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VECTOR ALGEBRA
Vector and Scalar
(i) Scalar quantity is a quantity having magnitude but no direction
...
g
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(ii) Vector quantity is a quantity having both direction and magnitude
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g
...


Representation of Vectors
A vector is often denoted by two letters with an arrow over it, i
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⃗⃗⃗⃗⃗
𝐀𝐁
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Its magnitude is given by the length AB and direction is from A
to B as indicated by the arrow, we write vector quantities also in single letter notation
like a, b, c and the corresponding letters a, b, c shows their magnitude
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Types of Vectors
1
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Thus,
the modulus of the null vector is zero and it is denoted by 0 or ⃗𝟎
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2
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If a is a vector whose
magnitude is a, then unit vector in the direction of a is denoted by 𝐚̂ and is obtained by
dividing the vector a by its magnitude |𝐚̂|
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Coinitial vectors:
Vectors having the same initial point are called coinitial vectors
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Collinear or parallel vectors:
The vectors which are parallel to the same straight line
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Coplanar vectors:
Three or more vectors are said to be coplanar when they are parallel to the same plane
otherwise they are said to be non-coplanar vector whatever their magnitudes be
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Coterminous vectors:
Vectors having the same terminal points are called coterminous vectors
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Negative of a vector:
The vector which has the same magnitude as the vector a but opposite in direction, is
called the negative of a and is denoted by –a
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Reciprocal of a vector:
A vector having the same direction as that of a given vector a but magnitude is equal to
the reciprocal of the given vector, a and is denoted by a-1
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There are three methods of addition of vectors

1
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Symbolically,
we have PQ+QR=PR
e
a
or
a+b= c
q
u
R
ot
Q
2
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Symbolically, m
we have OP+OR=OQ or a+b=c
e
nt
3
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OA+AB+BC+CD+DE=OE
Also, (a+b)+c=a+(b+c)=a+b+c

Properties of Vector Addition
(i) Vector addition is commutative
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e
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i
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(a+b)+c=a+(b+c)

(iii)Existence of additive identity, for every vector a, we have,
a+0=a=0+a, where 0 is the null vector
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e
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Thus, to subtract b
from a, reverse the direction of b and add to a as
shown in figure

Multiplication of Vectors
Let m be a scalar and a be a vector, then their product is defined by ma or am
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Properties of Multiplication of Vectors by a Scalar
For vectors a, b and scalars m, n we have
1
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3
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5
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7
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m(-a)=(-m)a=-(ma)
(-m)(-a)=ma
m(na)=(mn)a =n(ma)
(m+n)a=ma+na
m(a+b)=ma+mb
m(a-b)=ma-mb
1
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Example 1
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(i)
Since AD is parallel to BC, then AD=2BC
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[by(i)and (ii)]
⇨CD=b-a
Now, CE=CD+DE
⇨CE=b-a+(-a) ……
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If we say that P is the point r, then we mean that the position vector of P is r with respect
to the same origin O
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Soln : Let the position vector of the point (12,n) be 12î + nĵ
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If a, b and c are non-coplanar vectors, then
xa+yb+zc=0⇨ x=y=z=0

Coplanarity of Four points
Four points with position vectors a, b, c and d are coplanar, if and only if there exist
scalars x, y, z, w not all zero such that
xa+yb+zc+wd=0, where x+y+z+w=0
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Then, the distance between two points is
given by
PQ=√(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2

Section Formulae
Let A and B be two points with position vectors a and b, respectively and let C be a point
dividing
(i) AB internally in the ratio m:n
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Then, the position vector of C is given by
OC=

m𝐛−n𝐚
m−n

(iii) If C is the midpoint of AB, then it divides AB in the ratio 1:1
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Find the position vector of the points which divide the join of the points
2a-3b and 3a-2b internally and externally in the ratio 2:3
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Then position vector of P=
=
And position vector of Q=

3(2𝐚−3𝐛)+2(3𝐚−2𝐛)
3+2
12𝐚−13𝐛
5

3(2𝐚−3𝐛)−2(3𝐚−2𝐛)
3+2

= -5b
Therefore the position vector of P is

12𝐚−13𝐛
5

and position vector of Q is -5b

Centroid of a Triangle
If a, b and c are the position vectors of the vertices A, B and C of a ∆ABC, with respect to
an origin O, then the position vector of the centroid G of ∆ABC is given by
OG=

1
3

(a+b+c)
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b=|𝐚||𝐛|cos𝜃
⇨ a
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Properties of Scalar Product
(i) Scalar product is commutative
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b=b
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If a and b are any two non-zero
vectors and n is a scalar, then a
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b=n(a
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If a, b and c are any
three non-zero vectors, then a
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b+a
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Therefore î
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ĵ=k̂
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ĵ=ĵ
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î=0
(v) If two vectors a and b can be expressed in terms of unit vectors î, ĵ and k̂ as
a =a1î+a2ĵ+a3k̂ and b=b1î+b2ĵ+b3k̂
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b= (a1î+a2ĵ+a3k̂)
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[since î
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ĵ=k̂
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ĵ=ĵ
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î=0]
(vi) If 𝜃 is the angle between two vectors a and b with magnitude a and b respectively,
𝐚
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[since a
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(b) If 𝜃=00, then a ∥b
⇨ a=mb, where m is a scalar
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Let a=2ĵ -3k̂ , b=ĵ +3k̂ and c=-3î +3ĵ +k̂
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n̂
=b
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n̂ ?

Soln: Here, a=2ĵ -3k̂ , b=ĵ +3k̂ and c=-3î +3ĵ +k̂
Let the unit vector n̂ =xî +yĵ +zk̂
Now a
...
( xî +yĵ +zk̂ )=0
⇨2y-3z=0………
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n̂ =0 ⇨( ĵ +3k̂ )
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(ii)
Solving (i) and (ii), we get:
y=z=0…………
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[by (iii)]
Therefore n̂ =î
Now c
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î
= -3
Therefore c
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Therefore, work=Fd cos 𝜃 =F
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The moment of the force about O is defined to be the
vector,
M=r×F
where |𝐌|=rF sin𝜃

Example 6
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If the force applied to move the particle to 1/3 of the distance from A to B is
F=î +2ĵ -k̂ and for the remaining distance it is F/3,then what is the work done in
displacing the particle from A to B?

Soln: Let 𝐚⃗

and 𝐛 be the position vector of A and B respectively
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d1
=( î +2ĵ -k̂ )
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Let the force be F1
Therefore F1=1/3(î +2ĵ -k̂ )
Therefore work done= F1
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( 2î +10ĵ +4k̂ )
=1/3(2+20-4)
=1/3(18)
=6
Therefore total work done=F
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d2
=9+6
=15units

Vector or Cross-product of Two Vectors
The vector product (cross-product) of two vectors 𝐚⃗ and 𝐛 is given by
𝐚⃗ × 𝐛 = |𝐚⃗||𝐛|𝐬𝐢𝐧̂𝛉 n̂
𝐚⃗ × 𝐛 = ab𝐬𝐢𝐧̂𝛉 n̂
where , a and b are the magnitude of the vectors 𝐚⃗ and 𝐛, respectively and 𝜃 is the angle
between the vectors 𝐚⃗ and 𝐛
...

(i) If 𝐚⃗ and 𝐛 are parallel, then 𝜃 =0
Therefore 𝐚⃗ × 𝐛 = ⃗𝟎 ……
...
If 𝐚⃗ and 𝐛 are any two vectors and if a=|𝐚⃗| , b=|𝐛|,
then 𝐚⃗ × 𝐛 ≠ 𝐛 × 𝐚⃗
𝐚⃗ × 𝐛 = -(𝐛 × 𝐚⃗)
(ii) Vector product is associative with respect to a Scalar
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If 𝐚⃗ and 𝐛 and 𝐜 are
any three vectors, then 𝐚⃗ × (𝐛 + 𝐜) = (𝐚⃗ × 𝐛) + (𝐚⃗ × 𝐜)
(iv) We know that, the unit vectors î , ĵ , k̂ are mutually perpendicular, so we find that
î ×ĵ=(sin900)n̂ , where n̂ is a unit vector perpendicular to both î and ĵ and so n̂ =k̂
Therefore 𝑖̂ × 𝑗̂ = 𝑘̂, 𝑗̂ × 𝑘̂ = 𝑖̂, 𝑘̂ × 𝑖̂ = 𝑗̂
𝑗̂ × 𝑖̂ = −𝑘̂, 𝑘̂ × 𝑗̂ = −𝑖̂, 𝑖̂ × 𝑘̂ = −𝑗̂
Also, 𝑖̂ × 𝑖̂ = 𝑗̂ × 𝑗̂ = 𝑘̂ × 𝑘̂ = 0
(v) If 𝑎 =a1î +a2ĵ +a3k̂ and 𝑏⃗=b1î +b2ĵ +b3k̂ be two vectors , then
𝑖̂
⃗
𝑎 × 𝑏 = |𝑎1
𝑏1

𝑗̂
𝑎2
𝑏2

𝑘̂
𝑎3 |
𝑏3

(vi) If 𝑎 and 𝑏⃗ are the adjacent sides of a parallelogram,
then area of parallelogram=|𝑎 × 𝑏⃗|
(vii) If 𝑎 and 𝑏⃗ are the diagonals of a parallelogram,
then area of parallelogram=

1
|𝑎 × 𝑏⃗|
2

(viii)If 𝑎 and 𝑏⃗ are the adjacent sides of a triangle,

1
then area of triangle= |𝑎 × 𝑏⃗|
2

(ix) If 𝑎 and 𝑏⃗ are non-zero, non-parallel vectors, then vector of magnitude 𝜆 normal to
the plane 𝑎and 𝑏⃗ is
= ±𝜆 (

⃗
𝑎⃗×𝑏
)
⃗|
|𝑎⃗×𝑏

Example 6
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Soln : Since the vector area of the parallelogram determined by two vectors is their
cross-product
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(𝑏⃗ × 𝑐 ) = |𝑏1
𝑐1

𝑎2
𝑏2
𝑐2

𝑎3
𝑏3 | = [𝑎
𝑐3

𝑏⃗

𝑐]

Geometrically, this represents the volume of the parallelepiped, whose edges are along 𝑎
, 𝑏⃗, and 𝑐
...
𝑎
𝑎
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(iv) The scalar triple product of three vectors is zero, if two of them are parallel or
collinear
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Vector Triple Product
If 𝑎, 𝑏⃗ and 𝑐 are three vector quantities, then 𝑎 × (𝑏⃗ × 𝑐 ), represents the vector triple
product and is given by
𝑎 × (𝑏⃗ × 𝑐 ) = (𝑎
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Properties of Vector Triple Product
(i) Vector triple product is a vector quantity
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e
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(iii) The vector triple product 𝑎 × (𝑏⃗ × 𝑐 ) is a linear combination of those two vectors
which are within brackets
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(v) The formula 𝑎 × (𝑏⃗ × 𝑐 ) = {(𝑎
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If it is not, we first shift on left by using the
properties of cross product and then apply the same formula
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𝑏⃗)𝑐 }
= (𝑎
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If (𝟑𝒂
A
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[since 𝑎 × 𝑎 = ⃗0 and 𝑎 × 𝑏⃗ = −(𝑏⃗ × 𝑎)]
⇨ 10(𝑎 × 𝑏⃗) = 𝑘(𝑎 × 𝑏⃗)
On comparing both sides, we get
k=10

⃗ , P is 𝒂
⃗ + 𝟐𝒃
⃗ and P divides AB in the ratio 2:3
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Point A is 𝒂
B?
A
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Q3
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Now ABCDEF is a regular hexagon
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(ii)
Therefore FA=DC
=AC-AD
=a+b-2b…………
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If |𝒂| = 𝟑, |𝒃| = 𝟒 and |𝒄| = 𝟓 such that each is perpendicular to sum of the other
two, then |𝒂 + 𝒃 + 𝒄| is?
A
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(𝑏 + 𝑐) = 0 , 𝑏
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(𝑎 + 𝑏) = 0
⇨ 2 ∑ 𝑎
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[where ∑ 𝑎
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𝑏 ……………[where ∑ 𝑎2 = 𝑎2 + 𝑏 2 + 𝑐 2 ]
= 9 + 16 + 25 + 0
= 50
Therefore, |𝑎 + 𝑏 + 𝑐| = √50
= 5√2

⃗ and ⃗𝒃 and |𝒂
⃗ × ⃗𝒃| = |𝒂
⃗
...
If 𝜃 is the angle between vectors 𝒂
of 𝜃?

⃗ × ⃗𝒃| = |𝒂
⃗
...
Now |𝒂
⇨ |𝑎𝑏 𝑠𝑖𝑛𝜃 𝑛̂| = |𝑎𝑏 𝑐𝑜𝑠𝜃|
⇨ 𝑎𝑏 𝑠𝑖𝑛𝜃 = 𝑎𝑏 |𝑐𝑜𝑠𝜃|
⇨ 𝑠𝑖𝑛𝜃 = ±𝑐𝑜𝑠𝜃
⇨ 𝑡𝑎𝑛𝜃 = ±1
⇨ 𝜃 = tan−1 (±1)
⇨ 𝜃 = 450 and 1350

Q6
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Let 𝑎 = 2𝑖̂ − 3𝑗̂ + 4𝑘̂ , 𝑏⃗ = 𝑖̂ + 2𝑗̂ − 𝑘̂ and 𝑐 = 𝑚𝑖̂ − 𝑗̂ + 2𝑘̂
Now if the vectors are coplanar, then [𝑎 𝑏⃗ 𝑐 ] = 0
2
⇨ |1
𝑚

−3
2
−1

4
−1| = 0
2

⇨ 2(4 − 1) + 3(2 + 𝑚) + 4(−1 − 2𝑚) = 0
⇨ 6 + 6 + 3𝑚 − 4 − 8𝑚 = 0
⇨ 8 − 5𝑚 = 0
⇨ 𝑚 = 8/5
Therefore for m=8/5, the given vectors are coplanar
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What is the projection of the vector î -2ĵ +k̂ on the vector 4î -4ĵ +7k̂ ?
A
...
𝑏
Now, the projection of 𝑎 on 𝑏⃗ = ⃗
|𝑏|

=
=
=

̂ )
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What is the area of the triangle, whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3,
1)
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Let A,B and C be the vertices of the triangle
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If P, Q, R are the mid-points of the sides AB, BC, CA
respectively of a ∆ABC and ⃗𝒂, ⃗𝒑 and ⃗𝒒 are the position
vector of A, P, Q respectively, then what is the position
vector of R?
A
...

Since P, Q, R are the mid-points of the sides AB, BC, CA
Then 𝑝 =

⃗
𝑎⃗+𝑏
2

⇨ 𝑏⃗ = 2𝑝 − 𝑎……
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What is a vector of unit length orthogonal to both the vectors 𝒊̂ + 𝒋̂ + 𝒌
̂?
𝟑𝒋̂ − 𝒌
A

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