;−Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408462;⋯(&#3627408470;&#3627408470;)
Let Ԧ&#3627408483;=Ԧ&#3627408464;×Ԧ&#3627408465;

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Vector Product of Four Vectors:
??????=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;⋯&#3627408470;
??????=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408463;−Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408462;⋯(&#3627408470;&#3627408470;)From &#3627408470;and &#3627408470;&#3627408470;
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408463;−Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408462;⋯(&#3627408470;&#3627408470;&#3627408470;)
Note:
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;=0
Planes containing the vectors Ԧ&#3627408462;& Ԧ&#3627408463;and Ԧ&#3627408464;& Ԧ&#3627408465;are parallel.
Ԧ&#3627408462;×Ԧ&#3627408463;⋅Ԧ&#3627408464;×Ԧ&#3627408465;=0
Two planes are perpendicular.

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Vector Product of Four Vectors:
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408463;−Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;Ԧ&#3627408462;⋯(&#3627408470;&#3627408470;&#3627408470;)
Note:
IfԦ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are four vectors & no threeof them are coplanarthen each one of
them can be expressed as alinear combination of others.
IfԦ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are&#3627408451;.??????.offourpointsthenthesefourpoints
areinsameplaneif
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408462;Ԧ&#3627408464;Ԧ&#3627408465;−Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;

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A
B
D
C
|Ԧ&#3627408463;|=1,|Ԧ&#3627408464;|=|Ԧ&#3627408462;|
|Ԧ&#3627408464;|=1,|Ԧ&#3627408462;|=|Ԧ&#3627408463;|
|Ԧ&#3627408463;|=2,|Ԧ&#3627408464;|=2|Ԧ&#3627408462;|
|Ԧ&#3627408462;|=1,|Ԧ&#3627408463;|=|Ԧ&#3627408464;|
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are three vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;,Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;, then:

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If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are three vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;,Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;, then:
Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;⋯&#3627408470;⇒Ԧ&#3627408464;⊥Ԧ&#3627408462;andԦ&#3627408464;⊥Ԧ&#3627408463;⇒Ԧ&#3627408464;⋅Ԧ&#3627408462;=Ԧ&#3627408464;⋅Ԧ&#3627408463;=0
Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;⋯&#3627408470;&#3627408470;⇒Ԧ&#3627408462;⊥Ԧ&#3627408463;andԦ&#3627408462;⊥Ԧ&#3627408464;⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;=Ԧ&#3627408462;⋅Ԧ&#3627408464;=0
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408464;×Ԧ&#3627408462;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408463;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408463;Ԧ&#3627408464;=Ԧ&#3627408464;×Ԧ&#3627408462;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408463;=Ԧ&#3627408464;×Ԧ&#3627408462;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408463;=Ԧ&#3627408464;×Ԧ&#3627408462;
∴Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are mutually perpendicular
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408465;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408463;=Ԧ&#3627408464;Ԧ&#3627408462;
Solution:

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If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are three vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;,Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;, then:
Solution:
⇒|Ԧ&#3627408463;|
2
=1⇒Ԧ&#3627408463;=1
⇒Ԧ&#3627408462;Ԧ&#3627408463;=Ԧ&#3627408464;⇒Ԧ&#3627408462;=Ԧ&#3627408464;
Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;⇒Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408463;=Ԧ&#3627408464;Ԧ&#3627408462;

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A
B
D
C
|Ԧ&#3627408463;|=1,|Ԧ&#3627408464;|=|Ԧ&#3627408462;|
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are three vectors such that Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;,Ԧ&#3627408463;×Ԧ&#3627408464;=Ԧ&#3627408462;, then:

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Linearly Dependent and
independent vectors
Session 08
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Theorem in Space (Fundamental theorem in Space) :
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are 3non-zero, non coplanar vectors then any vector
can be expressed as a linear combination.
Ԧ&#3627408479;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;ofԦ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;→Linear combination of Ƹ&#3627408470;,Ƹ&#3627408471;& ෠&#3627408472;
Example :
⇒Ԧ&#3627408462;×Ԧ&#3627408463;,Ԧ&#3627408463;×Ԧ&#3627408464;,Ԧ&#3627408464;×Ԧ&#3627408462;are also non coplanar
Ԧ&#3627408462;=&#3627408485;Ԧ&#3627408462;×Ԧ&#3627408463;+&#3627408486;Ԧ&#3627408463;×Ԧ&#3627408464;+&#3627408487;Ԧ&#3627408464;×Ԧ&#3627408462;⋯1
Express the non coplanar vectors Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;in term of Ԧ&#3627408463;×Ԧ&#3627408464;,Ԧ&#3627408464;×Ԧ&#3627408462;,Ԧ&#3627408462;×Ԧ&#3627408463;
Vector Ԧ&#3627408462;is represented as a linear combination of
Ԧ&#3627408462;×Ԧ&#3627408463;,Ԧ&#3627408463;×Ԧ&#3627408464;&Ԧ&#3627408464;×Ԧ&#3627408462;(non coplanar)

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Example :Ԧ&#3627408462;=&#3627408485;Ԧ&#3627408462;×Ԧ&#3627408463;+&#3627408486;Ԧ&#3627408463;×Ԧ&#3627408464;+&#3627408487;Ԧ&#3627408464;×Ԧ&#3627408462;⋯1
Ԧ&#3627408462;⋅Ԧ&#3627408463;=&#3627408485;ത&#3627408462;ത&#3627408462;ത&#3627408463;+&#3627408486;ത&#3627408463;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ത&#3627408463;ҧ&#3627408464;ത&#3627408462;⇒&#3627408487;=
&#3627408462;⋅&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408462;
Ԧ&#3627408462;⋅Ԧ&#3627408462;=&#3627408485;ത&#3627408462;ത&#3627408462;ത&#3627408463;+&#3627408486;ത&#3627408462;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ത&#3627408462;ҧ&#3627408464;ത&#3627408462;
Ԧ&#3627408462;
2
=&#3627408486;ത&#3627408462;ത&#3627408463;ҧ&#3627408464;⇒&#3627408486;=
&#3627408462;
2
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408463;
Ԧ&#3627408462;⋅Ԧ&#3627408464;=&#3627408485;ҧ&#3627408464;ത&#3627408462;ത&#3627408463;+&#3627408486;ҧ&#3627408464;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ҧ&#3627408464;ҧ&#3627408464;ത&#3627408462;⇒&#3627408485;=
&#3627408462;⋅Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408464;
Ԧ&#3627408462;=
&#3627408462;⋅Ԧ&#3627408464;&#3627408462;×&#3627408463;+&#3627408462;
2
&#3627408463;×Ԧ&#3627408464;+&#3627408462;⋅&#3627408463;Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Theorem in Space (Fundamental theorem in Space) :

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Example :Ԧ&#3627408462;=&#3627408485;Ԧ&#3627408462;×Ԧ&#3627408463;+&#3627408486;Ԧ&#3627408463;×Ԧ&#3627408464;+&#3627408487;Ԧ&#3627408464;×Ԧ&#3627408462;⋯1
Ԧ&#3627408462;⋅Ԧ&#3627408463;=&#3627408485;ത&#3627408462;ത&#3627408462;ത&#3627408463;+&#3627408486;ത&#3627408463;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ത&#3627408463;ҧ&#3627408464;ത&#3627408462;⇒&#3627408487;=
&#3627408462;⋅&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408462;
Ԧ&#3627408462;⋅Ԧ&#3627408462;=&#3627408485;ത&#3627408462;ത&#3627408462;ത&#3627408463;+&#3627408486;ത&#3627408462;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ത&#3627408462;ҧ&#3627408464;ത&#3627408462;
Ԧ&#3627408462;
2
=&#3627408486;ത&#3627408462;ത&#3627408463;ҧ&#3627408464;⇒&#3627408486;=
&#3627408462;
2
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408463;
Ԧ&#3627408462;⋅Ԧ&#3627408464;=&#3627408485;ҧ&#3627408464;ത&#3627408462;ത&#3627408463;+&#3627408486;ҧ&#3627408464;ത&#3627408463;ҧ&#3627408464;+&#3627408487;ҧ&#3627408464;ҧ&#3627408464;ത&#3627408462;⇒&#3627408485;=
&#3627408462;⋅Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Dot with Ԧ&#3627408464;
∴Ԧ&#3627408462;=
&#3627408462;⋅Ԧ&#3627408464;&#3627408462;×&#3627408463;+&#3627408462;
2
&#3627408463;×Ԧ&#3627408464;+&#3627408462;⋅&#3627408463;Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Put values of &#3627408485;,&#3627408486;and &#3627408487;in (1)
Theorem in Space (Fundamental theorem in Space) :

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Let &#3627408450;&#3627408436;&#3627408437;&#3627408438;be a regular tetrahedron of side length unity where &#3627408451;is a point at
a unit distance from the origin and &#3627408450;&#3627408451;is equally inclined to &#3627408450;&#3627408436;,&#3627408450;&#3627408437;,&#3627408450;&#3627408438;at
an angle &#3627409148;. Find cos
2
&#3627409148;.
Solution :

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&#3627408450;
&#3627408438;
&#3627408437;
&#3627408436;&#3627408451;
&#3627409148;
ො&#3627408462;
෠&#3627408463;
Ƹ&#3627408464;
Ƹ&#3627408477;
Let &#3627408450;&#3627408436;&#3627408437;&#3627408438;be a regular tetrahedron of side length unity where &#3627408451;is a point at
a unit distance from the origin and &#3627408450;&#3627408451;is equally inclined to &#3627408450;&#3627408436;,&#3627408450;&#3627408437;,&#3627408450;&#3627408438;at
an angle &#3627409148;. Find cos
2
&#3627409148;.
ො&#3627408462;⋅෠&#3627408463;=෠&#3627408463;⋅Ƹ&#3627408464;=Ƹ&#3627408464;⋅ො&#3627408462;=1×1×cos
??????
3
=
1
2
Ƹ&#3627408477;=&#3627408485;ො&#3627408462;+&#3627408486;෠&#3627408463;+&#3627408487;Ƹ&#3627408464;→&#3627408477;can be represented as a
linear combination of ො&#3627408462;,෠&#3627408463;&Ƹ&#3627408464;
⇒cos&#3627409148;=&#3627408485;+
1
2
&#3627408486;+
1
2
&#3627408487;Dot product with ො&#3627408462;
Dot product with ෠&#3627408463;⇒cos&#3627409148;=
1
2
&#3627408485;+&#3627408486;+
1
2
&#3627408487;
⇒cos&#3627409148;=
1
2
&#3627408485;+
1
2
&#3627408486;+&#3627408487;Dot product with Ƹ&#3627408464;
⋯(&#3627408470;)
⋯(&#3627408470;&#3627408470;)
⋯(&#3627408470;&#3627408470;&#3627408470;)
Adding &#3627408470;,(&#3627408470;&#3627408470;)& &#3627408470;&#3627408470;&#3627408470;⇒3cos&#3627409148;=2&#3627408485;+&#3627408486;+&#3627408487;
⇒1=&#3627408485;+&#3627408486;+&#3627408487;cos&#3627409148;Dot product with Ƹ&#3627408477;
⇒sec&#3627409148;=&#3627408485;+&#3627408486;+&#3627408487;
⇒2sec&#3627409148;=3cos&#3627409148;⇒cos
2
&#3627409148;=
2
3
ො&#3627408462;,෠&#3627408463;&Ƹ&#3627408464;are 3non zero, non coplanar vectorsSolution :

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Linearly Dependent and Independently Vectors :
If ??????
1,??????
2,⋯,??????
&#3627408475;are vectors and &#3627409158;
1,&#3627409158;
2,⋯,&#3627409158;
&#3627408475;are scalars.
And if the linear combination &#3627409158;
1??????
1+&#3627409158;
2??????
2+⋯+&#3627409158;
&#3627408475;??????
&#3627408475;=0
necessarily implies &#3627409158;
1=&#3627409158;
2=⋯=&#3627409158;
&#3627408475;=0
∴??????
1,??????
2,⋯,??????
&#3627408475;are Linearly Independent set of vectors.
If ??????
1,??????
2,⋯,??????
&#3627408475;are vectors and &#3627409158;
1,&#3627409158;
2,⋯,&#3627409158;
&#3627408475;are scalars.
And if the linear combination &#3627409158;
1??????
1+&#3627409158;
2??????
2+⋯+&#3627409158;
&#3627408475;??????
&#3627408475;=0
At least one of scalar &#3627409158;
1=&#3627409158;
2=⋯=&#3627409158;
&#3627408475;≠0
∴??????
1,??????
2,⋯,??????
&#3627408475;are Linearly Dependent set of vectors.

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Linearly Dependent and Independently Vectors :
&#3627408485;ƶ&#3627408470;+&#3627408486;ƶ&#3627408471;=0
where &#3627408485;,&#3627408486;=0
∴Ƹ&#3627408470;and Ƹ&#3627408471;are Linearly Independent.
Two vectors Ƹ&#3627408470;and 2Ƹ&#3627408470;
&#3627409158;
1
ƶ&#3627408470;+2&#3627409158;
2
ƶ&#3627408470;=0
⇒&#3627409158;
1=2and &#3627409158;
2=−1
∴ƶ&#3627408470;and 2Ƹ&#3627408470;are Linearly Dependent.

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Linearly Dependent and Independently Vectors :
If Ԧ&#3627408462;||Ԧ&#3627408463;⇒Linearly Dependent
Ԧ&#3627408462;=&#3627409158;Ԧ&#3627408463;or Ԧ&#3627408462;−&#3627409158;Ԧ&#3627408463;=0or &#3627409158;
1Ԧ&#3627408462;+&#3627409158;
2
Ԧ&#3627408463;=0
For two non-zero vectors
If Ԧ&#3627408462;∦Ԧ&#3627408463;⇒Linearly Independent
For three non-zero vectors
If Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;≠0⇒Linearly Independent, Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧc, are non co-planar.
If Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=0⇒Linearly Dependent and Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are coplanar vectors.
Ԧ&#3627408464;=&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;(Theorem in Plane)
&#3627409158;
1Ԧ&#3627408462;+&#3627409158;
2
Ԧ&#3627408463;+&#3627409158;
3Ԧ&#3627408464;=0(Linear Dependent)
&#3627409158;
1,&#3627409158;
2,&#3627409158;
3≠0

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Linearly Dependent and Independently Vectors :
Four or more vectors is a Linearly Dependent System .
Consider four vectors Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;
Proof :
Case 1∶Three of them are coplanar.
Ԧ&#3627408462;=&#3627408485;&#3627408463;+&#3627408486;&#3627408464;+0Ԧ&#3627408465;
&#3627408485;&#3627408463;+&#3627408486;&#3627408464;+0Ԧ&#3627408465;−Ԧ&#3627408462;=0
Hence Linear Dependent.
Case 2∶Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;are non-coplanar.
Ԧ&#3627408462;=&#3627408485;&#3627408463;+&#3627408486;&#3627408464;+&#3627408487;Ԧ&#3627408465;(Theorem in Space)
Hence Linear Dependent

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Let Ԧ&#3627408462;=&#3627408467;&#3627408485;ƶ&#3627408470;−&#3627408467;

&#3627408485;ƶ&#3627408471;,Ԧ&#3627408463;=&#3627408468;&#3627408485;ƶ&#3627408470;+&#3627408468;

&#3627408485;ƶ&#3627408471;be two non-zero vectors and
ℎ&#3627408485;be the anti derivative of &#3627408467;&#3627408485;×&#3627408468;&#3627408485;.If ℎ1=1,ℎ2=3,ℎ4=7,
then Ԧ&#3627408462;and Ԧ&#3627408463;are linearly dependent for
A
B
D
C
at least one &#3627408485;∈1,2
at least one &#3627408485;∈1,4
at least two &#3627408485;∈1,4
at least one &#3627408485;∈2,4

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Solution :
Let Ԧ&#3627408462;=&#3627408467;&#3627408485;ƶ&#3627408470;−&#3627408467;

&#3627408485;ƶ&#3627408471;,Ԧ&#3627408463;=&#3627408468;&#3627408485;ƶ&#3627408470;+&#3627408468;

&#3627408485;ƶ&#3627408471;be two non-zero vectors and ℎ&#3627408485;
be the anti derivative of &#3627408467;&#3627408485;×&#3627408468;&#3627408485;.If ℎ1=1,ℎ2=3,ℎ4=7,then Ԧ&#3627408462;
and Ԧ&#3627408463;are linearly dependent for


&#3627408485;=&#3627408468;&#3627408485;×&#3627408467;&#3627408485;
Applying LMVT on ℎ&#3627408485;in 1,2i.e. &#3627408438;
1∈1,2


&#3627408438;
1 =3−1=2=
ℎ2−ℎ1
2−1
for &#3627408438;
1∈(1,2)
→ℎ&#3627408485;is differentiable
Applying LMVT on ℎ&#3627408485;in 2,4


&#3627408438;
2 =2=
ℎ4−ℎ2
4−2
for &#3627408438;
2∈(2,4)
Applying Rolle’s Theorem on ℎ

&#3627408485;on &#3627408438;
1,&#3627408438;
2, ℎ

&#3627408438;
1=ℎ′(&#3627408438;
2)

′′
&#3627408485;=0for some &#3627408438;∈1,4

′′
&#3627408485;=&#3627408468;

&#3627408485;&#3627408467;&#3627408485;+&#3627408467;

&#3627408485;&#3627408468;&#3627408485;=0for some &#3627408485;∈1,4

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Solution :
Let Ԧ&#3627408462;=&#3627408467;&#3627408485;ƶ&#3627408470;−&#3627408467;

&#3627408485;ƶ&#3627408471;,Ԧ&#3627408463;=&#3627408468;&#3627408485;ƶ&#3627408470;+&#3627408468;

&#3627408485;ƶ&#3627408471;be two non-zero vectors and ℎ&#3627408485;
be the anti derivative of &#3627408467;&#3627408485;×&#3627408468;&#3627408485;.If ℎ1=1,ℎ2=3,ℎ4=7,then Ԧ&#3627408462;
and Ԧ&#3627408463;are linearly dependent for

′′
&#3627408485;=&#3627408468;

&#3627408485;&#3627408467;&#3627408485;+&#3627408467;

&#3627408485;&#3627408468;&#3627408485;=0for some &#3627408485;∈1,4
So, Ԧ&#3627408462;∥Ԧ&#3627408463;⇒Linearly Dependent
then,
&#3627408467;&#3627408485;
&#3627408468;&#3627408485;
=
−&#3627408467;

&#3627408485;
&#3627408468;

&#3627408485;
→Ratio of Ƹ&#3627408470;&Ƹ&#3627408471;are equal
for atleastone &#3627408485;∈1,4
Coefficients of Ԧ&#3627408462;and Ԧ&#3627408463;are equal.

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Let Ԧ&#3627408462;=&#3627408467;&#3627408485;ƶ&#3627408470;−&#3627408467;

&#3627408485;ƶ&#3627408471;,Ԧ&#3627408463;=&#3627408468;&#3627408485;ƶ&#3627408470;+&#3627408468;

&#3627408485;ƶ&#3627408471;be two non-zero vectors and ℎ&#3627408485;
be the anti derivative of &#3627408467;&#3627408485;×&#3627408468;&#3627408485;.If ℎ1=1,ℎ2=3,ℎ4=7,then Ԧ&#3627408462;
and Ԧ&#3627408463;are linearly dependent for
A
B
D
C
at least one &#3627408485;∈1,2
at least one &#3627408485;∈1,4
at least two &#3627408485;∈1,4
at least one &#3627408485;∈2,4

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Coplanarity of Four Points :
Method 1:&#3627408436;&#3627408437;,&#3627408436;&#3627408438;&&#3627408436;&#3627408439;are coplanar.
there exists scalars &#3627408485;,&#3627408486;,&#3627408487;and &#3627408481;not all
simultaneously zero
satisfying&#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;+&#3627408481;Ԧ&#3627408465;=0.
&#3627408436;Ԧ&#3627408462;
&#3627408439;Ԧ&#3627408465;
&#3627408438;Ԧ&#3627408464;
&#3627408437;&#3627408463;
&#3627408436;&#3627408438;&#3627408436;&#3627408437;&#3627408436;&#3627408439;=0
Method 2:4points Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;,Ԧ&#3627408465;are coplanar if
where &#3627408485;+&#3627408486;+&#3627408487;+&#3627408481;=0
For &#3627408485;Ԧ&#3627408462;+&#3627408486;Ԧ&#3627408463;+&#3627408487;Ԧ&#3627408464;=0
Ԧ&#3627408464;=
&#3627408485;&#3627408462;+&#3627408486;&#3627408463;
&#3627408485;+&#3627408486;
→Section formula
Collinearity of Ԧ&#3627408462;,b&Ԧ&#3627408464;,&#3627408485;+&#3627408486;+&#3627408487;=0⇒&#3627408487;=−&#3627408485;+&#3627408486;

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Reciprocal System of Vectors :
Ԧ&#3627408462;⋅Ԧ&#3627408462;

=Ԧ&#3627408463;⋅Ԧ&#3627408463;

=Ԧ&#3627408464;⋅Ԧ&#3627408464;

=1
If Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408462;

,Ԧ&#3627408463;

,Ԧ&#3627408464;

are 2sets of non-coplanar vectors such that
then Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;and Ԧ&#3627408462;

,Ԧ&#3627408463;

,Ԧ&#3627408464;

are said to constitute a reciprocal system of vectors.
Ԧ&#3627408462;⋅Ԧ&#3627408462;

=Ԧ&#3627408463;⋅Ԧ&#3627408463;

=Ԧ&#3627408464;⋅Ԧ&#3627408464;

=1
Ԧ&#3627408462;⋅Ԧ&#3627408463;

=Ԧ&#3627408462;⋅Ԧ&#3627408464;

=Ԧ&#3627408463;⋅Ԧ&#3627408462;

=Ԧ&#3627408463;⋅Ԧ&#3627408464;

=Ԧ&#3627408464;⋅Ԧ&#3627408462;

=Ԧ&#3627408464;⋅Ԧ&#3627408463;

=0
Reciprocal system of vectors exists only in case of dot product.
It is possible to define Ԧ&#3627408462;

,Ԧ&#3627408463;

,Ԧ&#3627408464;

in terms of Ԧ&#3627408462;,Ԧ&#3627408463;,Ԧ&#3627408464;as
Ԧ&#3627408462;

=
&#3627408463;×Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
;Ԧ&#3627408463;

=
Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
;Ԧ&#3627408464;

=
&#3627408462;×&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;≠0

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Reciprocal System of Vectors :
Ԧ&#3627408462;×Ԧ&#3627408462;

+Ԧ&#3627408463;×Ԧ&#3627408463;

+Ԧ&#3627408464;×Ԧ&#3627408464;

=0
Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;⋅Ԧ&#3627408462;

+Ԧ&#3627408463;

+Ԧ&#3627408464;

=3
If Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=??????then Ԧ&#3627408462;
′Ԧ&#3627408463;

Ԧ&#3627408464;

=
1
??????
Note :
Ԧ&#3627408462;×Ԧ&#3627408462;

+Ԧ&#3627408463;×Ԧ&#3627408463;

+Ԧ&#3627408464;×Ԧ&#3627408464;

=
&#3627408462;×&#3627408463;×Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
+
&#3627408463;×Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
+
Ԧ&#3627408464;×&#3627408462;×&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
&#3627408462;⋅Ԧ&#3627408464;&#3627408463;−&#3627408462;⋅&#3627408463;Ԧ&#3627408464;+&#3627408463;⋅&#3627408462;Ԧ&#3627408464;−&#3627408463;⋅Ԧ&#3627408464;&#3627408462;+Ԧ&#3627408464;⋅&#3627408463;&#3627408462;−Ԧ&#3627408464;⋅&#3627408462;&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=0
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;
′Ԧ&#3627408463;

Ԧ&#3627408464;

=1

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Reciprocal System of Vectors :
If Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;=??????then Ԧ&#3627408462;
′Ԧ&#3627408463;

Ԧ&#3627408464;

=
1
??????
Proof :
⇒Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;Ԧ&#3627408462;
′Ԧ&#3627408463;

Ԧ&#3627408464;

=1
&#3627408462;′=
&#3627408463;×Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
&#3627408463;×Ԧ&#3627408464;
??????
;&#3627408463;′=
Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
Ԧ&#3627408464;×&#3627408462;
??????
;&#3627408464;′=
&#3627408462;×&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
&#3627408462;×&#3627408463;
??????
=
1
??????
3
Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;Ԧ&#3627408462;×Ԧ&#3627408463;
=
1
??????
3
Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
1
??????
3
×??????
2
∵Ԧ&#3627408462;×Ԧ&#3627408463;Ԧ&#3627408463;×Ԧ&#3627408464;Ԧ&#3627408464;×Ԧ&#3627408462;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
2
=
1
??????
Ԧ&#3627408462;
′Ԧ&#3627408463;

Ԧ&#3627408464;

=
&#3627408463;×Ԧ&#3627408464;
??????
Ԧ&#3627408464;×&#3627408462;
??????
&#3627408462;×&#3627408463;
??????

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Ԧ&#3627408462;

×Ԧ&#3627408463;

+Ԧ&#3627408463;

×Ԧ&#3627408464;

+Ԧ&#3627408464;

×Ԧ&#3627408462;

=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
,Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;≠0
&#3627408455;
1=Ԧ&#3627408462;

×Ԧ&#3627408463;

=
&#3627408463;×Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
×
Ԧ&#3627408464;×&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
&#3627408463;Ԧ&#3627408464;&#3627408462;Ԧ&#3627408464;−&#3627408463;Ԧ&#3627408464;Ԧ&#3627408464;&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
2
Reciprocal System of Vectors :
Note :
Ԧ&#3627408462;×Ԧ&#3627408463;×Ԧ&#3627408464;×Ԧ&#3627408465;=Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408465;Ԧ&#3627408464;−Ԧ&#3627408465;Ԧ&#3627408462;Ԧ&#3627408463;Ԧ&#3627408464;
&#3627408455;
2=Ԧ&#3627408463;

×Ԧ&#3627408464;

=
&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
,
&#3627408455;
3=&#3627408464;′×&#3627408462;′=
&#3627408463;
&#3627408462;&#3627408463;Ԧ&#3627408464;
⇒&#3627408455;
1+&#3627408455;
2+&#3627408455;
3=
&#3627408462;+&#3627408463;+Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;
=
&#3627408462;&#3627408463;Ԧ&#3627408464;&#3627408462;
&#3627408462;&#3627408463;Ԧ&#3627408464;
2=
Ԧ&#3627408464;
&#3627408462;&#3627408463;Ԧ&#3627408464;

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Solving Simultaneous Vector Equations For Unknown Vectors
➢Satisfy a given relationship with some known vectors.
➢There is no general method for solving such equations, however dot or cross with
known or unknown vectors or dot with Ԧ&#3627408462;×Ԧ&#3627408463;,generally isolates the unknown vectors.
➢Use of linear combination also proves to be advantageous.

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Find Ԧ&#3627408479;such that &#3627408481;Ԧ&#3627408479;+Ԧ&#3627408479;×Ԧ&#3627408462;=Ԧ&#3627408463;, where Ԧ&#3627408462;and Ԧ&#3627408463;are non collinear vectors.
Solution:
&#3627408481;Ԧ&#3627408479;+Ԧ&#3627408479;×Ԧ&#3627408462;=Ԧ&#3627408463;⋯&#3627408470;
&#3627408481;Ԧ&#3627408479;⋅Ԧ&#3627408462;+(Ԧ&#3627408479;×Ԧ&#3627408462;)⋅Ԧ&#3627408462;=Ԧ&#3627408463;⋅Ԧ&#3627408462;
Ԧ&#3627408462;and Ԧ&#3627408463;are non collinear vectors.
Ԧ&#3627408479;=?
⇒Ԧ&#3627408479;⋅Ԧ&#3627408462;=
Ԧ&#3627408463;⋅Ԧ&#3627408462;
&#3627408481;
⋯&#3627408470;&#3627408470;Ԧ&#3627408479;×Ԧ&#3627408462;=Ԧ&#3627408463;−&#3627408481;Ԧ&#3627408479;⋯&#3627408470;&#3627408470;&#3627408470;
Taking cross product with Ԧ&#3627408462;in eq &#3627408470;
&#3627408481;Ԧ&#3627408479;×Ԧ&#3627408462;+Ԧ&#3627408479;×Ԧ&#3627408462;×Ԧ&#3627408462;=Ԧ&#3627408463;×Ԧ&#3627408462;
⇒&#3627408481;Ԧ&#3627408463;−&#3627408481;Ԧ&#3627408479;+Ԧ&#3627408462;⋅Ԧ&#3627408479;Ԧ&#3627408462;−Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408479;=Ԧ&#3627408463;×Ԧ&#3627408462;
Taking dot product with Ԧ&#3627408462;in eq &#3627408470;

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Find Ԧ&#3627408479;such that &#3627408481;Ԧ&#3627408479;+Ԧ&#3627408479;×Ԧ&#3627408462;=Ԧ&#3627408463;, where Ԧ&#3627408462;and Ԧ&#3627408463;are non collinear vectors.
Solution:
&#3627408481;Ԧ&#3627408479;+Ԧ&#3627408479;×Ԧ&#3627408462;=Ԧ&#3627408463;⋯&#3627408470; Ԧ&#3627408462;and Ԧ&#3627408463;are non collinear vectors.
Ԧ&#3627408479;=?
⇒&#3627408481;Ԧ&#3627408463;−&#3627408481;Ԧ&#3627408479;+Ԧ&#3627408462;⋅Ԧ&#3627408479;Ԧ&#3627408462;−Ԧ&#3627408462;⋅Ԧ&#3627408462;Ԧ&#3627408479;=Ԧ&#3627408463;×Ԧ&#3627408462;
⇒&#3627408481;Ԧ&#3627408463;−&#3627408481;
2
+&#3627408462;
2
Ԧ&#3627408479;+
&#3627408463;⋅&#3627408462;
&#3627408481;
Ԧ&#3627408462;=Ԧ&#3627408463;×Ԧ&#3627408462;
⇒&#3627408481;
2
+&#3627408462;
2
Ԧ&#3627408479;=&#3627408481;Ԧ&#3627408463;+
&#3627408463;⋅&#3627408462;
&#3627408481;
Ԧ&#3627408462;+Ԧ&#3627408462;×Ԧ&#3627408463;
⇒Ԧ&#3627408479;=
1
&#3627408481;
2
+&#3627408462;
2
&#3627408481;Ԧ&#3627408463;+
&#3627408463;⋅&#3627408462;
&#3627408481;
Ԧ&#3627408462;+Ԧ&#3627408462;×Ԧ&#3627408463;

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Solve the following simultaneous equations for Ԧ&#3627408485;and Ԧ&#3627408486;:
Ԧ&#3627408485;+Ԧ&#3627408486;=Ԧ&#3627408462;,Ԧ&#3627408485;×Ԧ&#3627408486;=Ԧ&#3627408463;and Ԧ&#3627408485;⋅Ԧ&#3627408462;=1
Solution:

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Solution:
Solve the following simultaneous equations for Ԧ&#3627408485;and Ԧ&#3627408486;:
Ԧ&#3627408485;+Ԧ&#3627408486;=Ԧ&#3627408462;,Ԧ&#3627408485;×Ԧ&#3627408486;=Ԧ&#3627408463;and Ԧ&#3627408485;⋅Ԧ&#3627408462;=1
Putting the value of Ԧ&#3627408486;from &#3627408470;in &#3627408470;&#3627408470;
⇒Ԧ&#3627408485;×Ԧ&#3627408462;−Ԧ&#3627408485;=Ԧ&#3627408463;⇒Ԧ&#3627408485;×Ԧ&#3627408462;=Ԧ&#3627408463;
⇒Ԧ&#3627408462;×Ԧ&#3627408485;×Ԧ&#3627408462;=Ԧ&#3627408462;×Ԧ&#3627408463;
⇒&#3627408462;
2
Ԧ&#3627408485;−Ԧ&#3627408462;⋅Ԧ&#3627408485;Ԧ&#3627408462;=Ԧ&#3627408462;×Ԧ&#3627408463;
⇒&#3627408462;
2
Ԧ&#3627408485;−Ԧ&#3627408462;=Ԧ&#3627408462;×Ԧ&#3627408463;
⇒Ԧ&#3627408485;=
1
&#3627408462;
2
Ԧ&#3627408462;+Ԧ&#3627408462;×Ԧ&#3627408463;∴Ԧ&#3627408486;=Ԧ&#3627408462;−
1
&#3627408462;
2
Ԧ&#3627408462;+Ԧ&#3627408462;×Ԧ&#3627408463;
Ԧ&#3627408485;×Ԧ&#3627408486;=Ԧ&#3627408463;⋯&#3627408470;&#3627408470;Ԧ&#3627408485;⋅Ԧ&#3627408462;=1⋯&#3627408470;&#3627408470;&#3627408470;Ԧ&#3627408485;+Ԧ&#3627408486;=Ԧ&#3627408462;⋯&#3627408470;

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Parametric Vector Equation of a Straight Line
Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408463;→vector equation of required straight line.
Vector equation of a straight line passing through a
given point &#3627408436;(Ԧ&#3627408462;)and parallel to a given vector &#3627408437;(Ԧ&#3627408463;)
Ԧ&#3627408479;=&#3627408485;Ƹ&#3627408470;+&#3627408486;Ƹ&#3627408471;+&#3627408487;෠&#3627408472;
⇒Ԧ&#3627408479;−Ԧ&#3627408462;Ԧ&#3627408463;⇒Ԧ&#3627408479;−Ԧ&#3627408462;=&#3627409158;Ԧ&#3627408463;
where &#3627409158;is a scalar and for different values of &#3627409158;, we
get different positionsof point &#3627408453;.
&#3627408436;(Ԧ&#3627408462;) &#3627408437;(Ԧ&#3627408463;)
&#3627408453;(Ԧ&#3627408479;)
⇒&#3627408436;&#3627408453;&#3627408463;
Ԧ&#3627408462;→point through which line is passing.
Ԧ&#3627408463;→Direction to which line is parallel.
∴Ԧ&#3627408479;=Ƹ&#3627408470;+Ƹ&#3627408471;+&#3627409159;2Ƹ&#3627408470;−Ƹ&#3627408471;+෠&#3627408472;

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➢However, in space we can have two neither parallel nor intersecting
lines. Such non coplanar lines are known as skew lines. If two lines are
parallel and have a common point then they are coincident.
➢Two lines in a plane are either intersecting or parallel conversely two
intersecting or parallel lines must be in the same plane.
Parametric Vector Equation of a Straight Line

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➢However, in space we can have two neither parallel nor intersecting
lines. Such non coplanar lines are known as skew lines. If two lines are
parallel and have a common point then they are coincident.
Parametric Vector Equation of a Straight Line

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A vector in the direction of the angle bisector between the two
vectors Ԧ&#3627408462;andԦ&#3627408463;.
Solution:

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A vector in the direction of the angle bisector between the two
vectors Ԧ&#3627408462;andԦ&#3627408463;.
Solution:
Ԧ&#3627408462;
&#3627408463;
&#3627409158;ො&#3627408462;+෠&#3627408463;
ො&#3627408462;
෠&#3627408463;
&#3627409158;ො&#3627408462;−෠&#3627408463;

Ԧ&#3627408462;+Ԧ&#3627408463;
Ԧ&#3627408462;+Ԧ&#3627408463;
Vectorinthedirectionoftheanglebisector

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Shortest between two skew
lines
Session 09
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Vector Equation of Angle Bisectors between Two Straight Lines
Angle bisector also pass through &#3627408436;Ԧ&#3627408462;
Line 1: Ԧ&#3627408479;= Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408463;⋯(&#3627408470;)
&#3627408463;
&#3627408451;(෠&#3627408463;)
Ԧ&#3627408464;
&#3627408452;(Ƹ&#3627408464;)
&#3627408448;
&#3627408436;(Ԧ&#3627408462;)
Line 2: Ԧ&#3627408479;= Ԧ&#3627408462;+&#3627409159;Ԧ&#3627408464;⋯(&#3627408470;&#3627408470;)
Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627408481;෠&#3627408463;+Ƹ&#3627408464;
Ԧ&#3627408462;→Point, ෠&#3627408463;+Ƹ&#3627408464;→Direction of angle bisector
&#3627408481;is scalar

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Vector Equation of Angle Bisectors between Two Straight Lines
Line 2: Ԧ&#3627408479;= Ԧ&#3627408462;+&#3627409159;Ԧ&#3627408464;⋯(&#3627408470;&#3627408470;)
Line 1: Ԧ&#3627408479;= Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408463;⋯(&#3627408470;)
&#3627408463;
&#3627408451;(෠&#3627408463;)
Ԧ&#3627408464;
&#3627408452;(Ƹ&#3627408464;)
&#3627408448;
&#3627408436;(Ԧ&#3627408462;)
Vector in the direction of external
angle bisector =෠&#3627408463;−ොc
Angle bisector also pass through &#3627408436;Ԧ&#3627408462;
Equation of external angle bisector: Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627408480;෠&#3627408463;−Ƹ&#3627408464;
&#3627408480;isscalar

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A vector Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;(&#3627409148;,&#3627409149;,∈ℝ)lies in the plane of the vectors,
Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;and Ԧ&#3627408464;=ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;. If Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;, then :
JEE MAINS Mar 2021
A
B
D
C
Ԧ&#3627408462;⋅ƶ&#3627408472;+2=0
Ԧ&#3627408462;⋅ƶ&#3627408472;+4=0
Ԧ&#3627408462;⋅ƶ&#3627408472;−2=0
Ԧ&#3627408462;⋅ƶ&#3627408472;+5=0

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Solution:
A vector Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;(&#3627409148;,&#3627409149;,∈ℝ)lies in the plane of the vectors,
Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;and Ԧ&#3627408464;=ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;. If Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;, then :
JEE MAINS Mar 2021
Ԧ&#3627408462;=&#3627409158;
ƶ&#3627408470;+ƶ&#3627408471;
2
+
ƶ&#3627408470;+ƶ&#3627408471;+4ƶ&#3627408472;
32
=
&#3627409158;
32
[3ƶ&#3627408470;+3ƶ&#3627408471;+ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;]
=
&#3627409158;
32
[4ƶ&#3627408470;+2ƶ&#3627408471;+4ƶ&#3627408472;]
Compare with Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;
Angle bisector :Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;+Ԧ&#3627408464;)or Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;−Ԧ&#3627408464;)
Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;

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Solution:
A vector Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;(&#3627409148;,&#3627409149;,∈ℝ)lies in the plane of the vectors,
Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;and Ԧ&#3627408464;=ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;. If Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;, then :
JEE MAINS Mar 2021
Ԧ&#3627408462;=&#3627409158;
ƶ&#3627408470;+ƶ&#3627408471;
2
+
ƶ&#3627408470;+ƶ&#3627408471;+4ƶ&#3627408472;
32
=
&#3627409158;
32
[4ƶ&#3627408470;+2ƶ&#3627408471;+4ƶ&#3627408472;]
Compare with Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;
2&#3627409158;
32
=2
⇒Ԧ&#3627408462;=4ƶ&#3627408470;+2ƶ&#3627408471;+4ƶ&#3627408472;
⇒&#3627409158;=32
→Notinoption
Ԧ&#3627408462;=&#3627409158;
ƶ&#3627408470;+ƶ&#3627408471;
2

ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;
32
=
&#3627409158;
32
(3ƶ&#3627408470;+3ƶ&#3627408471;−ƶ&#3627408470;+ƶ&#3627408471;−4ƶ&#3627408472;)
Angle bisector :Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;+Ԧ&#3627408464;)
Angle bisector :Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;+Ԧ&#3627408464;)or Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;−Ԧ&#3627408464;)

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Solution:
A vector Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;(&#3627409148;,&#3627409149;,∈ℝ)lies in the plane of the vectors,
Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;and Ԧ&#3627408464;=ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;. If Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;, then :
JEE MAINS Mar 2021
Ԧ&#3627408462;=&#3627409158;
ƶ&#3627408470;+ƶ&#3627408471;
2

ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;
32
=
&#3627409158;
32
(3ƶ&#3627408470;+3ƶ&#3627408471;−ƶ&#3627408470;+ƶ&#3627408471;−4ƶ&#3627408472;)
=
&#3627409158;
32
(2ƶ&#3627408470;+4ƶ&#3627408471;−4ƶ&#3627408472;)
ComparewithԦ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;

4&#3627409158;
32
=2⇒&#3627409158;=
32
2
⇒Ԧ&#3627408462;=ƶ&#3627408470;+2ƶ&#3627408471;−2ƶ&#3627408472;⇒Ԧ&#3627408462;⋅ƶ&#3627408472;+2=0
Angle bisector :Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;+Ԧ&#3627408464;)
Angle bisector :Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;+Ԧ&#3627408464;)or Ԧ&#3627408462;=&#3627409158;(Ԧ&#3627408463;−Ԧ&#3627408464;)

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A vector Ԧ&#3627408462;=&#3627409148;ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409149;ƶ&#3627408472;(&#3627409148;,&#3627409149;,∈ℝ)lies in the plane of the vectors,
Ԧ&#3627408463;=ƶ&#3627408470;+ƶ&#3627408471;and Ԧ&#3627408464;=ƶ&#3627408470;−ƶ&#3627408471;+4ƶ&#3627408472;. If Ԧ&#3627408462;bisects the angle between Ԧ&#3627408463;and Ԧ&#3627408464;, then :
JEE MAINS Mar 2021
B
D
C
Ԧ&#3627408462;⋅ƶ&#3627408472;+2=0
Ԧ&#3627408462;⋅ƶ&#3627408472;+4=0
Ԧ&#3627408462;⋅ƶ&#3627408472;−2=0
Ԧ&#3627408462;⋅ƶ&#3627408472;+5=0
A

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2lines in a plane if not ∥must intersect & vice versa.
Conversely, 2intersecting or parallel lines must be
coplanar.


Shortest distance between 2skew lines
In space, however we come across situation when
two lines neither intersect nor ∥are known as skew
lines or non coplanar lines.

Note:
Shortest distance always lie along the common
normal.

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Method I
Two ways to determine the shortest distance
&#3627408447;
1∶Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408477;
&#3627408447;
2∶Ԧ&#3627408479;=Ԧ&#3627408463;+&#3627409159;Ԧ&#3627408478;
&#3627408475;=Ԧ&#3627408477;×Ԧ&#3627408478;
&#3627408436;&#3627408437;=Ԧ&#3627408463;−Ԧ&#3627408462;
Shortest distance=|Projection of &#3627408436;&#3627408437;on &#3627408475;|
=
&#3627408436;&#3627408437;⋅&#3627408475;
&#3627408475;
=
&#3627408463;−&#3627408462;⋅Ԧ&#3627408477;×&#3627408478;
Ԧ&#3627408477;×&#3627408478;
&#3627408436;(Ԧ&#3627408462;)
&#3627408437;(Ԧ&#3627408463;)
Ԧ&#3627408477;
Ԧ&#3627408478;
&#3627408447;
1
&#3627408447;
2
&#3627408449;
1
&#3627408449;
2

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Method II
For a fixed &#3627409158;:&#3627408449;
1=Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408477;
&#3627408449;
2−&#3627408449;
1=Ԧ&#3627408463;−Ԧ&#3627408462;+&#3627409159;Ԧ&#3627408478;−&#3627409158;Ԧ&#3627408477;
(Two linear equations to get the unique values of &#3627409158;and &#3627409159;)
Let &#3627408449;
1&#3627408449;
2be the line perpendicular to &#3627408447;
1&&#3627408447;
2
&#3627408436;(Ԧ&#3627408462;)
&#3627408437;(Ԧ&#3627408463;)
Ԧ&#3627408477;
Ԧ&#3627408478;
&#3627408447;
1
&#3627408447;
2
&#3627408449;
1
&#3627408449;
2
For a fixed &#3627409159;:&#3627408449;
2=Ԧ&#3627408463;+&#3627409159;Ԧ&#3627408478;
&#3627408449;
2−&#3627408449;
1⋅Ԧ&#3627408477;=0
&#3627408449;
2−&#3627408449;
1⋅Ԧ&#3627408478;=0
Two equation in &#3627409158;&&#3627409159;
&#3627408454;.&#3627408439;=&#3627408449;
2−&#3627408449;
1

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We can also determine the equation to the line of shortest
distance and the shortest distance between lines :
&#3627408465;=&#3627408449;
2−&#3627408449;
1
Once position vectors of &#3627408449;
1and &#3627408449;
2are known,
Method II
&#3627408449;
1=Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408477;
Let &#3627408449;
1&#3627408449;
2be the line perpendicular to &#3627408447;
1&&#3627408447;
2
&#3627408449;
2=Ԧ&#3627408463;+&#3627409159;Ԧ&#3627408478;
&#3627408449;
2−&#3627408449;
1⋅Ԧ&#3627408477;=0
&#3627408449;
2−&#3627408449;
1⋅Ԧ&#3627408478;=0
&#3627408436;(Ԧ&#3627408462;)
&#3627408437;(Ԧ&#3627408463;)
Ԧ&#3627408477;
Ԧ&#3627408478;
&#3627408447;
1
&#3627408447;
2
&#3627408449;
1
&#3627408449;
2

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Find the shortest distance between the two lines whose vector equations are
given by: Ԧ&#3627408479;=ƶ&#3627408470;+2ƶ&#3627408471;+3ƶ&#3627408472;+&#3627409158;2ƶ&#3627408470;+3ƶ&#3627408471;+4ƶ&#3627408472;and Ԧ&#3627408479;=2ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;+&#3627409159;3ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;
Solution:

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&#3627408463;
1×&#3627408463;
2=−1
2
+2
2
+−1
2
=6
&#3627408462;
2−&#3627408462;
1=ƶ&#3627408470;+2ƶ&#3627408471;+2ƶ&#3627408472;
Find the shortest distance between the two lines whose vector equations are
given by: Ԧ&#3627408479;=ƶ&#3627408470;+2ƶ&#3627408471;+3ƶ&#3627408472;+&#3627409158;2ƶ&#3627408470;+3ƶ&#3627408471;+4ƶ&#3627408472;and Ԧ&#3627408479;=2ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;+&#3627409159;3ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;
Solution:Ԧ&#3627408479;=ƶ&#3627408470;+2ƶ&#3627408471;+3ƶ&#3627408472;+&#3627409158;2ƶ&#3627408470;+3ƶ&#3627408471;+4ƶ&#3627408472;
Ԧ&#3627408479;=2ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;+&#3627409159;3ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;
&#3627408463;
1×&#3627408463;
2=
Ƹ&#3627408470;Ƹ&#3627408471;෠&#3627408472;
234
345
=Ƹ&#3627408470;15−16−Ƹ&#3627408471;10−12+෠&#3627408472;8−9=−Ƹ&#3627408470;+2Ƹ&#3627408471;−෠&#3627408472;
&#3627408463;
1
&#3627408462;
1
&#3627408462;
2 &#3627408463;
2
&#3627408454;&#3627408439;=
&#3627408462;2−&#3627408462;1⋅&#3627408463;1×&#3627408463;2
&#3627408463;1×&#3627408463;2

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=1×−1+2×2+2×−1=1
&#3627408462;
2−&#3627408462;
1⋅&#3627408463;
2×&#3627408463;
1=ƶ&#3627408470;+2ƶ&#3627408471;+2ƶ&#3627408472;⋅−ƶ&#3627408470;+2ƶ&#3627408471;−ƶ&#3627408472;
&#3627408462;
2−&#3627408462;
1=ƶ&#3627408470;+2ƶ&#3627408471;+2ƶ&#3627408472;
&#3627408463;
1×&#3627408463;
2=−Ƹ&#3627408470;+2Ƹ&#3627408471;−෠&#3627408472;
⇒&#3627408465;=
1
6
=
1
6
Find the shortest distance between the two lines whose vector equations are
given by: Ԧ&#3627408479;=ƶ&#3627408470;+2ƶ&#3627408471;+3ƶ&#3627408472;+&#3627409158;2ƶ&#3627408470;+3ƶ&#3627408471;+4ƶ&#3627408472;and Ԧ&#3627408479;=2ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;+&#3627409159;3ƶ&#3627408470;+4ƶ&#3627408471;+5ƶ&#3627408472;
Solution:
&#3627408454;&#3627408439;=
&#3627408462;2−&#3627408462;1⋅&#3627408463;1×&#3627408463;2
&#3627408463;1×&#3627408463;2
&#3627408463;
1×&#3627408463;
2=6

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Given a tetrahedron &#3627408450;&#3627408436;&#3627408437;&#3627408438;where &#3627408436;&#3627408437;=12,&#3627408438;&#3627408450;=6. If the shortest
distance between&#3627408436;&#3627408437;and&#3627408438;&#3627408450;is8and angle between them is
??????
6
,
then find the volume of&#3627408450;&#3627408436;&#3627408437;&#3627408438;.
Solution:

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Ԧ&#3627408463;−Ԧ&#3627408462;=12 Ԧ&#3627408464;=6
Vector equation of &#3627408438;&#3627408450;:Ԧ&#3627408479;=&#3627409158;Ԧ&#3627408464;
Vector equation of &#3627408436;&#3627408437;:Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627409159;Ԧ&#3627408463;−Ԧ&#3627408462;
&#3627408450;
&#3627408438;
&#3627408437;
Ԧ&#3627408462;
Ԧ&#3627408464;
12
6
8=
&#3627408462;−0⋅Ԧ&#3627408464;×&#3627408463;−&#3627408462;
Ԧ&#3627408464;×&#3627408463;−&#3627408462;
=
&#3627408462;⋅Ԧ&#3627408464;×&#3627408463;−Ԧ&#3627408464;×&#3627408462;
Ԧ&#3627408464;&#3627408463;−&#3627408462;sin
??????
6
⇒8=
&#3627408462;⋅Ԧ&#3627408464;×&#3627408463;−0
Ԧ&#3627408464;&#3627408463;−&#3627408462;sin??????
=
&#3627408462;&#3627408463;Ԧ&#3627408464;
6×6

&#3627408462;&#3627408463;Ԧ&#3627408464;
6
=8×6=48
∴Required volume =48units
Given a tetrahedron &#3627408450;&#3627408436;&#3627408437;&#3627408438;where &#3627408436;&#3627408437;=12,&#3627408438;&#3627408450;=6. If the shortest
distance between&#3627408436;&#3627408437;and&#3627408438;&#3627408450;is8and angle between them is
??????
6
,
then find the volume of&#3627408450;&#3627408436;&#3627408437;&#3627408438;.
Solution:
Ԧ&#3627408463;
&#3627408436;

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Shortest distance between 2skew lines

Note
To determine whether lines are skew or intersecting :
If shortest distance =0
⇒Then the lines are intersecting and hence coplanar.

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Determine whether the following pair of lines intersect, Ԧ&#3627408479;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;(2ƶ&#3627408470;+ƶ&#3627408472;)
and Ԧ&#3627408479;=4ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409159;(ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;).
Solution:
&#3627408454;&#3627408439;=
&#3627408462;2−&#3627408462;1⋅&#3627408463;1×&#3627408463;2
&#3627408463;1×&#3627408463;2
The two lines will intersect if and only if &#3627408465;=0.
Ԧ&#3627408463;
1×Ԧ&#3627408463;
2=
ƶ&#3627408470;ƶ&#3627408471;ƶ&#3627408472;
201
11−1
=−ƶ&#3627408470;+3ƶ&#3627408471;+2ƶ&#3627408472;
Ԧ&#3627408462;
2−Ԧ&#3627408462;
1=3ƶ&#3627408470;+ƶ&#3627408471;
∴Ԧ&#3627408462;
2−Ԧ&#3627408462;
&#3627408473;⋅Ԧ&#3627408463;
&#3627408473;×Ԧ&#3627408463;
2=3ƶ&#3627408470;+ƶ&#3627408471;⋅−ƶ&#3627408470;+3ƶ&#3627408471;+2ƶ&#3627408472;
=−13+31+20=0
&#3627408447;
1∶Ԧ&#3627408479;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;2ƶ&#3627408470;+ƶ&#3627408472;&#3627408447;
2∶Ԧ&#3627408479;=4ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409159;ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;.
&#3627408462;
1 &#3627408463;
1
&#3627408462;
2 &#3627408463;
2

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&#3627408462;
2−&#3627408462;
1⋅Ԧ&#3627408463;
1×Ԧ&#3627408463;
2=
ƶ3ƶ10
201
11−1
=3−1−1−3+0=0
∴Shortest distance =0
Hence, the given lines intersect.
⇒Ԧ&#3627408462;
2−Ԧ&#3627408462;
1⋅Ԧ&#3627408463;
1×Ԧ&#3627408463;
2=0
Determine whether the following pair of lines intersect, Ԧ&#3627408479;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;(2ƶ&#3627408470;+ƶ&#3627408472;)
and Ԧ&#3627408479;=4ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409159;(ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;).
Solution:
Ԧ&#3627408462;
2−Ԧ&#3627408462;
1=3ƶ&#3627408470;+ƶ&#3627408471;
&#3627408447;
1∶Ԧ&#3627408479;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;2ƶ&#3627408470;+ƶ&#3627408472;&#3627408447;
2∶Ԧ&#3627408479;=4ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409159;ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;.
&#3627408462;
1 &#3627408463;
1
&#3627408462;
2 &#3627408463;
2
The two lines will intersect if and only if &#3627408465;=0.
&#3627408454;&#3627408439;=
&#3627408462;2−&#3627408462;1⋅&#3627408463;1×&#3627408463;2
&#3627408463;1×&#3627408463;2

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Shortest distance between 2skew lines

Note
To determine whether lines are skew or intersecting
and find the intersection point if exists :
If all the equation satisfies, then they are intersecting
and there exists a point of intersection.
Take points on line &#3627408447;
1and line &#3627408447;
2,and equate them.
By equating the coefficients of Ƹ&#3627408470;,Ƹ&#3627408471;and ෠&#3627408472;.

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Determine whether the following pair of lines intersect, and find the point
of intersection if it exists. Ԧ&#3627408479;=ƶ&#3627408470;+ƶ&#3627408471;+&#3627409158;(2ƶ&#3627408470;+ƶ&#3627408472;)and Ԧ&#3627408479;=4ƶ&#3627408470;+2ƶ&#3627408471;+&#3627409159;(ƶ&#3627408470;+ƶ&#3627408471;−ƶ&#3627408472;).
Solution:
&#3627408462;
2=4+&#3627409159;Ƹ&#3627408470;+2+&#3627409159;Ƹ&#3627408471;−&#3627409159;෠&#3627408472;
2&#3627409158;+1=4+&#3627409159;
1=2+&#3627409159;
Then, Point of intersection =3,1,1
Since line intersects, &#3627408462;
1=&#3627408462;
2
A general point on line &#3627408447;
1for a particular &#3627409158;
&#3627408462;
1=2&#3627409158;+1Ƹ&#3627408470;+Ƹ&#3627408471;+&#3627409158;෠&#3627408472;
If shortest distance =0,then lines are intersecting.
A general point on line &#3627408447;
2for a particular &#3627409159;
&#3627409158;=−&#3627409159;
&#3627409158;=1,&#3627409159;=−1
&#3627408447;
1
&#3627408447;
2

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Computing distance between two parallel lines
&#3627408447;
1∶Ԧ&#3627408479;=Ԧ&#3627408462;+&#3627409158;Ԧ&#3627408463;, &#3627408447;
2∶Ԧ&#3627408479;=Ԧ&#3627408464;+&#3627409159;Ԧ&#3627408463;
&#3627408437;Ԧ&#3627408463;
&#3627408436;Ԧ&#3627408462;
&#3627408447;
1
&#3627408447;
2
Ԧ&#3627408463;
&#3627408438;Ԧ&#3627408464;&#3627408439;
Ԧ&#3627408463;
Projection of &#3627408436;&#3627408438;on Ԧ&#3627408463;=&#3627408436;&#3627408438;⋅෠&#3627408463;=&#3627408438;&#3627408439;
⇒&#3627408438;&#3627408439;=&#3627408436;&#3627408438;⋅෠&#3627408463;
In Δ&#3627408436;&#3627408438;&#3627408439;
&#3627408436;&#3627408439;
2
=&#3627408436;&#3627408438;
2
−&#3627408438;&#3627408439;
2
&#3627408436;&#3627408439;=Shortest Distance =Ԧ&#3627408462;−Ԧ&#3627408464;
2
−&#3627408436;&#3627408438;⋅෠&#3627408463;
2

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Let Ԧ&#3627408462;andԦ&#3627408463;be two non-zero vectors perpendicular to each other and |Ԧ&#3627408462;|=|Ԧ&#3627408463;|.
If|Ԧ&#3627408462;×Ԧ&#3627408463;|=|Ԧ&#3627408462;|, then the angle between the vectors (Ԧ&#3627408462;+Ԧ&#3627408463;+(Ԧ&#3627408462;×Ԧ&#3627408463;))and Ԧ&#3627408462;is
equal to:
Solution:
Ԧ&#3627408462;=Ԧ&#3627408463;,Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408462;,Ԧ&#3627408462;⊥Ԧ&#3627408463;⇒|Ԧ&#3627408462;×Ԧ&#3627408463;|=|Ԧ&#3627408462;|
⇒??????=cos
−1
1
3
cos??????=
(ƶ&#3627408470;+&#3627408471;+ƶ&#3627408472;)∙ƶ&#3627408470;
31
=
1
3
Let Ԧ&#3627408462;=ƶ&#3627408470;,Ԧ&#3627408463;=ƶ&#3627408471;⇒Ԧ&#3627408462;×Ԧ&#3627408463;=ƶ&#3627408472;
Ԧ&#3627408462;and Ԧ&#3627408463;are mutually perpendicular unit vectors.
⇒Ԧ&#3627408462;|Ԧ&#3627408463;|sin90

=|Ԧ&#3627408462;|⇒|Ԧ&#3627408463;|=1=|Ԧ&#3627408462;|
JEE Main Mar 2021
A
B
C
D
sin
−1
1
3
cos
−1
1
3
cos
−1
1
2
sin
−1
1
6
Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408462;×Ԧ&#3627408463;=Ƹ&#3627408470;+Ƹ&#3627408471;+෠&#3627408472;

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JEE Main Jan 2020
A
B
D
C
3
2
,3Ԧ&#3627408462;×Ԧ&#3627408463;
3
2
,3Ԧ&#3627408462;×Ԧ&#3627408464;

3
2
,3Ԧ&#3627408462;×Ԧ&#3627408464;

3
2
,3Ԧ&#3627408462;×Ԧ&#3627408463;
Let Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;be three-unit vectors such that Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;=0. If &#3627409158;=Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;
and Ԧ&#3627408465;=Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;, then the ordered pair, &#3627409158;,Ԧ&#3627408465;is equal to:

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JEE Main Jan 2020
Let Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;be three-unit vectors such that Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;=0. If &#3627409158;=Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;
and Ԧ&#3627408465;=Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;, then the ordered pair, &#3627409158;,Ԧ&#3627408465;is equal to:
Solution:
⇒3+2Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;=0
⇒Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;=
−3
2
⇒&#3627409158;=
−3
2
Cross product with Ԧ&#3627408462;=Ԧ&#3627408462;×Ԧ&#3627408463;=Ԧ&#3627408464;×Ԧ&#3627408462;
Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;=0⇒|Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;|=0⇒Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;
2
=0
&#3627409158;=Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;
Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;=0cross product with Ԧ&#3627408463;
Ԧ&#3627408462;×Ԧ&#3627408463;+0+Ԧ&#3627408464;×Ԧ&#3627408463;=0⇒Ԧ&#3627408462;×Ԧ&#3627408463;=−Ԧ&#3627408464;×Ԧ&#3627408463;=Ԧ&#3627408463;×Ԧ&#3627408464;
Ԧ&#3627408465;=3Ԧ&#3627408462;×Ԧ&#3627408463;

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JEE Main Jan 2020
A
B
D
C
3
2
,3Ԧ&#3627408462;×Ԧ&#3627408463;
3
2
,3Ԧ&#3627408462;×Ԧ&#3627408464;

3
2
,3Ԧ&#3627408462;×Ԧ&#3627408464;

3
2
,3Ԧ&#3627408462;×Ԧ&#3627408463;
Let Ԧ&#3627408462;,Ԧ&#3627408463;and Ԧ&#3627408464;be three-unit vectors such that Ԧ&#3627408462;+Ԧ&#3627408463;+Ԧ&#3627408464;=0. If &#3627409158;=Ԧ&#3627408462;⋅Ԧ&#3627408463;+Ԧ&#3627408463;⋅Ԧ&#3627408464;+Ԧ&#3627408464;⋅Ԧ&#3627408462;
and Ԧ&#3627408465;=Ԧ&#3627408462;×Ԧ&#3627408463;+Ԧ&#3627408463;×Ԧ&#3627408464;+Ԧ&#3627408464;×Ԧ&#3627408462;, then the ordered pair, &#3627409158;,Ԧ&#3627408465;is equal to:

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