mathongo.com-NCERT-Solutions-Class-12-Maths-Chapter-10-Vector-Algebra.pdf

Exercise 10.1
1. Represent graphically a displacement of 40 km, 30° east of
north.
Sol. Displacement 40 km, 30° East of North.
⇒ Displacement vector OA
→
(say)
such that | OA
→
| = 40 (given)
and vector OA
→
makes an angle
30° with North in East-North
quadrant.
Note. α° South of West ⇒ A
vector in South-West quadrant
making an angle of α° with West.
2. Check the following measures as scalars and vectors:
(i) 10 kg (ii) 2 meters north-west (iii) 40°
(iv) 40 Watt (v) 10– 19
coulomb (vi) 20 m/sec2
.
Sol. (i) 10 kg is a measure of mass and therefore a scalar.
(... 10 kg has no direction, it is magnitude only).
(ii) 2 meters North-West is a measure of velocity (i.e., has
magnitude and direction both) and hence is a vector.
(iii) 40° is a measure of angle i.e., is magnitude only and,
therefore, a scalar.
(iv) 40 Watt is a measure of power (i.e., 40 watt has no
direction) and, therefore, a scalar.
(v) 10– 19
coulomb is a measure of electric charge (i.e., is
magnitude only) and, therefore, a scalar.
(vi) 20 m/sec2
is a measure of acceleration i.e., is a measure of
rate of change of velocity and hence is a vector.
3. Classify the following as scalar and vector quantities:
(i) time period (ii) distance (iii) force
(iv) velocity (v) work done.
Sol. (i) Time-scalar (ii) Distance-scalar (iii) Force-vector
(iv) Velocity-vector (v) Work done-scalar.
4. In the adjoining figure, (a square),
identify the following vectors.
(i) Coinitial
(ii) Equal
(iii) Collinear but not equal.
Sol. (i) a
→
and d
→
have same initial point and,
therefore, coinitial vectors.
30°
N
S
E
W
O
A
Class 12 Chapter 10 - Vector Algebra
MathonGo 1

(ii) b
→
and d
→
have same direction and same magnitude. Therefore,
b
→
and d
→
are equal vectors.
(iii) a
→
and c
→
have parallel supports, so that they are collinear.
Since they have opposite directions, they are not equal.
Hence a
→
and c
→
are collinear but not equal.
5. Answer the following as true or false.
(i) a
→
and – a
→
are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are
equal.
Sol. (i) True.
(ii) False. (... a
→
and 2 a
→
are collinear vectors but | 2 a
→
| = 2 | a
→
|)
(iii) False.
(... | i
∧
| = | j
∧
| = 1 but i
∧
and j
∧
are vectors along x-axis
(OX) and y-axis (OY) respectively.)
(iv) False.
(... Vectors a
→
and – a
→
(= (– 1) a
→
= m a
→
) are collinear
vectors and | a
→
| = | – a
→
| but we know that a
→
≠ – a
→
because their directions are opposite).
Note. Two vectors
→
a and
→
b are said to be equal if
(i) |
→
a | = |
→
b | (ii)
→
a and
→
b have same (like) direction.
MathonGo 2

Exercise 10.2
1. Compute the magnitude of the following vectors:
→
a =
∧
i +
∧
j +
∧
k ,
→
b = 2
∧
i – 7
∧
j – 3
∧
k ,
→
c =
1
3
∧
i +
1
3
∧
j –
1
3
∧
k .
Sol. Given: a
→
= i
∧
+ j
∧
+ k
∧
.
Therefore, | a
→
| = 2 2 2
x y z
+ + = 1 1 1
+ + = 3 .
b
→
= 2 i
∧
– 7 j
∧
– 3 k
∧
.
Therefore, | b
→
| = 4 49 9
+ + = 62 .
c
→
=
1
3
i
∧
+
1
3
j
∧
–
1
3
k
∧
.
MathonGo 3

Therefore, | c
→
| =
2 2 2
1 1 1
3 3 3
−
     
+ +
     
     
=
1 1 1
3 3 3
+ + =
3
3
= 1 = 1.
2. Write two different vectors having same magnitude.
Sol. Let a
→
= i
∧
+ j
∧
+ k
∧
and b
→
= i
∧
+ j
∧
– k
∧
.
Clearly, a
→
≠ b
→
. (... Coefficients of i
∧
and j
∧
are same in
vectors a
→
and b
→
but coefficients of k
∧
in a
→
and b
→
are
unequal as 1 ≠ – 1).
But | a
→
| = 2 2 2
x y z
+ + = 1 1 1
+ + = 3
and | b
→
| = 1 1 1
+ + = 3 ∴ | a
→
| = | b
→
|.
Remark. In this way, we can construct an infinite number of
possible answers.
3. Write two different vectors having same direction.
Sol. Let a
→
= i
∧
+ 2 j
∧
+ 3 k
∧
...(i)
and b
→
= 2( i
∧
+ 2 j
∧
+ 3 k
∧
) ...(ii)
= 2 a
→
[By (i)]
∴ b
→
= m a
→
where m = 2 > 0.
∴ Vectors a
→
and b
→
have the same direction.
But b
→
≠ a
→
[... b
→
= 2 a
→
⇒| b
→
| = |2|| a
→
| = 2| a
→
| ≠ | a
→
|]
Remark. In this way, we can construct an infinite number of
possible answers.
4. Find the values of x and y so that the vectors 2
∧
i + 3
∧
j and
x
∧
i + y
∧
j are equal.
Sol. Given: 2 i
∧
+ 3 j
∧
= x i
∧
+ y j
∧
.
Comparing coefficients of i
∧
and j
∧
on both sides, we have
x = 2 and y = 3.
5. Find the scalar and vector components of the vector with
initial point (2, 1) and terminal point (– 5, 7).
Sol. Let AB
→
be the vector with initial point A(2, 1) and terminal
point B(– 5, 7).
⇒ P.V. (Position Vector) of point A is (2, 1) = 2 i
∧
+ j
∧
and P.V.
of point B is (– 5, 7) = – 5 i
∧
+ 7 j
∧
.
MathonGo 4

∴ AB
→
= P.V. of point B – P.V. of point A
= (– 5 i
∧
+ 7 j
∧
) – (2 i
∧
+ j
∧
) = – 5 i
∧
+ 7 j
∧
– 2 i
∧
– j
∧
⇒ AB
→
= – 7 i
∧
+ 6 j
∧
.
∴ By definition, scalar components of the vectors AB
→
are
coefficients of i
∧
and j
∧
in AB
→
i.e., – 7 and 6 and vector
components of the vector AB
→
are – 7 i
∧
and 6 j
∧
.
6. Find the sum of the vectors:
a
→
= i
∧
– 2 j
∧
+ k
∧
, b
→
= – 2 i
∧
+ 4 j
∧
+ 5
∧
k
and c
→
= i
∧
– 6 j
∧
– 7
∧
k .
Sol. Given: a
→
= i
∧
– 2 j
∧
+ k
∧
, b
→
= – 2 i
∧
+ 4 j
∧
+ 5 k
∧
and c
→
= i
∧
– 6 j
∧
– 7 k
∧
.
Adding a
→
+ b
→
+ c
→
= 0 i
∧
– 4 j
∧
– k
∧
= – 4 j
∧
– k
∧
.
7. Find the unit vector in the direction of the vector
→
a =
∧
i +
∧
j + 2
∧
k .
Sol. We know that a unit vector in the direction of the vector
a
→
= i
∧
+ j
∧
+ 2 k
∧
is a
∧
=
| |
a
a
→
→
=
2
1 1 4
i j k
∧ ∧ ∧
+ +
+ +
⇒ a
∧
=
2
6
i j k
∧ ∧ ∧
+ +
=
1
6
i
∧
+
1
6
j
∧
+
2
6
k
∧
.
8. Find the unit vector in the direction of the vector PQ
→
where P and Q are the points (1, 2, 3) and (4, 5, 6)
respectively.
Sol. Because points P and Q are P(1, 2, 3) and Q(4, 5, 6) (given),
therefore, position vector of point P = OP
→
= 1 i
∧
+ 2 j
∧
+ 3 k
∧
and position vector of point Q = OQ
→
= 4 i
∧
+ 5 j
∧
+ 6 k
∧
where O is the origin.
∴ PQ
→
= Position vector of point Q – Position vector of point P
= OQ
→
– OP
→
= 3 i
∧
+ 3 j
∧
+ 3 k
∧
Therefore, a unit vector in the direction of vector PQ
→
=
PQ
|PQ|
→
→
=
3 3 3
9 9 9 27 9 3
i j k
∧ ∧ ∧
+ +
+ + = = ×
MathonGo 5

=
3( )
3 3
i j k
∧ ∧ ∧
+ +
=
( )
3
i j k
∧ ∧ ∧
+ +
=
1
3
i
∧
+
1
3
j
∧
+
1
3
k
∧
.
9. For given vectors
→
a = 2
∧
i –
∧
j + 2
∧
k and
→
b = –
∧
i +
∧
j –
∧
k ;
find the unit vector in the direction of
→
a +
→
b .
Sol. Given: Vectors a
→
= 2 i
∧
– j
∧
+ 2 j
∧
and b
→
= – i
∧
+ j
∧
– k
∧
∴ a
→
+ b
→
= 2 i
∧
– j
∧
+ 2 k
∧
– i
∧
+ j
∧
– k
∧
= i
∧
+ 0 j
∧
+ k
∧
∴ | a
→
+ b
→
| = 2 2 2
(1) (0) (1)
+ + = 2
∴ A unit vector in the direction of a
→
+ b
→
is
| |
a b
a b
→ →
→ →
+
+
=
0
2
i j k
∧ ∧ ∧
+ +
=
2
i k
∧ ∧
+
=
1
2
i
∧
+
1
2
k
∧
.
10. Find a vector in the direction of vector 5
∧
i –
∧
j + 2
∧
k
which has magnitude 8 units.
Sol. Let a
→
= 5 i
∧
– j
∧
+ 2 k
∧
.
∴ A vector in the direction of vector a
→
which has magnitude 8
units
= 8 a
∧
= 8
| |
a
a
→
→
=
8(5 2 )
25 1 4
i j k
∧ ∧ ∧
− +
+ +
=
8
30
(5 i
∧
– j
∧
+ 2 k
∧
) =
40
30
i
∧
–
8
30
j
∧
+
16
30
k
∧
.
11. Show that the vectors 2
∧
i – 3
∧
j + 4
∧
k and – 4
∧
i + 6
∧
j – 8
∧
k
are collinear.
Sol. Let a
→
= 2 i
∧
– 3 j
∧
+ 4 k
∧
...(i)
and b
→
= – 4 i
∧
+ 6 j
∧
– 8 k
∧
= – 2(2 i
∧
– 3 j
∧
+ 4 k
∧
) = – 2 a
→
[By (i)]
⇒ b
→
= – 2 a
→
= m a
→
where m = – 2 < 0
∴ Vectors a
→
and b
→
are collinear (unlike because m = – 2 < 0).
12. Find the direction cosines of the vector
∧
i + 2
∧
j + 3
∧
k .
Sol. The given vector is ( a
→
) = i
∧
+ 2 j
∧
+ 3 k
∧
a
→
b
→
MathonGo 6

=
2 3
14
i j k
∧ ∧ ∧
+ +
=
1
14
i
∧
+
2
14
j
∧
+
3
14
k
∧
We know that direction cosines of a vector a
→
are coefficients of
i
∧
, j
∧
, k
∧
in a
∧
i.e.,
1
14
,
2
14
,
3
14
.
13. Find the direction cosines of the vector joining the points
A(1, 2, – 3) and B(– 1, – 2, 1) directed from A to B.
Sol. Given: Points A(1, 2, – 3) and B(– 1, – 2, 1).
A
(1, 2, – 3)
B
(– 1, – 2, 1)
⇒ P.V. (Position Vector, OA
→
) of point A is A(1, 2, – 3) = i
∧
+ 2 j
∧
– 3k
∧
and P.V. of point B is B(– 1, – 2, 1) = – i
∧
– 2 j
∧
+ k
∧
.
∴ Vector AB
→
(directed from A to B)
= – i
∧
– 2 j
∧
+ k
∧
– ( i
∧
+ 2 j
∧
– 3 k
∧
)
= – i
∧
– 2 j
∧
+ k
∧
– i
∧
– 2 j
∧
+ 3 k
∧
= – 2 i
∧
– 4 j
∧
+ 4 k
∧
∴ AB = | AB
→
| = 2 2 2
( 2) ( 4) 4
− + − + = 4 16 16
+ + = 6
∴ A unit vector along AB
→
=
AB
|AB|
→
→
=
2 4 4
6
i j k
∧ ∧ ∧
− − +
= –
2
6
i
∧
–
4
6
j
∧
+
4
6
k
∧
=
1
3
−
i
∧
–
2
3
j
∧
+
2
3
k
∧
.
We know that Direction Cosines of the vector AB
→
are the
coefficients of i
∧
, j
∧
, k
∧
in a unit vector along AB
→
i.e.,
1
3
−
,
− 2
3
,
2
3
.
14. Show that the vector
∧
i +
∧
j +
∧
k is equally inclined to the
axes OX, OY and OZ.
Sol. Let a
→
= i
∧
+ j
∧
+ k
∧
.
Let us find angle θ1 (say) between
vector a
→
and OX (⇒ i
∧
)
(... i
∧
represents OX in vector form)
∴ cos θ1 =
.
| || |
a i
a i
→ ∧
→ ∧
O
Y
X
Z
i
^
j
^
k
^
MathonGo 7

⇒ cos θ1 =
( ) . ( 0 0 )
| |
| 0 0 |
i j k i j k
i j k i j k
∧ ∧ ∧ ∧ ∧ ∧
∧ ∧ ∧ ∧ ∧ ∧
+ + + +
+ + + +
⇒ cos θ1 =
1(1) 1(0) 1(0)
1 1 1 1 0 0
+ +
+ + + +
=
1
3
⇒ θ1 = cos– 1 1
3
Similarly, angle θ2 between vectors a
→
and j
∧
(OY) is cos–1 1
3
and angle θ3 between vectors a
→
and k
∧
(OZ) is also cos– 1 1
3
.
∴ θ1 = θ2 = θ3.
∴Vectors a
→
= i
∧
+ j
∧
+ k
∧
is equally inclined to OX, OY and
OZ
15. Find the position vector of a point R which divides the
line joining two points P and Q whose position vectors are
∧
i + 2
∧
j –
∧
k and – i
∧
+
∧
j +
∧
k respectively, in the ratio 2 : 1
(i) internally (ii) externally.
Sol. P.V. of point P is a
→
= i
∧
+ 2 j
∧
– k
∧
and P.V. of point Q is b
→
= – i
∧
+ j
∧
+ k
∧
(i) Therefore P.V. of point R dividing PQ internally (i.e., R lies
within the segment PQ) in the ratio 2 : 1 (= m : n) (= PR : QR)
is
→ →
mb + na
m + n
=
2( ) 2
2 1
i j k i j k
∧ ∧ ∧ ∧ ∧ ∧
− + + + + −
+
=
2 2 2 2
3
i j k i j k
∧ ∧ ∧ ∧ ∧ ∧
− + + + + −
=
4
3
i j k
∧ ∧ ∧
− + +
=
1
3
−
i
∧
+
4
3
j
∧
+
1
3
k
∧
.
(ii) P.V. of point R dividing PQ externally (i.e., R lies outside PQ
and to the right of point Q because ratio 2 : 1 =
2
1
> 1 as PR is
2 times PQ i.e.,
PR
QR
=
2
1



is
→ →
mb – na
m – n
=
2( ) ( 2 )
2 1
i j k i j k
∧ ∧ ∧ ∧ ∧ ∧
− + + − + −
−
= – 2 i
∧
+ 2 j
∧
+ 2 k
∧
– i
∧
– 2 j
∧
+ k
∧
= – 3 i
∧
+ k
∧
.
Remark. In the above question 15(ii), had R been dividing PQ
externally in the ratio 1 : 2; then R will lie to the left of point P
and
PR
QR
=
1
2
.
P ( )
a
→
Q ( )
b
→
R
P ( )
a
→
Q ( )
b
→
R
2 : 1 = m : n
MathonGo 8

16. Find the position vector of the mid-point of the vector
joining the points P(2, 3, 4) and Q(4, 1, – 2).
Sol. Given: Point P is (2, 3, 4) and Q is (4, 1, – 2).
⇒ P.V. of point P(2, 3, 4) is a
→
= 2 i
∧
+ 3 j
∧
+ 4 k
∧
and P.V. of point Q(4, 1, – 2) is b
→
= 4 i
∧
+ j
∧
– 2 k
∧
.
∴ P.V. of mid-point R of PQ is
2
→ →
a + b
.
[By Formula of Internal division]
=
2 3 4 4 2
2
i j k i j k
∧ ∧ ∧ ∧ ∧ ∧
+ + + + −
=
6 4 2
2
i j k
∧ ∧ ∧
+ +
= 3 i
∧
+ 2 j
∧
+ k
∧
.
17. Show that the points A, B and C with position vectors,
→
a = 3
∧
i – 4
∧
j – 4
∧
k ,
→
b = 2
∧
i –
∧
j +
∧
k and
→
c =
∧
i – 3
∧
j –5
∧
k ,
respectively form the vertices of a right-angled triangle.
Sol. Given: P.V. of points A, B, C respectively are a
→
(= OA
→
) = 3 i
∧
– 4 j
∧
–4 k
∧
,
b
→
(= OB
→
) = 2 i
∧
– j
∧
+ k
∧
and c
→
(= OC
→
) = i
∧
– 3 j
∧
– 5 k
∧
, where
O is the origin.
Step I. ∴ AB
→
= 2 i
∧
– j
∧
+ k
∧
– (3 i
∧
– 4 j
∧
– 4 k
∧
) = 2 i
∧
– j
∧
+ k
∧
– 3 i
∧
+ 4 j
∧
+ 4 k
∧
or AB
→
= – i
∧
+ 3 j
∧
+ 5 k
∧
...(i)
BC
→
= P.V. of point C – P.V. of point B
= ( i
∧
– 3 j
∧
– 5 k
∧
) – (2 i
∧
– j
∧
+ k
∧
) = i
∧
– 3 j
∧
– 5 k
∧
– 2 i
∧
+ j
∧
– k
∧
= – i
∧
– 2 j
∧
– 6 k
∧
...(ii)
AC
→
= P.V. of point C – P.V. of point A
= i
∧
– 3 j
∧
– 5 k
∧
– (3 i
∧
– 4 j
∧
– 4 k
∧
) = i
∧
– 3 j
∧
– 5 k
∧
– 3 i
∧
+ 4 j
∧
+ 4 k
∧
= – 2 i
∧
+ j
∧
– k
∧
...(iii)
Adding (i) and (ii),
AB
→
+ BC
→
= – i
∧
+ 3 j
∧
+ 5 k
∧
– i
∧
– 2 j
∧
– 6 k
∧
– 2 i
∧
+ j
∧
– k
∧
= AC
→
[By (iii)]
∴ By Triangle Law of addition of Vectors, Points A, B, C are the
Vertices of a triangle or points A, B, C are collinear.
Step II.
From (i) AB = | AB
→
| = 1 9 25
+ + = 35
MathonGo 9

From (ii), BC = | BC
→
| = 1 4 36
+ + = 41
From (iii), AC = | AC
→
| = 4 1 1
+ + = 6
We can observe that (Longest side BC)2
= ( 41 )2
= 41 = 35 + 6
= AB2
+ AC2
∴ Points A, B, C are the vertices of a right-angled triangle.
18. In triangle ABC (Fig. below), which of the following is not
true:
(A) AB
→
+ BC
→
+ CA
→
= 0
→
(B) AB
→
+ BC
→
– AC
→
= 0
→
(C) AB
→
+ BC
→
– CA
→
= 0
→
(D) AB
→
– CB
→
+ CA
→
= 0
→
Sol. Option (C) is not true.
Because we know by Triangle Law of Addition of vectors that
AB
→
+ BC
→
= AC
→
, i.e., AB
→
+ BC
→
= – CA
→
⇒ AB
→
+ BC
→
– AC
→
= 0
→
⇒ AB
→
+ BC
→
+ CA
→
= 0
→
But for option (C), AB
→
+ BC
→
– CA
→
= AC
→
+ AC
→
= 2 AC
→
≠ 0
→
.
Option (D) is same as option (A).
19. If
→
a and
→
b are two collinear vectors, then which of the
following are incorrect:
(A)
→
b = λ
λ
λ
λ
λ
→
a , for some scalar λ
λ
λ
λ
λ. (B)
→
a = ±
±
±
±
±
→
b
(C) the respective components of
→
a and
→
b are
proportional
(D) both the vectors
→
a and
→
b have same direction, but
different magnitudes.
Sol. Option (D) is not true because two
collinear vectors can have different
directions and also different
magnitudes.
The options (A) and (C) are true by definition of collinear vectors.
Option (B) is a particular case of option (A) (taking λ = ± 1).
A B
C
a
→
b
→
MathonGo 10

Exercise 10.3
1. Find the angle between two vectors
→
a and
→
b with
magnitude 3 and 2, respectively having
→
a .
→
b = 6 .
Sol. Given: | a
→
| = 3 , | b
→
| = 2 and a
→
. b
→
= 6
MathonGo 11

Let θ be the angle between the vectors a
→
and b
→
. We know that
cos θ
θ
θ
θ
θ =
| || |
.
→ →
→ →
a b
a b
Putting values, cos θ =
6
3(2)
=
6
3 4
=
6
12
=
6
12
=
1
2
=
1
2
= cos
4
π
∴ θ =
4
π
.
2. Find the angle between the vectors
∧
i – 2
∧
j + 3
∧
k and
3
∧
i – 2
∧
j +
∧
k .
Sol. Given: Let a
→
= i
∧
– 2 j
∧
+ 3 k
∧
and b
→
= 3 i
∧
– 2 j
∧
+ k
∧
.
∴| a
→
| = 1 4 9
+ + = 14 |... |x i
∧
+ y j
∧
+ z k
∧
| = 2 2 2
x y z
+ +
and | b
→
| = 9 4 1
+ + = 14
Also, a
→
. b
→
= Product of coefficients of i
∧
+ Product of coefficient of j
∧
+ Product of coefficients of k
∧
= 1(3) + (– 2)(– 2) + 3(1) = 3 + 4 + 3 = 10
Let θ be the angle between the vectors a
→
and b
→
.
We know that cos θ =
.
| || |
a b
a b
→ →
→ →
=
10
14 14
=
10
14
=
5
7
∴ θ = cos– 1 5
7
.
3. Find the projection of the vector
∧
i –
∧
j on the vector
∧
i +
∧
j .
Sol. Let a
→
= i
∧
– j
∧
= i
∧
– j
∧
+ 0 k
∧
and b
→
= i
∧
+ j
∧
= i
∧
+ j
∧
+ 0 k
∧
Projection of vector
→
a and
→
b
= Length LM =
| |
.
→ →
→
a b
b
=
2 2 2
(1)(1) ( 1)(1) 0(0)
(1) (1) 0
+ − +
+ +
=
1 1 0
2
− +
=
0
2
= 0.
Remark. If projection of vector a
→
on b
→
is zero, then vector a
→
is perpendicular to
vector b
→
.
90° 90°
L M
A
B
a
→
b
→
90°
A
B
a
→
→
b
MathonGo 12

4. Find the projection of the vector
∧
i + 3
∧
j + 7
∧
k on the
vector 7
∧
i –
∧
j + 8
∧
k .
Sol. Let a
→
= i
∧
+ 3 j
∧
+ 7 k
∧
and b
→
= 7 i
∧
– j
∧
+ 8 k
∧
We know that projection of vector a
→
on vector b
→
=
.
| |
a b
b
→ →
→
=
2 2 2
1(7) 3( 1) 7(8)
(7) ( 1) (8)
+ − +
+ − +
=
7 3 56
49 1 64
− +
+ +
=
60
114
.
5. Show that each of the given three vectors is a unit
vector:
1
7
(2
∧
i + 3
∧
j + 6
∧
k ),
1
7
(3
∧
i – 6
∧
j + 2
∧
k ),
1
7
(6
∧
i + 2
∧
j – 3
∧
k ).
Also show that they are mutually perpendicular to each
other.
Sol. Let a
→
=
1
7
(2 i
∧
+ 3 j
∧
+ 6 k
∧
) =
2
7
i
∧
+
3
7
j
∧
+
6
7
k
∧
...(i)
b
→
=
1
7
(3 i
∧
– 6 j
∧
+ 2 k
∧
) =
3
7
i
∧
–
6
7
j
∧
+
2
7
k
∧
...(ii)
c
→
=
1
7
(6 i
∧
+ 2 j
∧
– 3 k
∧
) =
6
7
i
∧
+
2
7
j
∧
–
3
7
k
∧
...(iii)
∴ | a
→
| =
2 2 2
2 3 6
7 7 7
     
+ +
     
     
=
4 9 36
49 49 49
+ +
=
49
49
= 1 = 1
| b
→
| =
2 2 2
3 6 2
7 7 7
−
     
+ +
     
     
=
9 36 4
49 49 49
+ + =
49
49
= 1 = 1
| c
→
| =
2 2 2
6 2 3
7 7 7
−
     
+ +
     
     
=
36 4 9
49 49 49
+ + =
49
49
= 1 = 1
∴ Each of the three given vectors a
→
, b
→
, c
→
is a unit vector.
From (i) and (ii),
a
→
. b
→
=
2
7
 
 
 
.
3
7
 
 
 
+
3
7
 
 
 
6
7
−
 
 
 
+
6
7
 
 
 
2
7
 
 
 
[ a
→
. b
→
= a1b1 + a2b2 + a3b3]
MathonGo 13

=
6
49
–
18
49
+
12
49
=
6 18 12
49
− +
=
0
49
= 0
∴ a
→
and b
→
are perpendicular to each other.
From (ii) and (iii),
b
→
. c
→
=
3
7
 
 
 
6
7
 
 
 
+
6
7
−
 
 
 
2
7
 
 
 
+
2
7
3
7
−
 
 
 
=
18
49
–
12
49
–
6
49
=
18 12 6
49
− −
=
0
49
= 0
∴ b
→
and c
→
From (i) and (iii),
a
→
. c
→
=
2
7
6
7
 
 
 
+
3
7
2
7
 
 
 
+
6
7
 
 
 
3
7
−
 
 
 
=
12
49
+
6
49
–
18
49
=
12 6 18
49
+ −
=
0
49
= 0
∴ a
→
and c
→
Hence, a
→
, b
→
, c
→
are mutually perpendicular vectors.
6. Find |
→
a | and |
→
b |, if (
→
a +
→
b ) . (
→
a –
→
b ) = 8 and
|
→
a | = 8|
→
b |.
Sol. Given: ( a
→
+ b
→
) . ( a
→
– b
→
) = 8 and | a
→
| = 8 | b
→
| ...(i)
⇒ a
→
. a
→
– a
→
. b
→
+ b
→
. a
→
– b
→
. b
→
= 8
⇒ | a
→
|2
– a
→
. b
→
+ a
→
. b
→
– | b
→
|2
= 8
[... We know that a
→
. a
→
= | a
→
|2
and b
→
. b
→
=| b
→
|2
and
b
→
. a
→
= a
→
. b
→
]
⇒ | a
→
|2
– | b
→
|2
= 8 ...(ii)
Putting | a
→
| = 8| b
→
| from (i) in (ii), 64 | b
→
|2
– | b
→
|2
= 8
or (64 – 1) | b
→
|2
= 8 ⇒ 63 | b
→
|2
= 8
⇒ | b
→
|2
=
8
63
⇒ | b
→
| =
8
63
=
4 2
9 7
×
×
(... Length i.e., modulus of a vector is never negative.)
⇒ | b
→
| =
2
3
2
7
Putting this value of | b
→
| in (i),
MathonGo 14

| a
→
| = 8
2 2
3 7
 
 
 
=
16
3
2
7
.
7. Evaluate the product (3
→
a – 5
→
b ) . (2
→
a + 7
→
b ).
Sol. The given expression = (3 a
→
– 5 b
→
) . (2 a
→
+ 7 b
→
)
= (3 a
→
) . (2 a
→
) + (3 a
→
) . (7 b
→
) – (5 b
→
) . (2 a
→
) – (5 b
→
) . (7 b
→
)
= 6 a
→
. a
→
+ 21 a
→
. b
→
– 10 b
→
. a
→
– 35 b
→
. b
→
= 6| a
→
|2
+ 21 a
→
. b
→
– 10 a
→
. b
→
– 35| b
→
|2
[... a
→
. a
→
= | a
→
|2
and b
→
. b
→
= | b
→
|2
and b
→
. a
→
= a
→
. b
→
]
= 6| a
→
|2
+ 11 a
→
. b
→
– 35| b
→
|2
.
8. Find the magnitude of two vectors
→
a and
→
b , having the
same magnitude such that the angle between them is 60°
and their scalar product is
1
2
.
Sol. Given: | a
→
| = | b
→
| and angle θ (say) between a
→
and b
→
is
60° and their scalar (i.e., dot) product =
1
2
i.e., a
→
. b
→
=
1
2
⇒ | a
→
| | b
→
| cos θ =
1
2
[... a
→
. b
→
= | a
→
| | b
→
| cos θ]
Putting | b
→
| = | a
→
| (given) and θ = 60° (given), we have
| a
→
| | a
→
| cos 60° =
1
2
⇒ | a
→
|2 1
2
 
 
 
=
1
2
Multiplying by 2, | a
→
|2
= 1 ⇒ | a
→
| = 1 ...(i)
(... Length of a vector is never negative)
∴ | b
→
| = | a
→
| = 1 [By (i)]
∴ | a
→
| = 1 and | b
→
| = 1.
9. Find |
→
x |, if for a unit vector
→
a , (
→
x –
→
a ) . (
→
x +
→
a ) = 12.
Sol. Given: a
→
is a unit vector ⇒ | a
→
| = 1 ...(i)
Also given ( x
→
– a
→
) . ( x
→
+ a
→
) = 12
⇒ x
→
. x
→
+ x
→
. a
→
– a
→
. x
→
– a
→
. a
→
= 12
⇒ | x
→
|2
+ a
→
. x
→
– a
→
. x
→
– | a
→
|2
= 12
⇒ | x
→
|2
– | a
→
|2
= 12
MathonGo 15

Putting | a
→
| = 1 from (i), | x
→
|2
– 1 = 12
⇒ | x
→
|2
= 13 ⇒ | x
→
| = 13 .
(... Length of a vector is never negative.)
10. If
→
a = 2
∧
i + 2
∧
j + 3
∧
k ,
→
b = –
∧
i + 2
∧
j +
∧
k and
→
c = 3
∧
i +
∧
j
are such that
→
a + λ
λ
λ
λ
λ
→
b is perpendicular to c
→
, then find
the value of λ
λ
λ
λ
λ.
Sol. Given : a
→
= 2 i
∧
+ 2 j
∧
+ 3 k
∧
, b
→
= – i
∧
+ 2 j
∧
+ k
∧
and c
→
= 3 i
∧
+ j
∧
.
Now, a
→
+ λ b
→
= 2 i
∧
+ 2 j
∧
+ 3 k
∧
+ λ(– i
∧
+ 2 j
∧
+ k
∧
)
= 2 i
∧
+ 2 j
∧
+ 3 k
∧
– λ i
∧
+ 2λ j
∧
+ λ k
∧
⇒ a
→
+ λ b
→
= (2 – λ) j
∧
+ (2 + 2λ) j
∧
+ (3 + λ) k
∧
Again given c
→
= 3 i
∧
+ j
∧
= 3 i
∧
+ j
∧
+ 0 k
∧
.
Because vector a
→
+ λ a
→
is perpendicular to c
→
, therefore,
( a
→
+ λ b
→
) . c
→
= 0
i.e., Product of coefficients of i
∧
+ ......... = 0
⇒ (2 – λ)3 + (2 + 2λ)1 + (3 + λ)0 = 0
⇒ 6 – 3λ + 2 + 2λ = 0 ⇒ – λ + 8 = 0
⇒ – λ = – 8 ⇒ λ = 8.
11. Show that |
→
a |
→
b + |
→
b |
→
a is perpendicular to
|
→
a |
→
b – |
→
b |
→
a , for any two non-zero vectors
→
a and
→
b .
Sol. Let c
→
= | a
→
| b
→
+ | b
→
| a
→
= l b
→
+ m a
→
where l = | a
→
| and m = | b
→
|
Let d
→
= | a
→
| b
→
– | b
→
| a
→
= l b
→
– m a
→
Now, c
→
. d
→
= (l b
→
+ m a
→
) . (l b
→
– m a
→
)
= l2
b
→
. b
→
– lm b
→
. a
→
+ lm a
→
. b
→
– m2
a
→
a
→
= l2
| b
→
|2
– lm a
→
. b
→
+ lm a
→
. b
→
– m2
| a
→
|2
= l2
| b
→
|2
– m2
| a
→
|
Putting l = | a
→
| and m = | b
→
|,
= | a
→
|2
| b
→
|2
– | b
→
|2
| a
→
|2
= 0
i.e., c
→
. d
→
= 0
∴ Vectors c
→
and d
→
MathonGo 16

12. If
→
a .
→
a = 0 and
→
a .
→
b = 0, then what can be concluded
about the vector
→
b ?
Sol. Given: a
→
. a
→
= 0 ⇒ | a
→
|2
= 0 ⇒ | a
→
| = 0 ...(i)
(⇒ a
→
is a zero vector by definition of zero vector.)
Again given a
→
. b
→
= 0 ⇒ | a
→
|| b
→
| cos θ = 0
Putting | a
→
| = 0 from (i), we have 0| b
→
| cos θ = 0
i.e., 0 = 0 for all (any) vectors b
→
. ∴ b
→
can be any vector.
Note. ( a
→
+ b
→
+ c
→
)2
= ( a
→
+ ( b
→
+ c
→
))2
= a
→ 2
+ ( b
→
+ c
→
)2
+ 2 a
→
. ( b
→
+ c
→
)
[... ( A
→
+ B
→
)2
= A
→ 2
+ B
→
2
+ 2 A
→
. B
→
]
= a
→ 2
+ b
→ 2
+ c
→ 2
+ 2 b
→
. c
→
+ 2 a
→
. b
→
+ 2 a
→
. c
→
Using a
→
. c
→
= c
→
. a
→
or ( a
→
+ b
→
+ c
→
)2
= a
→ 2
+ b
→ 2
+ c
→ 2
+ 2( a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
)
13. If
→
a ,
→
b ,
→
c are unit vectors such that
→
a +
→
b +
→
c = 0
→
,
find the value of
→
a .
→
b +
→
b .
→
c +
→
c .
→
a .
Sol. Because a
→
, b
→
, c
→
are unit vectors, therefore,
| a
→
| = 1, | b
→
| = 1 and | c
→
| = 1. ...(i)
Again given a
→
+ b
→
+ c
→
= 0
→
Squaring both sides ( a
→
+ b
→
+ c
→
)2
= 0
Using formula of Note above
⇒ a
→ 2
+ b
→ 2
+ c
→ 2
+ 2( a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
) = 0
or | a
→
|2
+ | b
→
|2
+ | c
→
|2
+ 2( a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
) = 0
Putting | a
→
| = 1, | b
→
| = 1, | c
→
| = 1 from (i),
1 + 1 + 1 + 2( a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
) = 0
⇒ 2( a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
) = – 3
Dividing both sides by 2, a
→
. b
→
+ b
→
. c
→
+ c
→
. a
→
=
3
2
−
.
14. If either vector
→
a = 0
→
or
→
b = 0
→
, then
→
a .
→
b = 0. But
the converse need not be true. Justify your answer with an
example.
MathonGo 17

Sol. Case I. Vector a
→
= 0
→
. Therefore, by definition of zero vector,
| a
→
| = 0 ...(i)
∴ a
→
. b
→
= | a
→
|| b
→
| cos θ = 0 (| b
→
| cos θ) [By (i)]
= 0
Case II. Vector b
→
= 0
→
. Proceeding as above we can prove that
a
→
. b
→
= 0
But the converse is not true.
Let us justify it with an example.
Let a
→
= i
∧
+ j
∧
+ k
∧
. Therefore, | a
→
| = 2 2 2
1 1 1
+ + = 3 ≠ 0.
Therefore a
→
≠ 0
→
(By definition of Zero Vector)
Let b
→
= i
∧
+ j
∧
– 2 k
∧
.
Therefore, | b
→
| = 2 2 2
(1) (1) ( 2)
+ + − = 6 ≠ 0.
Therefore, b
→
≠ 0
→
.
But a
→
. b
→
= 1(1) + 1(1) + 1(– 2) = 1 + 1 – 2 = 0
So here a
→
. b
→
= 0 but neither a
→
= 0
→
nor b
→
= 0
→
.
15. If the vertices A, B, C of a triangle ABC are (1, 2, 3),
(– 1, 0, 0) and (0, 1, 2), respectively, then find ∠
∠
∠
∠
∠ABC.
Sol. Given: Vertices A, B, C of a triangle are A(1, 2, 3), B(– 1, 0, 0)
and C(0, 1, 2) respectively.
A(1, 2, 3)
B(– 1, 0, 0) C(0, 1, 2)
∴ Position vector (P.V.) of point A (=s OA
→
) = (1, 2, 3)
= i
∧
+ 2 j
∧
+ 3 k
∧
Position vector (P.V.) of point B (= OB
→
) = (– 1, 0, 0)
= – i
∧
+ 0 j
∧
+ 0 k
∧
and position vector (P.V.) of point C (= OC
→
) = (0, 1, 2)
= 0 i
∧
+ j
∧
+ 2 k
∧
We can see from the above figure that ∠ABC is the angle
between the vectors BA
→
and BC
→
MathonGo 18

Now BA
→
= P.V. of terminal point A – P.V. of initial point B
= i
∧
+ 2 j
∧
+ 3 k
∧
– (– i
∧
+ 0 j
∧
+ 0 k
∧
)
= i
∧
+ 2 j
∧
+ 3 k
∧
+ i
∧
– 0 j
∧
– 0 k
∧
= 2 i
∧
+ 2 j
∧
+ 3 k
∧
...(i)
and BC
→
= 0 i
∧
+ j
∧
+ 2 k
∧
– (– i
∧
+ 0 j
∧
+ 0 k
∧
)
= 0 i
∧
+ j
∧
+ 2 k
∧
+ i
∧
– 0 j
∧
– 0 k
∧
= i
∧
+ j
∧
+ 2 k
∧
...(ii)
We know that cos ∠ABC =
BA . BC
|BA|
|BC|
→ →
→ →
→
→
θ = →
→
.
cos
| |
| |
a b
a b
Using (i) and (ii)
=
2(1) 2(1) 3(2)
4 4 9 1 1 4
+ +
+ + + +
=
10
17 6
=
10
102
∴ ∠ABC = cos– 1 10
102
.
16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, – 1) are
collinear.
Sol. Given points are A(1, 2, 7), B(2, 6, 3) and C(3, 10, – 1).
⇒ P.V.’s OA
→
, OB
→
, OC
→
of points A, B, C are
OA
→
= (1, 2, 7) = i
∧
+ 2 j
∧
+ 7 k
∧
OB
→
= (2, 6, 3) = 2 i
∧
+ 6 j
∧
+ 3 k
∧
and OC
→
= (3, 10, – 1) = 3 i
∧
+ 10 j
∧
– k
∧
∴ AB
→
= P.V. of terminal point B – P.V. of initial point A
= 2 i
∧
+ 6 j
∧
+ 3 k
∧
– ( i
∧
+ 2 j
∧
+ 7 k
∧
)
= 2 i
∧
+ 6 j
∧
+ 3 k
∧
– i
∧
– 2 j
∧
– 7 k
∧
= i
∧
+ 4 j
∧
– 4 k
∧
...(i)
and AC
→
= 3 i
∧
+ 10 j
∧
– k
∧
– ( i
∧
+ 2 j
∧
+ 7 k
∧
)
= 3 i
∧
+ 10 j
∧
– k
∧
– i
∧
– 2 j
∧
– 7 k
∧
= 2 i
∧
+ 8 j
∧
– 8 k
∧
= 2( i
∧
+ 4 j
∧
– 4 k
∧
)
⇒ AC
→
= 2 AB
→
[By (i)]
MathonGo 19

⇒ Vectors AB
→
and AC
→
are collinear or parallel. |... a
→
= m b
→
⇒ Points A, B, C are collinear.
(... Vectors AB
→
and AC
→
have a common point A and hence can’t
be parallel.)
Remark. When we come to exercise 10.4 and learn that Exercise,
we have a second solution for proving points A, B, C to be
collinear:
Prove that AB
→
× AC
→
= 0
→
.
17. Show that the vectors 2
∧
i –
∧
j +
∧
k ,
∧
i – 3
∧
j – 5
∧
k and
3
∧
i – 4
∧
j – 4
∧
k form the vertices of a right angled triangle.
Sol. Let the given (position) vectors be P.V.’s of the points A, B, C
respectively.
P.V. of point A is 2 i
∧
– j
∧
+ k
∧
and
P.V. of point B is i
∧
– 3 j
∧
– 5 k
∧
and
P.V. of point C is 3 i
∧
– 4 j
∧
– 4 k
∧
.
∴ AB
→
= i
∧
– 3 j
∧
– 5 k
∧
– (2 i
∧
– j
∧
+ k
∧
) = i
∧
– 3 j
∧
– 5 k
∧
– 2 i
∧
+ j
∧
– k
∧
= – i
∧
– 2 j
∧
– 6 k
∧
...(i)
and BC
→
= 3 i
∧
– 4 j
∧
– 4 k
∧
– ( i
∧
– 3 j
∧
– 5 k
∧
) = 3 i
∧
– 4 j
∧
– 4 k
∧
– i
∧
+ 3 j
∧
+ 5 k
∧
= 2 i
∧
– j
∧
+ k
∧
...(ii)
and AC
→
= 3 i
∧
– 4 j
∧
– 4 k
∧
– (2 i
∧
– j
∧
+ k
∧
) = 3 i
∧
– 4 j
∧
– 4 k
∧
– 2 i
∧
+ j
∧
– k
∧
= i
∧
– 3 j
∧
– 5 k
∧
...(iii)
Adding (i) and (ii), we have
AB
→
+ BC
→
= – i
∧
– 2 j
∧
– 6 k
∧
+ 2 i
∧
– j
∧
+ k
∧
= i
∧
– 3 j
∧
– 5 k
∧
= AC
→
[By (iii)]
∴ By Triangle Law of addition of vectors, points A, B, C are the
vertices of a triangle ABC or points A, B, C are collinear.
Now from (i) and (ii), AB
→
. BC
→
= (– 1)(2) + (– 2)(– 1) + (– 6)(1)
= – 2 + 2 – 6 = – 6 ≠ 0
MathonGo 20

From (ii) and (iii), BC
→
. AC
→
= 2(1) + (– 1)(– 3) + 1(– 5)
= 2 + 3 – 5 = 0
⇒ BC
→
is perpendicular to AC
→
⇒ Angle C is 90°. ∴ ∆ABC is right angled at point C.
∴ Points A, B, C are the vertices of a right angled triangle.
18. If
→
a is a non-zero vector of magnitude ‘a’ and λ
λ
λ
λ
λ is a non-
zero scalar, then λ
λ
λ
λ
λ
→
a is a unit vector if
(A) λ
λ
λ
λ
λ = 1 (B) λ
λ
λ
λ
λ = – 1 (C) a = |λ
λ
λ
λ
λ| (D) a =
1
| |
λ
Sol. Given: a
→
is a non-zero vector of magnitude a
⇒ | a
→
| = 1 ...(i)
Also given: λ ≠ 0 and λ a
→
is a unit vector.
⇒ |λ a
→
| = 1 ⇒ |λ|| a
→
| = 1
⇒ |λ| a = 1 ⇒ a =
1
| |
λ
∴ Option (D) is the correct answer.
MathonGo 21

Exercise 10.4
1. Find |
→
a ×
→
b |, if
→
a =
∧
i – 7
∧
j + 7
∧
k and
→
b = 3
∧
i
– 2
∧
j + 2
∧
k .
Sol. Given: a
→
= i
∧
– 7 j
∧
+ 7 k
∧
and b
→
= 3 i
∧
– 2 j
∧
+ 2 k
∧
.
Therefore, a
→
× b
→
= 1 7 7
3 2 2
i j k
∧ ∧ ∧
−
−
[... If a
→
= a1 i
∧
+ a2 j
∧
+ a3 k
∧
and b
→
= b1 i
∧
+ b2 j
∧
+ b3 k
∧
;
then a
→
× b
→
= 1 2 3
1 2 3
i j k
a a a
b b b
∧ ∧ ∧ 






Expanding along first row,
a
→
× b
→
= i
∧ 7 7
2 2
−
−
– j
∧ 1 7
3 2
+ k
∧ 1 7
3 2
−
−
⇒ a
→
× b
→
= i
∧
(– 14 + 14) – j
∧
(2 – 21) + k
∧
(– 2 + 21)
MathonGo 22

= 0 i
∧
+ 19 j
∧
+ 19 k
∧
∴ | a
→
× b
→
| =
2 2 2
0 (19) (19)
+ + = 2
2(19) = 2 (19) = 19 2 .
Result: We know that n
→
= a
→
× b
→
is a vector perpendicular
to both the vectors a
→
and b
→
.
Therefore, a unit vector perpendicular
to both the vectors a
→
and b
→
is
∧
n = ±
→
→
→
→
a × b
|a × b|
.
A
A
|A|
→
∧
→
 
 
=
 
 
 
∵
2. Find a unit vector perpendicular to each of the vectors
→
a +
→
b and
→
a –
→
b where
→
a = 3
∧
i + 2
∧
j + 2
∧
k and
→
b =
∧
i + 2
∧
j – 2
∧
k .
Sol. Given: a
→
= 3 i
∧
+ 2 j
∧
+ 2 k
∧
and b
→
= i
∧
+ 2 j
∧
– 2 k
∧
Adding, c
→
= a
→
+ b
→
= 4 i
∧
+ 4 j
∧
+ 0 k
∧
Subtracting d
→
= a
→
– b
→
= 2 i
∧
+ 0 j
∧
+ 4 k
∧
Therefore, n
→
= c
→
× d
→
=
∧ ∧ ∧
4 4 0
2 0 4
i j k
Expanding along first row = i
∧
(16 – 0) – j
∧
(16 – 0) + k
∧
(0 – 8)
⇒ n
→
= 16 i
∧
– 16 j
∧
– 8 k
∧
∴| n
→
| = 2 2 2
(16) ( 16) ( 8)
+ − + − = 256 256 64
+ + = 576 = 24.
Therefore, a unit vector perpendicular to both a
→
and b
→
is
n
∧
= ±
| |
n
n
→
→
= ±
(16 16 8 )
24
i j k
∧ ∧ ∧
− −
= ±
16 16 8
24 24 24
i j k
∧ ∧ ∧
 
− −
 
 
= ±
2 2 1
3 3 3
i j k
∧ ∧ ∧
 
− −
 
 
.
3. If a unit vector
∧
a makes an angle
π
3
with
∧
i ,
π
4
with
∧
j
and an acute angle θ
θ
θ
θ
θ with
∧
k , then find θ
θ
θ
θ
θ and hence, the
components of
∧
a .
a
→
n
→
b
→
n
→
MathonGo 23

Sol. Let a
∧
= x i
∧
+ y j
∧
+ z k
∧
be a unit vector ...(i)
⇒ | a
∧
| = 1 ⇒
2 2 2
x y z
+ + = 1
Squaring both sides, x2
+ y2
+ z2
= 1 ...(ii)
Given: Angle between vectors a
∧
and i
∧
= i
∧
+ 0 j
∧
+ 0 k
∧
is
3
π
.
∴ cos
3
π
=
.
| |
| |
a i
a i
∧
∧
∧
∧
.
cos
| |
| |
a b
a b
 
→
 
→
 
θ = →
→
 
 
 
 
∵
⇒
1
2
=
(1) (0) (0)
(1)(1)
x y z
+ +
or
1
2
= x ...(iii)
Again, Given: Angle between vectors a
∧
and j
∧
= 0 i
∧
+ j
∧
+ 0 k
∧
is
4
π
.
∴ cos
4
π
=
.
| |
| |
a j
a j
∧
∧
∧
∧
⇒
1
2
=
(0) (1) (0)
(1)(1)
x y z
+ +
⇒
1
2
= y ...(iv)
Again, Given: Angle between vectors a
∧
and k
∧
= 0 i
∧
+ 0 j
∧
+ k
∧
is θ where θ is acute.
∴ cos θ =
.
| || |
a k
a k
∧
∧
∧
∧
=
(0) (0) (1)
(1)(1)
x y z
+ +
= z ...(v)
Putting values of x, y and z from (iii), (iv) and (v) in (ii),
1
4
+
1
2
+ cos2
θ = 1
⇒ cos2
θ = 1 –
1
4
–
1
2
=
4 1 2
4
− −
=
1
4
⇒ cos θ = ±
1
2
But θ is acute angle (given)
⇒ cos θ is positive and hence =
1
2
= cos
3
π
⇒ θ =
3
π
From (v), z = cos θ =
1
2
Putting values of x, y, z in (i), a
∧
=
1
2
i
∧
+
1
2
j
∧
+
1
2
k
∧
∴ Components of a
∧
are coefficients of i
∧
, j
∧
, k
∧
in a
∧
i.e.,
1
2
,
1
2
,
1
2
and acute angle θ =
3
π
.
MathonGo 24

4. Show that (
→
a –
→
b ) × (
→
a +
→
b ) = 2
→
a ×
→
b .
Sol. L.H.S. = ( a
→
– b
→
) × ( a
→
+ b
→
)
= a
→
× a
→
+ a
→
× b
→
– b
→
× a
→
– b
→
× b
→
= 0
→
+ a
→
× b
→
+ a
→
× b
→
– 0
→
[... a
→
× a
→
= 0
→
, b
→
× b
→
= 0
→
and b
→
× a
→
= – a
→
× b
→
]
= 2 a
→
× b
→
= R.H.S.
5. Find λ
λ
λ
λ
λ and µ
µ
µ
µ
µ if (2
∧
i + 6
∧
j + 27
∧
k ) × (
∧
i + λ
λ
λ
λ
λ
∧
j + µ
µ
µ
µ
µ
∧
k ) =
→
0 .
Sol. Given: (2 i
∧
+ 6 j
∧
+ 27 k
∧
) × ( i
∧
+ λ j
∧
+ µ k
∧
) = 0
→
⇒ 2 6 27
1
i j k
∧ ∧ ∧
λ µ
= 0
→
Expanding along first row,
i
∧
(6µ – 27λ) – j
∧
(2µ – 27) + k
∧
(2λ – 6) = 0
→
= 0 i
∧
+ 0 j
∧
+ 0 k
∧
∧
, j
∧
, k
∧
6µ – 27λ = 0 ...(i)
2µ – 27 = 0 ...(ii)
and 2λ – 6 = 0 ...(iii)
From (ii), 2µ = 27 ⇒ µ =
27
2
From (iii), 2λ = 6 ⇒ λ =
6
2
= 3
Putting λ = 3 and µ =
27
2
in (i), 6
27
2
 
 
 
– 27(3) = 0
or 81 – 81 = 0 or 0 = 0 which is true. ∴ λ = 3 and µ =
27
2
.
6. Given that
→
a .
→
b = 0 and
→
a ×
→
b =
→
0 . What can you
conclude about the vectors
→
a and
→
b ?
Sol. Given: a
→
. b
→
= 0 ⇒ | a
→
| | b
→
| cos θ = 0
⇒ Either | a
→
| = 0
or | b
→
| = 0 or cos θ = 0 (⇒ θ = 90°)
⇒ Either a
→
= 0
→
or b
→
= 0
→
or vector a
→
is perpendicular to b
→
. ...(i)
(... By definition, vector a
→
is zero vector if and only if | a
→
| = 0)
b
→
→
a
MathonGo 25

Again given a
→
× b
→
= 0
→
⇒ | a
→
× b
→
| = 0
⇒ | a
→
| | b
→
| sin θ = 0
[... | a
→
× b
→
| = | a
→
| | b
→
| sin θ]
⇒ Either | a
→
| = 0 or | b
→
| = 0 or sin θ = 0 (⇒ θ = 0)
a
→
b
→
⇒ Either a
→
= 0
→
or b
→
= 0
→
or vectors a
→
and b
→
are
collinear (or parallel) vectors. ...(ii)
We know from common sense that vectors a
→
and b
→
are
perpendicular to each other as well as are parallel (or collinear)
is impossible. ...(iii)
∴ From (i), (ii) and (iii), either a
→
= 0
→
or b
→
= 0
→
∴ a
→
. b
→
= 0 and a
→
× b
→
= 0
→
⇒ Either a
→
= 0
→
or b
→
= 0
→
.
7. Let the vectors
→
a ,
→
b ,
→
c be given as a1
∧
i + a2
∧
j + a3
∧
k ,
b1
∧
i + b2
∧
j + b3
∧
k , c1
∧
i + c2
∧
j + c3
∧
k .
Then show that
→
a × (
→
b +
→
c ) =
→
a ×
→
b +
→
a ×
→
c .
→
= a1 i
∧
+ a2 j
∧
+ a3 k
∧
, b
→
= b1 i
∧
+ b2 j
∧
+ b3 k
∧
,
c
→
= c1 i
∧
+ c2 j
∧
+ c3 k
∧
∴ b
→
+ c
→
= (b1 + c1) i
∧
+ (b2 + c2) j
∧
+ (b3 + c3) k
∧
L.H.S. = a
→
× ( b
→
+ c
→
) = 1 2 3
1 1 2 2 3 3
i j k
a a a
b c b c b c
∧ ∧ ∧
+ + +
= 1 2 3
1 2 3
i j k
a a a
b b b
∧ ∧ ∧
+ 1 2 3
1 2 3
i j k
a a a
c c c
∧ ∧ ∧
[By Property of Determinants]
= a
→
× b
→
+ a
→
× c
→
= R.H.S.
8. If either
→
a =
→
0 or
→
b =
→
0 , then
→
a ×
→
b =
→
0 . Is the
converse true? Justify your answer with an example.
MathonGo 26

Sol. Given: Either a
→
= 0
→
or b
→
= 0
→
.
∴ | a
→
| = | 0
→
| = 0 or | b
→
| = | 0
→
| = 0 ...(i)
∴ | a
→
× b
→
| = | a
→
|| b
→
| sin θ = 0 (sin θ) = 0 [By (i)]
∴ a
→
× b
→
= 0
→
(By definition of zero vector)
But the converse is not true.
Let a
→
= i
∧
+ j
∧
+ k
∧
∴ | a
→
| = 1 1 1
+ + = 3 ≠ 0.
∴ a
→
is a non-zero vector.
Let | b
→
| = 2( i
∧
+ j
∧
+ k
∧
) = 2 i
∧
+ 2 j
∧
+ 2 k
∧
∴ | b
→
| = 4 4 4
+ + or | b
→
| = 12 = 4 3
× = 2 3 ≠ 0.
∴ b
→
is a non-zero vector.
But a
→
× b
→
=
∧ ∧ ∧
1 1 1
2 2 2
i j k
Taking 2 common from R3, =
∧ ∧ ∧
1 1 1
1 1 1
i j k
= 0
→
(... R2 and R3 are identical)
9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5)
and C(1, 5, 5).
Sol. Vertices of ∆ABC are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
∴ Position Vector (P.V.) of point A is (1, 1, 2) = i
∧
+ j
∧
+ 2 k
∧
P.V. of point B is (2, 3, 5)
= 2 i
∧
+ 3 j
∧
+ 5 k
∧
P.V. of point C is (1, 5, 5)
= i
∧
+ 5 j
∧
+ 5 k
∧
∴ AB
→
= 2 i
∧
+ 3 j
∧
+ 5 k
∧
– ( i
∧
+ j
∧
+ 2 k
∧
)
= 2 i
∧
+ 3 j
∧
+ 5 k
∧
– i
∧
– j
∧
– 2 k
∧
= i
∧
+ 2 j
∧
+ 3 k
∧
and AC
→
= i
∧
+ 5 j
∧
+ 5 k
∧
– ( i
∧
+ j
∧
+ 2 k
∧
) = i
∧
+ 5 j
∧
+ 5 k
∧
– i
∧
– j
∧
– 2 k
∧
= 0 i
∧
+ 4 j
∧
+ 3 k
∧
A(1, 1, 2)
B(2, 3, 5) C(1, 5, 5)
MathonGo 27

∴
→
AB ×
→
AC = 1 2 3
0 4 3
i j k
∧ ∧ ∧
= i
∧
(6 – 12) – j
∧
(3 – 0) + k
∧
(4 – 0) = – 6 i
∧
– 3 j
∧
+ 4 k
∧
We know that area of triangle ABC
=
1
2
|
→
AB ×
→
AC | =
1
2
36 9 16
+ + | 2 2 2
x y z
+ +
=
1
2
61 sq. units.
10. Find the area of the parallelogram whose adjacent
sides are determined by the vectors
→
a =
∧
i –
∧
j + 3
∧
k
and
→
b = 2
∧
i – 7
∧
j +
∧
k .
Sol. Given: Vectors representing two adjacent sides of a
parallelogram are
a
→
= i
∧
– j
∧
+ 3 k
∧
and b
→
= 2 i
∧
– 7 j
∧
+ k
∧
.
∴ a
→
× b
→
= 1 1 3
2 7 1
i j k
∧ ∧ ∧
−
−
= i
∧
(– 1 + 21) – j
∧
(1 – 6) + k
∧
(– 7 + 2) = 20 i
∧
+ 5 j
∧
– 5 k
∧
We know that area of parallelogram = |
→
a ×
→
b |
= 400 25 25
+ + = 450 = 25 9 2
× ×
= 5(3) 2 = 15 2 square units.
Note. Area of parallelogram whose diagonal vectors are
→
α and
→
β is
1
2
|
→
α ×
→
β |.
11. Let the vectors
→
a and
→
b be such that | a
→
| = 3,
|
→
b | =
2
3
, then
→
a ×
→
b is a unit vector, if the angle
between
→
a and
→
b is
(A)
π
6
(B)
π
4
(C)
π
3
(D)
π
2
.
Sol. Given: | a
→
| = 3, | b
→
| =
2
3
and a
→
× b
→
is a unit vector.
b i j k
= 2 – 7 +
→ ^ ^ ^
a i j k
= – + 3
→ ^ ^ ^
MathonGo 28

⇒ | a
→
× b
→
| = 1 ⇒ | a
→
| | b
→
| sin θ = 1
where θ is the angle between vectors a
→
and b
→
.
Putting values of | a
→
| and | b
→
|, 3
2
3
 
 
 
sin θ = 1
⇒ 2 sin θ = 1 ⇒ sin θ =
1
2
= sin
4
π
⇒ θ =
4
π
∴ Option (B) is the correct answer.
12. Area of a rectangle having vertices A, B, C and D with
position vectors –
∧
i +
1
2
∧
j + 4
∧
k ,
∧
i +
1
2
∧
j + 4
∧
k ,
∧
i –
1
2
∧
j
+ 4
∧
k and –
∧
i –
1
2
∧
j + 4
∧
k , respectively, is
(A)
1
2
(B) 1 (C) 2 (D) 4
Sol. Given: ABCD is a rectangle.
We know that AB
→
= P.V. of point
B – P.V. of point A
= i
∧
+
1
2
j
∧
+ 4 k
∧
–
1
4
2
i j k
∧ ∧ ∧
 
− + +
 
 
= i
∧
+
1
2
j
∧
+ 4 k
∧
+ i
∧
–
1
2
j
∧
– 4 k
∧
= 2 i
∧
+ 0 j
∧
+ 0 k
∧
∴ AB = | AB
→
| = 4 0 0
+ + = 4 = 2
and AD
→
= P.V. of point D – P.V. of point A
= – i
∧
–
1
2
j
∧
+ 4 k
∧
–
1
4
2
i j k
∧ ∧ ∧
 
− + +
 
 
= – i
∧
–
1
2
j
∧
+ 4 k
∧
+ i
∧
–
1
2
j
∧
– 4 k
∧
= – j
∧
= 0 i
∧
– j
∧
+ 0 k
∧
∴ AD = | AD
→
| = 0 1 0
+ + = 1 = 1
∴ Area of rectangle ABCD = (AB)(AD) (= Length × Breadth)
= 2(1) = 2 sq. units
∴ Option (C) is the correct answer.
or Area of rectangle ABCD = AB AD
→ →
× .
A B
D C
MathonGo 29

MISCELLANEOUS EXERCISE
1. Write down a unit vector in XY-plane making an angle of 30°
with the positive direction of x-axis.
Sol. Let OP
→
be the unit vector in XY-plane such that ∠XOP = 30°
MathonGo 30

Therefore, | OP
→
| = 1
i.e., OP = 1 ...(i)
By Triangle Law of Addition of
vectors,
In ∆OMP, OP
→
= OM
→
+ MP
→
= (OM) i
∧
+ (MP) j
∧
[
| |
a
=
a
a
→
∧
→
= ⇒
∧
→ →
∵ a |a| a and unit vector along OX is i
∧
and along OY is j
∧
]
⇒ OP
→
= OP
OM
OP
i
∧
+ OP
MP
OP
j
∧
(Dividing and multiplying by OP in R.H.S.)
= (1) (cos 30°) i
∧
+ (1) (sin 30°) j
∧
[... By (i), OP = 1]
⇒ unit vector OP
→
= (cos 30) i + (sin 30°) j ... (ii)
⇒ OP
→
=
3
2
i
∧
+
1
2
j
∧
.
Remark: From Eqn. (ii) of above solution, we can generalise the
following result.
A unit vector along a line making an angle θ
θ
θ
θ
θ with positive
x-axis is (cos θ)
θ)
θ)
θ)
θ) i
∧
+ (sin θ)
θ)
θ)
θ)
θ) j
∧
2. Find the scalar components and magnitude of the vector
joining the points P(x1, y1, z1) and Q(x2, y2, z2).
Sol. Given points are P(x1, y1, z1) and Q(x2, y2, z2).
P
( , , )
x y z
1 1 1
Q
( , , )
x y z
2 2 2
⇒ P.V. (Position vector) of point P is
(x1, y1, z1) = x1 i
∧
+ y1 j
∧
+ z1 k
∧
and P.V. of point Q is (x2, y2, z2) = x2 i
∧
+ y2 j
∧
+ z2 k
∧
∴ Vector PQ
→
, the vector joining the points P and Q.
= P.V. of terminal point Q – P.V. of initial point P
= x2 i
∧
+ y2 j
∧
+ z2 k
∧
– (x1 i
∧
+ y1 j
∧
+ z1 k
∧
)
= x2 i
∧
+ y2 j
∧
+ z2 k
∧
– x1 i
∧
– y1 j
∧
– z1 k
∧
⇒ PQ
→
= (x2 – x1) i
∧
+ (y2 – y1) j
∧
+ (z2 – z1) k
∧
X
Y
Y′
X′
30°
O M
P
MathonGo 31

∴ Scalar components of the vector PQ
→
are the coefficients of
i
∧
, j
∧
, k
∧
in PQ
→
i.e., (x2 – x1), ( y2 – y1), (z2 – z1)
and magnitude of vector PQ
→
= 2 2 2
2 1 2 1 2 1
( ) ( ) ( )
x x y y z z
− + − + − . | 2 2 2
x y z
+ +
3. A girl walks 4 km towards west, then she walks 3 km in a
direction 30° east of north and stops. Determine the girl’s
displacement from her initial point of departure.
Sol. Let us take the initial point of departure as origin.
Let the girl walk a distance OA = 4 km towards west.
Through the point A draw a line AQ parallel to a line OP (which
is 30° east of North i.e., in East-North quadrant making an angle
of 30° with North)
Let the girl walk a distance AB = 3 km (given) along this
direction OQ
→
(given). ∴ OA
→
= 4 (– i
∧
)[...Vector OA
→
is along
OX′)]
= – 4 i
∧
...(i)
We know that (By Remark Q.N. 1 of this miscellaneous exercise)
a unit vector along AQ
→
(or AB
→
) making an angle θ = 60° with
positive x-axis is (cos θ) i
∧
+ (sin θ) j
∧
= (cos 60°) i
∧
+ (sin 60°) j
∧
=
1
2
i
∧
+
3
2
j
∧
.
∴ AB
→
= | AB
→
| (A unit vector along AB
→
) |... a
→
= | a
→
| a
∧
=
1 3 3 3 3
3
2 2 2 2
i j i j
∧ ∧ ∧ ∧
 
+ = +
 
 
... (ii)
MathonGo 32

∴ Girl’s displacement from her initial point O of departure (to
final point B) = OB
→
= OA
→
+ AB
→
(By Triangle Law of Addition
of vectors)
= – 4 i
∧
+
3 3 3
+
2 2
i j
∧ ∧
 
 
 
=
3
4
2
 
− +
 
 
i
∧
+
3 3
2
j
∧
[By (i)] [By (ii)]
=
5
2
−
i
∧
+
3 3
2
j
∧
.
4. If
→
a =
→
b +
→
c , then is it true that |
→
a | = |
→
b | + |
→
c |?
Justify your answer.
Sol. The result is not true (always).
Given: a
→
= b
→
+ c
→
.
∴ Either the vectors a
→
, b
→
, c
→
are collinear or vectors a
→
, b
→
,
c
→
form the sides of a triangle.
Case I. Vectors
→
a ,
→
b ,
→
c are collinear.
Let a
→
= AC
→
, b
→
= AB
→
and c
→
= BC
→
,
then a
→
= AC
→
= AB
→
+ BC
→
= b
→
+ c
→
.
Also, | a
→
| = AC = AB + BC = | b
→
| + | c
→
|.
Case II. Vectors
→
a ,
→
b ,
→
c form a
triangle.
Here also by Triangle Law of vectors,
a
→
= b
→
+ c
→
But | a
→
| < | b
→
| + | c
→
|
(... Each side of a triangle is less than sum of the other two sides)
∴ |( a
→
) = b
→
+ c
→
| = | b
→
| + | c
→
| is true only when vectors
b
→
and c
→
are collinear vectors.
5. Find the value of x for which x(
∧
i +
∧
j +
∧
k ) is a unit vector.
Sol. Because x( i
∧
+ j
∧
+ k
∧
) = x i
∧
+ x j
∧
+ x k
∧
is a unit vector (given)
Therefore, |x i
∧
+ x j
∧
+ x k
∧
| = 1
c
→
b
→ B C
A
A
B C
a
→
b
→
c
→
MathonGo 33

∴ 2 2 2
x x x
+ + = 1
[... x i
∧
+ y j
∧
+ z k
∧
= 2 2 2
x y z
+ + ]
Squaring both sides 3x2
= 1 or x2
=
1
3
∴ x = ±
1
3
.
6. Find a vector of magnitude 5 units and parallel to the
resultant of the vectors
→
a = 2
∧
i + 3
∧
j –
∧
k and
→
b =
∧
i – 2
∧
j +
∧
k .
→
= 2 i
∧
+ 3 j
∧
– k
∧
and b
→
= i
∧
– 2 j
∧
+ k
∧
.
Let vector c
→
be the resultant of vectors a
→
and b
→
.
∴
→
c =
→
a +
→
b = 2 i
∧
+ 3 j
∧
– k
∧
+ i
∧
– 2 j
∧
+ k
∧
.
= 3 i
∧
+ j
∧
+ 0 k
∧
.
∴ Required vector of magnitude 5 units and parallel (or
collinear) to resultant vector c
→
= a
→
+ b
→
is
5
∧
c = 5
| |
c
c
→
→
= 5
3 0
9 1 0
i j k
∧ ∧ ∧
 
+ +
 
 
+ +
 
=
5
10
(3 i
∧
+ j
∧
) =
5
10
10
10
(3 i
∧
+ j
∧
)
=
5
10
10 (3 i
∧
+ j
∧
) =
10
2
(3 i
∧
+ j
∧
)=
3
2
10 i
∧
+
10
2
j
∧
.
7. If
→
a =
∧
i +
∧
j +
∧
k ,
→
b = 2
∧
i –
∧
j + 3
∧
k and
→
c =
∧
i – 2
∧
j +
∧
k , find a unit vector parallel to the
vector 2
→
a –
→
b + 3
→
c .
→
= i
∧
+ j
∧
+ k
∧
, b
→
= 2 i
∧
– j
∧
+ 3 k
∧
and c
→
= i
∧
– 2 j
∧
+ k
∧
.
Let d
→
= 2 a
→
– b
→
+ 3 c
→
= 2( i
∧
+ j
∧
+ k
∧
) – (2 i
∧
– j
∧
+ 3 k
∧
) + 3( i
∧
– 2 j
∧
+ k
∧
)
= 2 i
∧
+ 2 j
∧
+ 2 k
∧
– 2 i
∧
+ j
∧
– 3 k
∧
+ 3 i
∧
– 6 j
∧
+ 3 k
∧
∴ d
∧
= 3 i
∧
– 3 j
∧
+ 2 k
∧
∴ A unit vector parallel to the vector
d
→
= 3 i
∧
– 3 j
∧
+ 2 k
∧
is
d
∧
=
| |
d
d
→
→
=
3 3 2
9 9 4 22
i j k
∧ ∧ ∧
− +
+ + =
=
3
22
i
∧
–
3
22
j
∧
+
2
22
k
∧
.
MathonGo 34

8. Show that the points A(1, – 2, – 8), B(5, 0, – 2) and
C(11, 3, 7) are collinear and find the ratio in which B dividesAC.
Sol. Given: Points A(1, – 2, – 8), B(5, 0, – 2) and C(11, 3, 7).
i.e., Position vectors of points A, B, C are
OA
→
(= A(1, – 2, – 8)) = i
∧
– 2 j
∧
– 8 k
∧
OB
→
(= B(5, 0, – 2)) = 5 i
∧
+ 0 j
∧
– 2 k
∧
= 5 i
∧
– 2 k
∧
and OC
→
(= C(11, 3, 7)) = 11 i
∧
+ 3 j
∧
+ 7 k
∧
∴ AB
→
= 5 i
∧
– 2 k
∧
– ( i
∧
– 2 j
∧
– 8 k
∧
) = 5 i
∧
– 2 k
∧
– i
∧
+ 2 j
∧
+ 8 k
∧
or AB
→
= 4 i
∧
+ 2 j
∧
+ 6 k
∧
∴ AB = | AB
→
| = 16 4 36
+ + = 56 = 4 14
× = 2 14
and BC
→
= 11 i
∧
+ 3 j
∧
+ 7 k
∧
– (5 i
∧
– 2 k
∧
) = 11 i
∧
+ 3 j
∧
+ 7 k
∧
– 5 i
∧
+ 2 k
∧
= 6 i
∧
+ 3 j
∧
+ 9 k
∧
∴ BC = | BC
→
| = 36 9 81
+ + = 126 = 9 14
× = 3 14
AC
→
= 11 i
∧
+ 3 j
∧
+ 7 k
∧
– ( i
∧
– 2 j
∧
– 8 k
∧
)
= 11 i
∧
+ 3 j
∧
+ 7 k
∧
– i
∧
+ 2 j
∧
+ 8 k
∧
= 10 i
∧
+ 5 j
∧
+ 15 k
∧
∴ AC = | AC
→
| = 100 25 225
+ + = 350 = 25 14
× = 5 14
Now, AB
→
+ BC
→
= 4 i
∧
+ 2 j
∧
+ 6 k
∧
+ 6 i
∧
+ 3 j
∧
+ 9 k
∧
= 10 i
∧
+ 5 j
∧
+ 15 k
∧
= AC
→
∴ Points A, B, C are either collinear or are the vertices of
∆ABC.
Again AB + BC = 2 14 + 3 14 = (2 + 3) 14 = 5 14 = AC
∴ Points A, B, C are collinear.
Now to find the ratio in which B divides AC
A(1, – 2, – 8) C(11, 3, 7)
→
B
λ : 1
(5, 0, – 2)
= a
→
= c
→
= b
Let the point B divides AC in the ratio λ : 1.
MathonGo 35

∴ By section formula, P.V. of point B is
1
1
c a
→ →
λ +
λ +
⇒ (5, 0, – 2) =
(11, 3, 7) (1, 2, 8)
1
λ + − −
λ +
Cross-multiplying,
(λ + 1)(5 i
∧
+ 0 j
∧
– 2 k
∧
) = λ(11 i
∧
+ 3 j
∧
+ 7 k
∧
) + ( i
∧
– 2 j
∧
– 8 k
∧
)
⇒ 5(λ + 1) i
∧
– 2(λ + 1) k
∧
= 11λ i
∧
+ 3λ j
∧
+ 7λ k
∧
+ i
∧
– 2 j
∧
– 8 k
∧
⇒ (5λ + 5) i
∧
– (2λ + 2) k
∧
= (11λ + 1) i
∧
+ (3λ – 2) j
∧
+ (7λ – 8) k
∧
∧
, j
∧
, k
∧
5λ + 5 = 11λ + 1, 0 = 3λ – 2, – (2λ + 2) = 7λ – 8
⇒ – 6λ = – 4, – 3λ = – 2, – 2λ – 2 = 7λ – 8 (⇒ – 9λ = – 6)
⇒ λ =
4
6
=
2
3
, λ =
2
3
, λ =
6
9
=
2
3
All three values of λ are same.
∴ Required ratio is λ : 1 =
2
3
: 1 = 2 : 3.
9. Find the position vector of a point R which divides the line
joining the two points P and Q whose position vectors are
(2
→
a +
→
b ) and (
→
a – 3
→
b ) externally in the ratio 1 : 2. Also,
show that P is the middle point of line segment RQ.
Sol. We know that position vector of the point R dividing the join of P
and Q externally in the ratio 1 : 2 = m : n is given by
→
c =
–
→ →
mb na
m – n
=
1( 3 ) 2(2 )
1 2
a b a b
→ →
→ →
− − +
−
=
3 4 2
1 2
a b a b
→ →
→ →
− − −
−
=
3 5
1
a b
→
→
− −
−
= 3 a
→
+ 5 b
→
Again position vector of the middle point of the line segment RQ
=
P.V. of point R + P.V. of point Q
2
=
3 5 3
2
a b a b
→
→
→ →
+ + −
=
4 2
2
a b
→
→
+
= 2 a
→
+ b
→
= P.V. of point P. (given)
∴ Point P is the middle point of the line segment RQ.
10. Two adjacent sides of a parallelogram are 2
∧
i – 4
∧
j + 5
∧
k
and
∧
i – 2
∧
j – 3
∧
k . Find the unit vector parallel to its
diagonal. Also, find its area.
Sol. Let ABCD be a parallelogram.
MathonGo 36

Given: The vectors representing two adjacent sides of this
parallelogram are say
a
→
= 2 i
∧
– 4 j
∧
+ 5 k
∧
and b
→
= i
∧
– 2 j
∧
– 3 k
∧
Formula: ∴ Vectors along the
diagonals AC
→
and DB
→
of the
parallelogram are
→
a +
→
b and
→
a –
→
b
i.e., a
→
+ b
→
= 2 i
∧
– 4 j
∧
+ 5 k
∧
+ i
∧
– 2 j
∧
– 3 k
∧
= 3 i
∧
– 6 j
∧
+ 2 k
∧
and a
→
– b
→
= 2 i
∧
– 4 j
∧
+ 5 k
∧
– ( i
∧
– 2 j
∧
– 3 k
∧
)
= 2 i
∧
– 4 j
∧
+ 5 k
∧
– i
∧
+ 2 j
∧
+ 3 k
∧
= i
∧
– 2 j
∧
+ 8 k
∧
∴ Unit vectors parallel to (or along) diagonals are
| |
a b
a b
→
→
+
→
→
+
and
| |
a b
a b
→
→
−
→
→
−
=
3 6 2
9 36 4 49 7
i j k
∧ ∧ ∧
− +
+ + = =
and
2 8
1 4 64 69
i j k
∧ ∧ ∧
− +
+ + =
Let us find area of parallelogram
a
→
× b
→
= 2 4 5
1 2 3
i j k
∧ ∧ ∧
−
− −
= i
∧
(12 + 10) – j
∧
(– 6 – 5) + k
∧
(– 4 + 4)
= 22 i
∧
+ 11 j
∧
+ 0 k
∧
We know that area of parallelogram = | a
→
× b
→
|
= 2 2 2
(22) (11) 0
+ + = 484 121
+ = 605
= 5 121
× = 121 5
× = 11 5 sq. units.
11. Show that the direction cosines of a vector equally inclined
to the axes OX, OY and OZ are
1
3
,
1
3
,
1
3
.
Sol. Let l, m, n be the direction cosines of a vector equally inclined to
the axes OX, OY, OZ.
∴ A unit vector along the given vector is
a
∧
= l i
∧
+ m j
∧
+ n k
∧
and | a
∧
| = 1
⇒ 2 2 2
l m n
+ + = 1 ∴ l2
+ m2
+ n2
= 1 ...(i)
Let the given vector (for which unit vector is a
∧
) make equal
angles (given) θ, θ, θ (say) with OX (⇒ i
∧
), OY (⇒
∧
j ) and OZ (⇒ k
∧
)
∴ The given vector is in positive octant OXYZ and hence θ is
acute. ...(ii)
b
→
a
→
a –
b
→
→
a + b
→
→
A B
D C
MathonGo 37

∴ For angle θ between a
∧
and i
∧
,
cos θ =
.
| || |
a i
a i
∧
∧
∧
∧
=
( ) . ( 0 0 )
(1)(1)
l i m j n k i j k
∧ ∧ ∧ ∧ ∧ ∧
+ + + +
or cos θ = l(1) + m(0) + n(0) = l
or l = cos θ ...(iii)
Similarly, for angle θ between a
∧
and j
∧
, m = cos θ ...(iv)
Similarly, for angle θ between a
∧
and k
∧
, n = cos θ ...(v)
Putting these values of l, m, n from (iii), (iv) and (v) in (i), we
have
cos2
θ + cos2
θ + cos2
θ = 1 ⇒ 3 cos2
θ = 1
⇒ cos2
θ =
1
3
⇒ cos θ = ±
1
3
= ±
1
3
∴ cos θ =
1
3
(... By (ii), θ is acute and hence cos θ is positive)
Putting cos θ =
1
3
in (ii), (iii) and (iv), direction cosines of the
required vector are l, m, n =
1
3
,
1
3
and
1
3
.
12. Let
→
a =
∧
i + 4
∧
j + 2
∧
k ,
→
b = 3
∧
i – 2
∧
j + 7
∧
k and
→
c = 2
∧
i –
∧
j + 4
∧
k . Find a vector
→
b which is
perpendicular to both
→
a and
→
b , and
→
c .
→
d = 15.
Sol. Given: Vectors are a
→
= i
∧
+ 4 j
∧
+ 2 k
∧
and b
→
= 3 i
∧
– 2 j
∧
+ 7 k
∧
By definition of cross-product of two
vectors, a
→
× b
→
is a vector
perpendicular to both a
→
and b
→
.
Hence, vector d
→
which is also perpendicular to both a
→
and b
→
is d
→
= λ( a
→
× b
→
) where λ = 1 or some other scalar.
Therefore, d
→
= λ 1 4 2
3 2 7
i j k
∧ ∧ ∧
−
Expanding along first row, = λ[ i
∧
(28 + 4) – j
∧
(7 – 6) +
k
∧
(– 2 – 12)]
or d
→
= λ[32 i
∧
– j
∧
– 14 k
∧
] ...(i)
d = a × b
λ( )
→
a
→
b
→
→ →
MathonGo 38

or d
→
= 32λ i
∧
– λ j
∧
– 14λ k
∧
To find λ
λ
λ
λ
λ: Given: c
→
= 2 i
∧
– j
∧
+ 4 k
∧
Also given c
→
. d
→
= 15
⇒ 2(32λ) + (– 1)(– λ) + 4(– 14λ) = 15
⇒ 64λ + λ – 56λ = 15 ⇒ 9λ = 15 ⇒ λ =
5
3
Putting λ =
5
3
in (i), required vector
d
→
=
5
3
(32 i
∧
– j
∧
– 14 k
∧
) =
1
3
(160 i
∧
– 5 j
∧
– 70 k
∧
).
13. The scalar product of the vector
∧
i +
∧
j +
∧
k with a unit
vector along the sum of the vectors 2
∧
i + 4
∧
j – 5
∧
k and
λ
λ
λ
λ
λ
∧
i + 2
∧
j + 3
∧
k is equal to one. Find the value of λ
λ
λ
λ
λ.
Sol. Given: Let a
→
= i
∧
+ j
∧
+ k
∧
...(i)
b
→
= 2 i
∧
+ 4 j
∧
– 5 k
∧
and c
→
= λ i
∧
+ 2 j
∧
+ 3 k
∧
∴ b
→
+ c
→
(= d
→
(say)) = (2 + λ) i
∧
+ 6 j
∧
– 2 k
∧
∴ d
∧
, a unit vector along b
→
+ c
→
= d
→
is
d
∧
=
| |
d
d
→
→ =
2 2
(2 ) 6 2
(2 ) 36 4 4 4 40
i j k
∧ ∧ ∧
+ λ + −
+ λ + + = + λ + λ +
or d
∧
=
2
(2 ) 6 2
4 44
i j k
∧ ∧ ∧
+ λ + −
λ + λ +
=
2
(2 )
4 44
+ λ
λ + λ +
i
∧
+
2
6
4 44
λ + λ +
j
∧
–
2
2
4 44
λ + λ +
k
∧
...(ii)
Given: Scalar (i.e., Dot) Product of a
→
and d
∧
i.e., = a
→
. d
∧
= 1
∴ From (i) and (ii),
2
1(2 )
4 44
+ λ
λ + λ +
+ 2
1(6)
4 44
λ + λ +
+ 2
1( 2)
4 44
−
λ + λ +
= 1
Multiplying by L.C.M. = 2
4 44
λ + λ + ,
2 + λ + 6 – 2 = 2
4 44
λ + λ + ⇒ λ + 6 = 2
4 44
λ + λ +
Squaring both sides (λ + 6)2
= λ2
+ 4λ + 44
⇒ λ2
+ 12λ + 36 = λ2
+ 4λ + 44
⇒ 8λ = 8 ⇒ λ = 1.
14. If
→
a ,
→
b ,
→
c are mutually perpendicular vectors of equal
magnitude, show that the vector
→
a +
→
b +
→
c is equally
inclined to
→
a ,
→
b ,
→
c .
MathonGo 39

Sol. Given: a
→
, b
→
, c
→
are mutually perpendicular vectors of equal
magnitude.
∴ a
→
. b
→
= b
→
. a
→
= 0, b
→
. c
→
= c
→
. b
→
= 0,
a
→
. a
→
= a
→
. c
→
= 0 ... (i)
and | a
→
| = | b
→
| = | c
→
| = λ (say) ... (ii)
Let vector d
→
= a
→
+ b
→
+ c
→
make angles θ1, θ2, θ3 with
vectors a
→
, b
→
, c
→
respectively.
∴ cos θ1 =
.
| |
d a
d a
→ →
→ →
+
=
( ) .
| || |
a b c a
a
a b c
→ → →
→
+ +
→ →
→ →
+ +
=
. . .
| || |
a a a b a c
a b c a
→ → →
→ →
→
+ +
→ →
→ →
+ +
=
2
| | 0 0
| |
| |
a
a b c a
→
+ +
→ →
→ →
+ +
[By (i)]
⇒ cos θ1 =
2
| |
| |
| |
a
a b c a
→
→ →
→ →
+ +
=
| |
| |
a
a b c
→
→
→ →
+ +
... (iii)
Let us now find |
→
a +
→
b +
→
c |.
We know that | a
→
+ b
→
+ c
→
|2
= ( a
→
+
→
b +
→
c )2
= a
→ 2
+ ( b
→
+ c
→
)2
+ 2 a
→
. ( b
→
+ c
→
)
[... ( A
→
+ B
→
)2
= A
→
2
+ B
→
2
+ 2 A
→
. B
→
]
= a
→ 2
+ b
→ 2
+ c
→ 2
+ 2 b
→
. c
→
+ 2 a
→
. b
→
+ 2 a
→
. c
→
= | a
→
|2
+ | b
→
|2
+ | c
→
|2
+ 2 b
→
. c
→
+ 2 a
→
. b
→
+ 2 a
→
. c
→
Putting values from (i) and (ii)
| a
→
+ b
→
+ c
→
|2
= λ2
+ λ2
+ λ2
+ 0 + 0 + 0 = 3λ2
∴ | a
→
+ b
→
+ c
→
| = 2
3λ = λ 3
Putting this value of | a
→
+ b
→
+ c
→
| = λ 3 and | a
→
| = λ
from (ii) in (iii), cos θ1 =
3
λ
λ
=
1
3
∴ θ1 = cos– 1 1
3
Similarly, θ2 = cos– 1 1
3
and θ3 = cos – 1 1
3
∴ θ1 = θ2 = θ3
1 1
cos
3
−
 
=
 
 
∴ Vector a
→
+ b
→
+ c
→
is equally inclined to the vectors a
→
,
b
→
and c
→
.





MathonGo 40

15. Prove that (
→
a +
→
b ) . (
→
a +
→
b ) = |
→
a |2
+ |
→
b |2
, if and
only if
→
a ,
→
b are perpendicular, given
→
a ≠
≠
≠
≠
≠
→
0 ,
→
b ≠
≠
≠
≠
≠
→
0 .
Sol. We know that ( a
→
+ b
→
) . ( a
→
+ b
→
)
= a
→
. a
→
+ a
→
. b
→
+ b
→
. a
→
+ b
→
. b
→
= | a
→
|2
+ a
→
. b
→
+ a
→
. b
→
+ | b
→
|2
= | a
→
|2
+ | b
→
|2
+ 2 a
→
. b
→
...(i)
For If part: Given:
→
a and
→
b are perpendicular
⇒ a
→
. b
→
= 0
Putting a
→
. b
→
= 0 in (i), we have
( a
→
+ b
→
) . ( a
→
+ b
→
) = | a
→
|2
+ | b
→
|2
For Only if part:
Given: ( a
→
+ b
→
) . ( a
→
+ b
→
) = | a
→
|2
+ | b
→
|2
Putting this value in L.H.S. eqn. (i), we have
| a
→
|2
+ | b
→
|2
= | a
→
|2
+ | b
→
|2
+ 2 a
→
. b
→
⇒ 0 = 2 a
→
. b
→
⇒ a
→
. b
→
=
0
2
= 0
But a
→
≠ 0
→
and b
→
≠ 0
→
(given).
∴ Vector a
→
and b
→
16. Choose the correct answer:
If θ
θ
θ
θ
θ is the angle between two vectors
→
a and
→
b , then
→
a .
→
b ≥
≥
≥
≥
≥ 0
only when
(A) 0 < θ
θ
θ
θ
θ <
π
2
(B) 0 ≤ θ ≤
≤ θ ≤
≤ θ ≤
≤ θ ≤
≤ θ ≤
π
2
(C) 0 < θ
θ
θ
θ
θ < π
π
π
π
π (D) 0 < θ ≤
θ ≤
θ ≤
θ ≤
θ ≤ π
π
π
π
π
Sol. Given: a
→
. b
→
≥ 0
⇒ | a
→
| | b
→
| cos θ ≥ 0 ⇒ cos θ ≥ 0
[... | a
→
| and | b
→
| being lengths of vectors are always ≥ 0]
and this is true only for option (B) out of the given
options For option (A) 0 , cos 0
2
π
 
< θ < θ >
 
 
∵ .
17. Choose the correct answer :
Let
→
a and
→
b be two unit vectors and θ
θ
θ
θ
θ is the angle
between them. Then
→
a +
→
b is a unit vector if
(A) θ
θ
θ
θ
θ =
π
4
(B) θ
θ
θ
θ
θ =
π
3
(C) θ
θ
θ
θ
θ =
π
2
(D) θ
θ
θ
θ
θ =
π
2
3
.
Sol. Given: a
→
, b
→
and a
→
+ b
→
are unit vectors
⇒ | a
→
| = 1, | b
→
| = 1 and | a
→
+ b
→
| = 1
Now, squaring both sides of | a
→
+ b
→
| = 1, we have
MathonGo 41

i
^
j
^
k
^
| a
→
+ b
→
|2
= 1 ⇒ ( a
→
+ b
→
)2
= 1
⇒ a
→ 2
+ b
→ 2
+ 2 a
→
. b
→
= 1
⇒ | a
→
|2
+ | b
→
|2
+ 2| a
→
| | b
→
| cos θ = 1
where θ is the given angle between vectors a
→
and b
→
.
Putting | a
→
| = 1 and | b
→
| = 1, we have 1 + 1 + 2 cos θ = 1
⇒ 2 cos θ = – 1 ⇒ cos θ =
1
2
−
= – cos 60°
⇒ cos θ = cos (180° – 60°) ⇒ cos θ = cos 120°
⇒ θ = 120° = 120 ×
180
π
=
2
3
π
∴ Option (D) is the correct answer.
Very Important Results
(1) i
∧
. i
∧
= | i
∧
|2
= 1, j
∧
. j
∧
= 1, k
∧
. k
∧
= 1.
(2) i
∧
× i
∧
= 0
→
, j
∧
× j
∧
= 0
→
and k
∧
× k
∧
= 0
→
.
(3) i
∧
. j
∧
= 0 = j
∧
. i
∧
, j
∧
. k
∧
= 0 = k
∧
. j
∧
, i
∧
. k
∧
= 0 = k
∧
. i
∧
.
(4) i
∧
× j
∧
= k
∧
, j
∧
× k
∧
= i
∧
and k
∧
× i
∧
= j
∧
.
18. Choose the correct answer:
The value of
∧
i . (
∧
j ×
∧
k ) +
∧
j . (
∧
i ×
∧
k ) +
∧
k . (
∧
i ×
∧
j ) is
(A) 0 (B) – 1 (C) 1 (D) 3
Sol. i
∧
. ( j
∧
× k
∧
) + j
∧
. ( i
∧
× k
∧
) + k
∧
. ( i
∧
× j
∧
)
= i
∧
. i
∧
+ j
∧
. (– j
∧
) + k
∧
. k
∧
(... i
∧
× k
∧
= – k
∧
× i
∧
= – j
∧
)
= 1 – 1 + 1 = 1
∴ Option (C) is the correct answer.
19. If θ
θ
θ
θ
θ be the angle between any two vectors
→
a and
→
b , then
|
→
a .
→
b | = |
→
a ×
→
b |, when θ
θ
θ
θ
θ is equal to
(A) 0 (B)
π
4
(C)
π
2
(D) π
π
π
π
π
Sol. Given: | a
→
. b
→
| = | a
→
× b
→
|
⇒ | a
→
| | b
→
| | cos θ | = | a
→
| | b
→
| sin θ
(... a
→
. b
→
= | a
→
|| b
→
| cos θ
⇒ | a
→
. b
→
| = | a
→
|| b
→
| | cos θ |)
Dividing both sides by | a
→
|| b
→
|, we have |cos θ| = sin θ
and this equation is true only for option (B) namely θ =
4
π
out of
the given options.
1 1
cos and also sin
4 4
2 2
π π
 
= =
 
 
∵
∴ Option (B) is the correct option.
MathonGo 42

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