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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra – Text

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Page 1

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Page 2

Class : 12th
Subject : Maths
Chapter : 10
Chapter Name : Vector Algebra

Q1 Represent graphically a displacement of 40 km, 30 ∘ east of north.

→
Here vector OP represent the displacement of 40 km, 30 ∘ East of north .

Page : 428 , Block Name : Exercise 10.1

Q2 Classify the following measures as scalars and vectors.
(i) 10kg (ii) 2 metres north-west (iii) 40 ∘
(iv) 40 watt (v)10 − 19 coulomb (vi) 20m / s 2

Answer. (i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40 ∘ is a scalar quantity as it involves only magnitude.
(iv) 40 watts is a scalar quantity as it involves only magnitude.
(v) 10 − 19 coulomb is a scalar quantity as it involves only magnitude.
(vi) 20m / s 2 is a vector quantity as it involves magnitude as well as direction.

Page : 428 , Block Name : Exercise 10.1

Q3 Classify the following as scalar and vector quantities.
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done

Answer. (i) Time period is a scalar quantity as it involves only magnitude.
(ii) Distance is a scalar quantity as it involves only magnitude.
(iii) Force is a vector quantity as it involves both magnitude and direction.
(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.
(v) Work done is a scalar quantity as it involves only magnitude.

Page : 428 , Block Name : Exercise 10.1

Q4 In Figure, identify the following vectors.

(i) Coinitial
(ii) Equal
(iii) Collinear but not equal

Page 3

→
Answer. (i) Vectors →a and d are coinitial because they have the same initial point.
→ →
(ii) vectors b and d are equal because they have the same magnitude and direction.
(iii) Vectors →
a and →
c are collinear but not equal. This is because although they are parallel, their directions are not the same.

Page : 428 , Block Name : Exercise 10.1

Q5 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.

Answer. (i) True.
→
Vectors → a and − a are parallel to the same line.
(ii) False.
Collinear vectors are those vectors that are parallel to the same line.
(iii) False.

Page : 428 , Block Name : Exercise 10.1

Q1 Compute the magnitude of the following vectors:
→ 1 1 1
→ →
a = î + ĵ + k̂; b = 2î − 7ĵ − 3k̂; c = î + ĵ − k̂
√3 √3 √3

Answer. The given vectors are:
→ 1 1 1
→ →
a = î + ĵ + k̂; b = 2î − 7ĵ − 3k̂; c = î + ĵ − k̂
√3 √3 √3
2 2 2
→
|a| = √(1) + (1) + (1) = √3
→
| b | = √(2) 2 + ( − 7) 2 + ( − 3) 2

= √4 + 49 + 9
= √62

|→
c| = √( √ ) ( √ ) ( √ )
1
3
2
+
1
3
2
+ −
1
3
2

√
1 1 1
= + + =1
3 3 3

Page : 440 , Block Name : Exercise 10.2

Q2 Write two different vectors having same magnitude.

→
Consider →
a = (î − 2ĵ + 3k̂) and b = (2î + ĵ − 3k̂)
It can be observed that | →
a| = √1 2 + ( − 2) 2 + 3 2 = √1 + 4 + 9 = √14 and
→
b| = √2 2 + 1 2 + ( − 3) 2 = √4 + 1 + 9 = √14
→
Hence, →a and b are two different vectors having the same magnitude. The vectors are
different because they have different directions.

Page : 440 , Block Name : Exercise 10.2

Q3 Write two different vectors having same direction.

Consider →
p = (î + ĵ + k̂) and →
q = (2î + 2ĵ + 2k̂)
The direction cosines of →
p are given by,
1 1 1 1 1 1
I= = ,m = = , and n = =
√1 2 + 1 2 + 1 2 √3 √1 2 + 1 2 + 1 2 √3 √1 2 + 1 2 + 1 2 √3
The direction cosines of →
q are given by

Page 4

2 2 1 2 2 1
l= = = ,m = = =
2√ 3 √3 2√ 3 √3
√2 2 + 2 2 + 2 2 √2 2 + 2 2 + 2 2
2 2 1
and n = = =
2√ 3 √3
√ 22 + 22 + 22

The direction cosines of →
p and →
q are the same. Hence, the two vectors have the same
direction.

Page : 440 , Block Name : Exercise 10.2

Q4 Find the values of x and y so that the vectors 2î + 3ĵ and xî + yŷ are equal.

Answer. The two vector 2î + 3ĵ and xî + yĵ will be equal if their corresponding components are equal.
Hence, the required values of x and y are 2 and 3 respectively.

Page : 440 , Block Name : Exercise 10.2

Q5 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point( -5, 7).

Answer. The vector with the initial point p (2, 1) and terminal point Q ( 5, 7) can be given by,
→
PQ = ( − 5 − 2)î + (7 − 1)ĵ
→
⇒ PQ = − 7î + 6ĵ
Hence, the required scalar components are —7 and 6 while the vector components are
− 7î and 6ĵ

Page : 440 , Block Name : Exercise 10.2

→
Q6 Find the sum of the vectors →
a = î − 2ĵ + k̇, b = − 2î + 4ĵ + 5k̂ and →
c = î − 6ĵ − 7k̂

→
Answer. The given vector are →
a = î − 2ĵ + k̂, b = − 2î + 4ĵ + 5k̂ and →
c = î − 6ĵ − 7k̂
→
∴→
a+b+→
c = (1 − 2 + 1)î + ( − 2 + 4 − 6)ĵ + (1 + 5 − 7)k̂
= 0 ⋅ î − 4ĵ − 1 ⋅ k̂
= − 4ĵ − k̂

Page : 440 , Block Name : Exercise 10.2

Q7 Find the unit vector in the direction of the vector →
a = î + ĵ + 2k̂

¯ a→
The unit vector â in the direction of vector a = î + ĵ + 2k̂ is given by â = | a |

|→
a| = √1 2 + 1 2 + 2 2 = √1 + 1 + 4 = √6
a→ î + ĵ + 2k̂ 1 1 2
∴ â = | a | =
→ = î + ĵ + k̇
√6 √6 √6 √6

Page : 440 , Block Name : Exercise 10.2

→
Q8 Find the unit vector in the direction of vector PQ , where p and Q are the points (1, 2, 3) and (4, S, 6), respectively.

The given points are P(1, 2, 3) and Q(4, 5, 6).
→
∴ PQ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)k̂ = 3î + 3ĵ + 3k̂
→
| PQ | = √3 2 + 3 2 + 3 2 = √9 + 9 + 9 = √27 = 3√3
¯
Hence, the unit vector in the direction of PQ is
→
PQ 3î + 3ĵ + 3k̂ 1 1 1
→ = = î + ĵ + k̇
3√ 3 √3 √3 √3
| PQ |

Page : 440 , Block Name : Exercise 10.2

Page 5

Q9
¯
→
For given vectors, a = 2î − ȷ̂ + 2k̂ and b = − î + ĵ − k̂ , find the unit vector in the direction
→
of the vector →
a+b

→
The given vectors are →
a = 2î − ĵ + 2k̂ and b = − î + ĵ − k̂
→
a = 2î − ĵ + 2k̂
→
b = − î + ĵ − k̂
→
∴→
a + b = (2 − 1)î + ( − 1 + 1)ĵ + (2 − 1)k̂ = 1î + 0ĵ + 1k̂ = î + k̂
→
|→
a + b| = √1 2 + 1 2 = √2
¯
→
Hence, the unit vector in the direction of a + b) is
→
( a→ + b ) î + k̂ 1 1
→ = = 2 î + k̂
√2 √2
→
|a+b|

Page : 440 , Block Name : Exercise 10.2

Q10 Find a vector in the direction of vector 5î − ĵ + 2k̂ which has magnitude 8 units.

Let →
a = 5î − ĵ + 2k̂
∴ |→
a| = √5 2 + ( − 1) 2 + 2 2 = √25 + 1 + 4 = √30
a→ 5î − ĵ + 2k̂
∴ â = | a | = →
√30
Hence, the vector in the direction of vector 5î − ĵ + 2k̂ which has magnitude 8 units is
given by,

8â = 8
( ) 5iˆ − jˆ + 2k̂

√30
=
40

√30
î −
8

√30
ĵ +
16

√30
k̂

( ) →
¯
→
5 i − j + 2k
=8
√30

40 → 8 → 16 →
= i − j + k
√30 √30 √30

Page : 440 , Block Name : Exercise 10.2

Q11 Show that the vectors 2î − 3ĵ + 4k̂ and − 4î + 6ĵ − 8k̂ are collinear.

→
Let →
a = 2î − 3ĵ + 4k̂ and b = − 4î + 6ĵ − 8k̂
→
It is observed that b = − 4î + 6ĵ − 8k̂ = − 2(2î − 3ĵ + 4k̂) = − 2→
a
→
∴ b = λ→
a
where,
λ= −2
Hence, the given vectors are collinear.

Page : 440 , Block Name : Exercise 10.2

Q12 Find the direction cosines of the vector î + 2ĵ + 3k̂

Let →
a = î + 2ĵ + 3k̂
∴ |→
a| = √1 2 + 2 2 + 3 2 = √1 + 4 + 9 = √14
Hence, the direction cosines of →
a are
( 1
,
√14 √14 √14
2
,
3
)

Page 6

Page : 440 , Block Name : Exercise 10.2

Q13 Find the direction cosines of the vector joining the points A (I, 2, -3 ) and B (-1 , -2 , -3 ) directed from A to B.

The given points are A(1, 2, − 3) and B( − 1, − 2, 1) .
→
∴ AB = ( − 1 − 1)î + ( − 2 − 2)ĵ + {1 − ( − 3)}k̂
→
⇒ AB = − 2î − 4ĵ + 4k̂
→
∴ | AB | = − √( − 2) 2 + ( − 4) 2 + 4 2 = √4 + 16 + 16 = √36 = 6
Hence, the direction cosines of `→
a are
( 1

√14
,
2

√14
,
3

√14 )
Page : 440 , Block Name : Exercise 10.2

Q14 Show that the vector î + ĵ + k̂ is equally inclined to the axes OX, OM, and OZ.

Let →
a = î + ĵ + k̂
Then,
|→
a| = √1 2 + 1 2 + 1 2 = √3
Therefore, the dlrectlon coslnes of →

1
aare

1
( 1

√3
,

1
1

√3
,
1

√3 )
Then we have cosα = , cosβ = , cosγ = Hence, the given vector is equally inclined to axes OX, OM, and OZ.
√3 √3 √3

Page : 440 , Block Name : Exercise 10.2

Q15 Find the position vector of a point R which divides the line joining two points P and Q î + 2ĵ − k̂ and − î + ȷ̂ + k̂ whose position vectors are
respectively, in the ratio 2:1
(i) internally
(ii) externally

Answer. The position vector of point R dividing the line segment joining two points
P and Q in the ratio m: n is given by:
1. Internally:
→
mb + na→
m+n

ii. Externally:
→
mb − na→
m−n
Position vectors of P and Q are given as:
→ →
OP = î + 2ĵ − k̂ and OQ = − î + ĵ + k̂
(i) The position vector Of point R Which divides the line joining two points p and Q
internally in the ratio 2:1 is given by,
¯ 2( − î + ĵ + k̂) + 1( î + 2 ĵ − k̂) ( − 2 î + 2 ĵ + 2k̂) + ( î + 2 ĵ − k̂)
OR = =
2+1 3
− î + 4 ĵ + k̂ 1 4 1
= = − î + ĵ + k̂
3 3 3 3
(ii) The position vector of point R which divides the line joining two points P and Q
externally in the ratio 2:1 is given by,
→ 2( − î + ĵ + k̇) − 1( î + 2 ĵ − k̇)
OR = = ( − 2i + 2ĵ + 2k̂) − (i + 2ĵ − k̇)
2−1
= − 3i + 3k̂

Page : 440 , Block Name : Exercise 10.2

Q16 Find the position vector of the midpoint of the vector joining the points P (2, 3, 4) and Q (4 , 1 , -2 )

Answer. The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, - 2) is given by,

Page 7

→ (2 î + 3 ĵ + 4k̂) + (4 î + ĵ − 2k̂) (2 + 4) î + (3 + 1) ĵ + (4 − 2)k̂
OR = =
2 2
6 î + 4 ĵ + 2k̂
= = 3î + 2ĵ + k̂
2

Page : 441 , Block Name : Exercise 10.2

Q17
¯
Show that the points A, B and C with position vectors, a = 3î − 4ĵ − 4k̂
→
b = 2î − ĵ + k̂ and →
c = î − 3ĵ − 5k̂ , respectively form the vertices of a right angled triangle.

Answer. position vectors Of points A, B, and c are respectively given as:
→ →
a = 3î − 4ĵ − 4k̂, b = 2î − ĵ + k̂ and →
c = î − 3ĵ − 5k̂
→ →
a = 3î − 4ĵ − 4k̂, b = 2î − ĵ + k̂ and →
c = î − 3ĵ − 5k̂
→
→
∴ AB = b − → a = (2 − 3)î + ( − 1 + 4)ĵ + (1 + 4)k̂ = − î + 3ĵ + 5k̂
¯
→
BC = →
c − b = (1 − 2)î + ( − 3 + 1)ĵ + ( − 5 − 1)k̂ = − î − 2ĵ − 6k̂
→
CA = →
a−→ c = (3 − 1)î + ( − 4 + 3)ĵ + ( − 4 + 5)k̂ = 2î − ĵ + k̂
→
∴ | AB | 2 = ( − 1) 2 + 3 2 + 5 2 = 1 + 9 + 25 = 35
| BC | 2 = ( − 1) 2 + ( − 2) 2 + ( − 6) 2 = 1 + 4 + 36 = 41
| CA | 2 = 2 2 + ( − 1) 2 + 1 2 = 4 + 1 + 1 = 6
→ → →
∴ | AB | 2 + | CA | 2 = 36 + 6 = 41 = | BC | 2
Hence, ABC is a right-angled triangle.

Page : 441 , Block Name : Exercise 10.2

Q18 In triangle ABC 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

On applying the triangle law of addition in the given triangle, we have:
→ → →
AB + BC = AC
→
⇒ AB + BC = − CA
¯ → ¯
→
⇒ AB + BC + CA = 0
∴ The equation given in alternative A is true.
→ → →
AB + BC = AC
→ → →
→
⇒ AB + BC − AC = 0
∴ The equation given in altermative B is true.
From equation (2), we have:

Page 8

Now, consider the equation given in altemative C:
→ → →
→
AB + BC − CA = 0
→ → ¯
⇒ AB + BC = CA
→ →
AC = CA
→ →
⇒ AC = − AC
→ →
→
⇒ AC + AC = 0
→
→
⇒ 2AC = 0
→
→
⇒ AC = 0 , which is not true.
Hence, the equation given in altemative C is incorrect.
The correct answer is C.

Page : 441 , Block Name : Exercise 10.2

Q19
→
If →
a and b are two collinear vectors, then which of the following are incorrect:
→
A. b = λ→
a , for some scalar 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

→
If →
a and b are two collinear vectors, then they are parallel.
Therefore, we have:
→
b = λ→
a (For some scalar λ)
→
If λ = ± 1, then →
a = ±b
→
If →
a = a 1î + a 2ĵ + a 3k̂ and b = b 1î + b 2ĵ + b 3k̂, then
→
b = λ→
a.

( )
⇒ b 1î + b 2ĵ + b 3k̂ = λ a 1î + a 2ĵ + a 3k̂

⇒ b î + b ĵ + b k̂ = (λa )î + (λa )ĵ + (λa )k̂
1 2 3 1 2 3

⇒ b 1 = λa 1, b 2 = λa 2, b 3 = λa 3
b1 b2 b3
⇒ a = a = a =λ
1 2 3
→
Thus, the respective components of →
a and b are proportional.
→
Thus, the respective components of →
a and b are proportional.
→
However, vectors →
a and b can have different directions.
Hence, the statement given in D is incorrect.
The correct answer is D.

Page : 441 , Block Name : Exercise 10.2

Q1
→
a and b with magnitudes √3 and 2, respectively
Find the angle between two vectors →
→
a ⋅ b = √6
having →

Page 9

→ →
a | = √3, | b | = 2 and, →
|→ a ⋅ b = √6
¯ ¯
→
Now, we know that a ⋅ b = | →
a | | b | cosθ
∴ √6 = √3 × 2 × cosθ

√6
⇒ cosθ =
√3 × 2
1
⇒ cosθ =
√2
π
⇒θ= 4
→ π
Hence , the angle between the given vectors →
a and b is 4

Page : 447 , Block Name : Exercise 10.3

Q2 Find the angle between the given vector î − 2ĵ + 3k̂ and 3î − 2ĵ + k̂

→
The qiven vectors are â = î − 2ĵ + 3k̂ and b = 3î − 2ĵ + k̂
|→
a| = √1 2 + ( − 2) 2 + 3 2 = √1 + 4 + 9 = √14
→
|b| = √3 2 + ( − 2) 2 + 1 2 = √9 + 4 + 1 = √14
→
a ⋅ b = (î − 2ĵ + 3k̂)(3î − 2ĵ + k̂)
Now, →
=3+4+3
= 10
\begin{array}{l}{\text { Also, we know that } \vec{a} \cdot \vec{b}}=|\vec{a}||\vec{b}| \cos \theta} \\ {\therefore 10=\sqrt{14} \sqrt{14} \cos \theta} \\ {\Rightarrow \cos \theta=\frac{1

Page : 447 , Block Name : Exercise 10.3

Q3 Find the projection of the vector î − ĵ on the vector î + ĵ

¯
→
Let a = î − ĵ and b = î + ĵ
→
Now, projection of vector →
a on b is given by
1 → 1 1
→ (→
a b) = {1.1 + ( − 1)(1)} = (1 − 1) = 0
|b| √1 + 1 √2
→
Hence the projection of vector →
a on b is 0.

Page : 447 , Block Name : Exercise 10.3

Q4 Find the projection of the vector î + 3ĵ + 7k̂ on the vector 7î − ȷ̂ + 8k̂

¯
Let a = î + 3ĵ + 7k̂ and b̂ = 7î − ĵ + 8k̂
→
Now, projection of vector →
a on b is given by
→
Now, projection of vector →
aon b is given by
1 → 1 7 − 3 + 56 60
→ a ⋅ b) =
(→ {1(7) + 3( − 1) + 7(8)} = =
|b|
√ 72 + ( − 1 )2 + 82 √49 + 1 + 64 √114

Page : 447 , Block Name : Exercise 10.3

Q5 Show that each of the given three vectors is a unit vector:
1 1 1
7
(2î + 3ĵ + 6k̂) ⋅ 7 (3î − 6ĵ + 2k̂), 7 (6î + 2ĵ − 3k̂). Also, show that they are mutually perpendicular to each other.

1 2 3 6
Let →
a = 7 (2î + 3ĵ + 6k̂) = 7 î + 7 ĵ + 7 k̂
→ 1 3 6 2
b = 7 (3î − 6ĵ + 2k̂) = 7 î − 7 ĵ + 7 k̂
¯ 1 6 2 3
c = 7 (6î + 2ĵ − 3k̂) = 7 î + 7 ĵ − 7 k̂

Page 10

|→
a| =
√( ) ( ) ( ) √
2
7
2
+
3
7
2
+
6
7
2
=
4
49
9
+ 49 + 49 = 1
36

→
|b| =
√( ) ( ) ( ) √
3
7
2
+
6
− 7
2
+
2
7
2
=
9
49
36
+ 49 + 49 = 1
9

|→
c| =
√( ) ( ) ( ) √
6
7
2
+
2
7
2
+ − 7
3 2
=
36
49
4
+ 49 + 49 = 1
9

Thus, each of the given three vectors is a unit vector.
→
a⋅b = 7 × 7 + 7 ×
→
2 3 3
( ) −6
7
6 2 6
+ 7 × 7 = 49 − 49 + 49 = 0
18 12

→
b⋅→
c = 7 × 7 +
3 6
( ) −6
7 ( ) 2
× 7 + 7 ×
2 −3
7
18 12
= 49 − 49 − 49 = 0
6

c ⋅→
→
6
a = 7 × 7 + 7 × 7 +
2 2
( ) 3 −3
7
6 12
× 7 = 49 + 49 − 49 = 0
6 18

Hence, the given three vectors are mutually perpendicular to each other.

Page : 447 , Block Name : Exercise 10.3

Q6 Find
→ → → →
|→ a + b) ⋅ (→
a | and | b | , if (→ a − b) = 8 and | →
a | = 8| b |

→ →
a ⋅ b) ⋅ (→
(→ a − b) = 8
→ → → →
⇒→
a⋅→ ab + b ⋅ →
a−→ a−b⋅b =8
→
a|2− |b|2 = 8
⇒ |→
→ →
⇒ (8 | b | ) 2 − | b | 2 = 8
→ →
⇒ 64 | b | 2 − | b | 2 = 8
→
⇒ 63 | b | 2 = 8
→ 8
⇒ | b | 2 = 63
8

√
→
⇒ |b| = 63

2√ 2
→
⇒ |b| =
3√ 7

8 × 2√ 2 16√2
→
|→
a | = 8| b | = =
3√ 7 3√ 7

Page : 448 , Block Name : Exercise 10.3

→ →
Q7 Evaluate the product (3→
a − 5 b) ⋅ (2→
a + 7 b)

→ →
a − 5 b) ⋅ (2→
(3→ a + 7 b)
→ → → →
a ⋅ 2→
= 3→ a ⋅ 7 b − 5 b ⋅ 2→
a + 3→ a − 5b ⋅ 7b
→ → → →
a⋅→
= 6→ a ⋅ b − 10→
a + 21→ a ⋅ b − 35 b ⋅ b

a | 2 + | 1→
= 6|→ a ⋅ b − 35 | b
→ →
|2
Page : 448 , Block Name : Exercise 10.3

→ 1
a and b , having the same magnitude and such that the angle between them is 60 ∘ and their scalar product is 2
Q8 Find the magnitude of two vectors →
.

Page 11

→
Let θ be the angle between the vectors →
a and b .
1
a ⋅ b = 2 , and θ = 60 ∘ .
→ →
|→
a | = |b |, →
¯
→ →
We know that a ⋅ b = | →
a | | b | cosθ
¯
→ →
we know that a ⋅ b = | →
a | | b | cosθ
1
∴ 2 = |→ a | cos60 ∘
a | |→
1 1
a|2× 2
⇒ 2 = |→
1 2
⇒ |a| = 1
→
⇒ |→
a| = |b| = 1

Page : 448 , Block Name : Exercise 10.3

Q9
Find | →
x | , if for a unit vector →
a, (→ a) ⋅ (→
x −→ x +→
a) = 12

(→ a) ⋅ (→
x −→ x +→
a) = 12
¯
⇒→
x ⋅→ x ⋅→
x +→ a⋅→
a−→ x −a⋅→
a = 12
x |2 − |→
⇒ |→ a | 2 = 12
x | 2 − 1 = 12
⇒ |→ [|→
a | = 1 as →
a is a unit vector ]
x | 2 = 13
⇒ |→
x | = √13
∴ |→

Page : 448 , Block Name : Exercise 10.3

Q10
→ →
If →
a = 2î + 2ĵ + 3k̂, b = − î + 2ĵ + k̂ and →
c = 3î + ĵ are such that →
a + λ b is perpendicular to →
c,
then find the value of λ .

¯
→
The given vectors are a = 2î + 2ĵ + 3k̂, b = − î + 2ĵ + k̂, and →
c = 3î + ĵ
Now,
→ →
a + λ b = (2î + 2ĵ + 3k̂) + λ( − î + 2ĵ + k̂) = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂
¯
→
If (→
a + λ b) is perpendicular to c, then
→
a + λ b) ⋅ →
(→ c =0
⇒ [(2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂] ⋅ (3î + ĵ) = 0
⇒ (2 − λ)3 + (2 + 2λ)1 + (3 + λ)0 = 0
⇒ 6 − 3λ + 2 + 2λ = 0
⇒ −λ+8=0
⇒λ=8
Hence, the required value of λ is 8 .

Page : 448 , Block Name : Exercise 10.3

→ → → → →
Q11 Show that | → a is perpendicular to 0 | →
a |b + |b |→ a, for any two non-zero vectors →
a |b − |b |→ a and b

→ → → →
(|→ a) ⋅ ( | →
a |b + |b |→ a |b − |b |→
a)
→ →
a | 2b ⋅ b − | →
= |→ a| |b|b ⋅→
→ →
a ⋅ b − |b →
a + |b | |→ a⋅→
a
→ → →
|2
→ →
a |2|b |2 − |b |2|→
= |→ a|2
=0

Page : 448 , Block Name : Exercise 10.3

Q12
¯
→ →
Ifa ⋅ → a ⋅ b = 0, then what can be concluded about the vector b
a = 0 and →

Page 12

→
a⋅→
It is given that → a⋅b =0
a = 0 and →
Now,
a⋅→
→
a |2 = 0 ⇒ |→
a = 0 ⇒ |→ a| = 0
∴→
a is a zero vector.
→ →
a ⋅ b = 0 can be any vector.
Hence, vector b satisfying →

Page : 448 , Block Name : Exercise 10.3

Q13
¯
→ →
If →
a, b, →
c are unit vectors such that →
a+b+→
c = 0, find the value of
→ →
a⋅b+b⋅→
→
c ⋅→
c +→ c ⋅→
a

→ → → →
(|→ a) ⋅ ( | →
a |b + |b |→ a |b − |b |→
a)
→
a | 2b ⋅ b − | →
= |→
→
a| |b|b ⋅→
→ →
a ⋅ b − |b →
a + |b | |→ a⋅→
a
→ → →
|2
→ →
a |2|b |2 − |b |2|→
= |→ a|2
=0

Page : 448 , Block Name : Exercise 10.3

Q14
→ → → →
If either vector → a ⋅ b = 0 . But the converse need not be true. Justify
a = 0 or b = 0 , then →
your answer with an example.

→
Consider →
a = 2î + 4ĵ + 3k̂ and b = 3î + 3ĵ − 6k̂
Then,
→
a ⋅ b = 2.3 + 4.3 + 3( − 6) = 6 + 12 − 18 = 0
→

we now observe that:
|→
a| = √2 2 + 4 2 + 3 2 = √29
∴ a = √3 2 + 3 2 + ( − 6) 2 = √54
→

→ →
∴b =3
Hence, the converse of the given statement need not be true.

Page : 448 , Block Name : Exercise 10.3

Q15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (—1, 0, 0), (0, 1, 2), respectively, then nd DABC. [ABC is the angle between the vectors
→ →
BA and BC ]

→
BA = {1 − ( − 1)}î + (2 − 0)ĵ + (3 − 0)k̂ = 2î + 2ĵ + 3k̂
¯
BC = {0 − ( − 1)}î + (1 − 0)ĵ + (2 − 0)k̂ = î + ĵ + 2k̂
→ →
∴ BA ⋅ BC = (2î + 2ĵ + 3k̂) ⋅ (î + ĵ + 2k̇) = 2 × 1 + 2 × 1 + 3 × 2 = 2 + 2 + 6 = 10
Now, it is known that:
→ →
BA ⋅ BC = | BA‖BC | cos(∠ABC)
∴ 10 = √17 × √6cos(∠ABC)
10
⇒ cos(∠ABC) =
√17 × √6

⇒ ∠ABC = cos − 1
( )
10

√102

Page : 448 , Block Name : Exercise 10.3

Page 13

Q16 Show that the points A (1, 2, 7), 3 (2, 6, 3) and C (3, 10, —1) are collinear.

→
| AB = (2 − 1)î + (6 − 2)ĵ + (3 − 7)k̂ = î + 4ĵ − 4k̂
→
∴ AC = (3 − 2)î + (10 − 6)ĵ + ( − 1 − 3)k̂ = î + 4ĵ − 4k̂
→
AC = (3 − 1)î + (10 − 2)ĵ + ( − 1 − 3)k̂ = 2î + 8ĵ − 8k̂
→
| AB | = √1 2 + 4 2 + ( − 4) 2 = √1 + 16 + 16 = √33
→
| AC | = √1 2 + 4 2 + 8 2 = √4 + 16 + 16 = √33
→ → →
∴ | AC | = | AB | + | BC |
→ → →
∴ | AC | = | AB | + | BC |
Hence, the given points A, B , and C are collinear.

Page : 448 , Block Name : Exercise 10.3

Q17 Show that the vector 2î − ĵ + k̂, î − 3ĵ − 5k̂ and 3î − 4ĵ − 4k̂ from the vertical of a right angled triangle.

Answer. Let vectors 2î − ĵ + k̂, î − 3ĵ − 5k̂ and 3î − 4ĵ − 4k̂ be position vector of point A,B,C respectively.
→
∴ AB = (1 − 2)î + ( − 3 + 1)ĵ + ( − 5 − 1)k̂ = − î − 2ĵ − 6k̂
¯
BC = (3 − 1)î + ( − 4 + 3)ĵ + ( − 4 + 5)k̂ = 2î − ĵ + k̂
¯
AC = (2 − 3)î + ( − 1 + 4)ĵ + (1 + 4)k̂ = − î + 3ĵ + 5k̂
→
√
| AB | = ( − 1) 2 + ( − 2) 2 + ( − 6) 2 = √1 + 4 + 36 = √41

| BC | = √2 2 + ( − 1) 2 + l 2 = √4 + 1 + 1 = √6

→
| AC | = √( − 1) 2 + 3 2 + 5 2 = √1 + 9 + 25 = √35
→
∴ | BC | 2 + | AC | 2 = 6 + 35 = 41 = | AB | 2

Page : 448 , Block Name : Exercise 10.3

Q18
→

a is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λ a is unit vector if
→

(A) = 1
(B) λ = − 1
(C) a = | λ |
1
(D) a =
|λ|

Vector λ→
a is a unit vector if
Now,
| λ→
a| = 1
⇒ |λ| |→
a| = 1
1
⇒ |→
a | = |λ|
1
⇒ a = |λ|
1
a = |λ| 1
a = |λ|
→
Hence, vector λ a is a unit vector if

Page : 448 , Block Name : Exercise 10.3

Q1
→ →
Find | →
a × b | , if →
a = î − 7ĵ + 7k̂ and b = 3î − 2ĵ + 2k̂

Page 14

→ →
a = î − 7ĵ + 7k̂ and b = 3î − 2ĵ + 2k̂

→ →
î
a×b = 1
| |
3
ĵ
−7 7
−2 2
k̂

= î( − 14 + 14) − ĵ(2 − 21) + k̂( − 2 + 21) = 19 ĵ + 19k̂
→
∴ |→
a × b| = √(19) 2 + (19) 2 = √2 × (19) 2 = 19√2
Page : 454 , Block Name : Exercise 10.4

Q2
→ →
Find a unit vector perpendicular to each of the vector →
a + b and →
a − b , where
→ →
a = 3î + 2ĵ + 2k̂ and b = î + 2ĵ − 2k̂

We have,
→ →
a = 3î + 2ĵ + 2k̂ and b = î + 2ĵ − 2k̂
→ →
∴→
a + b = 4î + 4ĵ, →
a − b = 2î + 4k̂

(→
a + b) × (→
→
a − b) = 4

→
→

→
| |î

2
ĵ
4
0
k̂
0 = î(16) − ĵ(16) + k̂( − 8) = 16î − 16ĵ − 8k̂
4

∴ (→
a + b) × (→
a − b) = √16 2 + ( − 16) 2 + ( − 8) 2
= √2 2 × 8 2 + 2 2 × 8 2 + 8 2

= 8√2 2 + 2 2 + 1 = 8√9 = 8 × 3 = 24
→ →
Hence, the unit vector perpendicular to each of the vectors →
a + b and →
a − b is given by the
relation,
→ →
( a→ + b ) × ( a→ − b ) 16î − 16ĵ − 8k̂
= ± → → → → = ± 24
(a+b) × (a−b)
2î − 2ĵ − k̂ 2 2 1
= ± 3
= ± 3 î ∓ 3 ĵ ∓ 3 k̂

Page : 454 , Block Name : Exercise 10.4

Q3
π
a makes an angles 3 with i , 4 with ȷ̂ and an acute angle θ with k̂ , then
If a unit vector →
find θ and hence, the compounds of →
a.

Let unit vector →
a have a 1, a 2, a 3 components. ( )
¯
◻a = a 1î + a 2ĵ + a 3k̂
since →
a is a unit vector, | →
a| = 1
π a1
cos 3 = | a | →

1
⇒ 2 = a1
π a2
cos 4 = | a | →

1
⇒ = a2
√2
a3
Also, cosθ = | a | →

Page 15

Now
|a| = 1
2 2 2
⇒
√a + a + a = 1
1 2 3

⇒
() ( )
1
2
2
+
√2
1 2
+ cos 2θ = 1

3
⇒ 4 + cos 2θ = 1
3
⇒ 4 + cos 2θ = 1
3
⇒ 4 + cos 2θ = 1
3 1
⇒ cos 2θ = 1 − 4 = 4
1 π
⇒ cosθ = 2 ⇒ θ = 3
π 1
∴ a 3 = cos 3 = 2

( )
π ¯ 1 1 1
θ = 3 and the components of a are 2 , , 2
√2

Page : 454 , Block Name : Exercise 10.4

Q4
Show that
→ → →
(→
a − b) × (→
a + b) = 2(→
a × b)

→
(→
a − b) × (→
a + b̂)
→ → →
= (→
a − b) × →
a + (→
a − b) × b
→ → → →
=→
a×→
a−b×→
a+→
a×b−b×b
→ → → →
= 0+→
a×b+→
a×b−0
→
= 2→
a×b

Page : 454 , Block Name : Exercise 10.4

→
Q5 Find λ and μ if (2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0

→
(2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0

ĵ
⇒ 2
| | 1
k̂
6
λ
27 = 0î + 0ĵ + 0k̂
μ

⇒ î(6μ − 27λ) − ĵ(2μ − 27) + k̂(2λ − 6) = 0î + 0ĵ + 0k̂
On comparing the corresponding components, we have:
6μ − 27λ = 0
2μ − 27 = 0
2λ − 6 = 0
Now,
27
2λ − 6 = 0 ⇒ λ = 2
27
2μ − 27 = 0 ⇒ μ = 2

Hence,

Page : 454 , Block Name : Exercise 10.4

¯
→ → → →
Q6 Given that →
a ⋅ b = 0 and →
a × b = 0 . What can you conclude about the vectors a and b

Page 16

→
a⋅b =0
→

Then,
→ → →
(i) Either | →
a | = 0 or | b | = 0 a ⊥ b( in case →
or → a and b are non-zero )
→ →
a×b =0
¯
→ →
(ii) Either | a | = 0 or | = 0, or →
a‖ b( in case →
a and b are non-zero )
→
But, →a and b cannot be perpendicular and parallel simultaneously.
¯
Hence, | →
a | = 0 or | b | = 0

Page : 454 , Block Name : Exercise 10.4

Q7
→
Let the vectors →
a, b, →
c given as a 1î + a 2ĵ + a 3k̂, b 1î + b 2ĵ + b 3k̂, c 1î + c 2ĵ + c 3k̂ . Then show
→ →
that = →
a × (b + →
c) = →
a×b+→
a×→
c

we have,
→ →
a = a 1î + a 2ĵ + a 3k̂, b = b 1î + b 2ĵ + b 3k̂, →
c = c 1î + c 2ĵ + c 3k̂
→
(b + → ( ) (
c ) = b 1 + c 1 î + b 2 + c 2 ĵ + b 3 + c 3 k̂ ) ( )

| |
î ĵ k̂
→
Now, →
a × (b + →
c) a1 a2 a3
b1 + c1 b2 + c2 b3 + c3

[ ( ) (
= i a2 b3 + c3 − a3 b2 + c2 )] − ĵ [a1 (b3 + c3 ) − a3 (b1 + c1 )] + k̂ [a1 (b2 + c2 ) − a2 (b1 + c1 )]
= î [a 2b 3 + a 2c 3 − a 3b 2 − a 3c 2 ] + ĵ [ − a 1b 3 − a 1c 3 + a 3c 1 ] + k̂ [a 1b 2 + a 1c 2 − a 2b 1 − a 2c 1 ]…(1)

| |
î ĵ k̂
→
a × b = a1
→ a2 a3
b1 b2 b3

[ ] [
= î a 2b 3 − a 3b 2 + ĵ b 1a 3 − a 1b 3 + k̂ a 1b 2 − a 2b 1 ] [ ]
(→
→
a × b) + (→
a×→ [
c ) = î a 2b 3 + a 2c 3 − a 3b 2 − a 3c 2 + ĵ ba 3 + a 3c 1 − a 1b 3 − a 1c 3 ] [ ]
[
+ k̂ a 1b 2 + a 1c 2 − a 2b 1 − a 2c 1 ]
Now, from (1) and (4), we have:
→ → →
a × (b + →
c) = →
a×b+→
a×→
c
Hence, the given result is proved.

Page : 454 , Block Name : Exercise 10.4

Q8
→ → → → →
If either →
a = 0 or b = 0, then →
a × b = 0 . Is the converse true? Justify your answer with an
example.

¯ ¯
→
Take any parallel non-zero vectors so that a × b = 0
→
Let →
a = 2î + 3ĵ + 4k̂, b = 4î + 6ĵ + 8k̂
Then,

→ →
î
a×b = 2
| |
4
ĵ
3
6
k̂
4 = î(24 − 24) − ĵ(16 − 16) + k̂(12 − 12) = 0î + 0ȷ̂ + 0k̂ = 0
8
→

Page 17

It can now be observed that:
|→
a| = √2 2 + 3 2 + 4 2 = √29
→
∴→
a ≠0
→
|b| = √4 2 + 6 2 + 8 2 = √116
→ →
∴b ≠0
Hence, the converse of the given statement need not be true.

Page : 454 , Block Name : Exercise 10.4

Q9
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and
C (1, 5, 5) .

The vertices of triangle ABC are given as A(1, 1, 2), B(2, 3, 5), and
C(1, 5, 5).
→ →
The adjacent sides AB and BC of ΔABC are given as:
→
AB = (2 − 1)î + (3 − 1)ĵ + (5 − 2)k̂ = î + 2ĵ + 3k̂
→
BC = (1 − 2)î + (5 − 3)ĵ + (5 − 5)k̂ = − î + 2ĵ

→ →
AB × BC =

→ →
| |
−1
1
î
2
2
ĵ k̂
3 = î( − 6) − ĵ(3) + k̂(2 + 2) = − 6î − 3ĵ + 4k̂
0

∴ | AB × BC | = √( − 6) 2 + ( − 3) 2 + 4 2 = √36 + 9 + 16 = √61
Hence, the area of ΔABC square units.

Page : 454 , Block Name : Exercise 10.4

Q10
Find the area of the parallelogram whose adjacent sides are determined by the vector
→ →
a = î − ĵ + 3k̂ and b = 2î − 7ĵ + k̂

¯ ¯
→ →
The area of the parallelogram whose adjacent sides are a and b is | a × b |
Adjacent sides are given as:
→ →
a = î − ĵ + 3k̂ and b = 2î − 7ĵ + k̂

∴→
a×b = 1

→
→

| |
2
î ĵ
−1
−7
k̂
3 = î( − 1 + 21) − ĵ(1 − 6) + k̂( − 7 + 2) = 20î + 5ĵ − 5k̂
1
→
|a × b| = √ 20 2 + 5 2 + 5 2 = √400 + 25 + 25 = 15√2
15√2

Page : 455 , Block Name : Exercise 10.4

Q11
→ →
√2 →
Let the vectors →
a and b be such that | →
a | = 3 and | b | = 3 , then →
a × b is a unit vector, if
→
the anale between →
a and b is
π π π π
(A) 6 (B) 4 (C) 3 (D) 2

Answer. It is given that
→
√2
|→
a | = 3 and | b | = 3 → → →
a × b = |→
a | | b | sinθn̂
we know that

Page 18

→ →
Now, →
a × b is a unit vector if | →
a × b| = 1
→
|→
a × b| = 1
→
⇒ |→
a | | b | sinθn̂ | = 1
→
⇒ |→
a | | b | sinθ | = 1
√2
⇒ 3 × 3 × sinθ = 1
1
⇒ sinθ =
√2
π
⇒θ= 4
→ → π
Hence, →
a × b is a unit vector if the angle between →
a and b is 4

The correct answer is B .

Page : 455 , Block Name : Exercise 10.4

Q12
Area of a rectangle having vertices A, B, C, and D with position vectors
1 1 1 respectively
− î + 2 ĵ + 4k̂, î + 2 ĵ + 4k̂, î − 2 ĵ + 4k̂
1
(A) 2 (B)1

(C) 2 (D) 4

→ 1 → 1 → 1 → 1
OA = − î + 2 ĵ + 4k̂, OB = î + 2 ĵ + 4k̂, OC = î − 2 ĵ + 4k̂, OD = − î − 2 ĵ + 4k̂

( )
→ 1 1
AB = (1 + 1)î + 2
− 2 ĵ + (4 − 4)k̂ = 2î

( )
→ 1 1
BC = (1 − 1)î + − 2 − 2 ĵ + (4 − 4)k̂ = − ĵ

→ → î
∴ AB × BC = 2

→ →
| |
0 −1
ĵ
0
k̂
0 = k̂( − 2) = − 2k̂
0

| AB × AC | = √( − 2) 2 = 2
Now, it is known that the area of a parallelogram whose adjacent sides are
→ → →
a and b is | →
a × b|
→ →
Hence, the area of the given rectangle is | AB × BC | = 2 square units.
The correct answer is C .

Page : 455 , Block Name : Exercise 10.4

Q1 Write down a unit vector in KY-plane, making an angle of 30 ∘ with the positive direction of x-axis.

If →r is a unit vector in the XY-plane, then →r = cosθî + sinθĵ
Here, θ is the angle made by the unit vector with the positive direction of the x -axis.
Therefore, for θ = 30 ∘ :
√3 1
r = cos30 ∘ î + sin30 ∘ ĵ = 2 î + 2 ĵ
→ 1
î + ĵ 2
Hence, the required unit vector is 2

Page : 458 , Block Name : Miscellaneous Exercise

Q2
Find the scalar components and magnitude of the vector joining the points

( )
P x 1, y 1, z 1 and Q x 2, y 2, z 2 ( )

Page 19

The vector joining the points x 1, y 1, z 1 and Q x 2, y 2, z 2 ( )
can be obtained by ( )
¯
PQ = Position vector of Q − Position vector of P

( ) (
= x 2 − x 1 î + y 2 − y 1 ĵ + z 2 − z 1 k̂ ) ( )
→
| PQ | = √ (x − x ) + (y − y ) + (z − z )
2 1
2
2 1
2
2 1
2

Hence, the scalar components and the magnitude of the vector joining the given points

are respectively {(x2 − x1 ), (y2 − y1 ), (z2 − z1 )} and √ (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 ) 2
Page : 458 , Block Name : Miscellaneous Exercise

Q3 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.

Answer. Let O and B be the initial and nal positions of the girl respectively.
Then, the girl's position can be shown as:

→
OA = − 4î
→ → →
AB = î | AB | cos60 ∘ + ĵ | AB | sin60 ∘
1 √3
= î3 × 2 + ĵ3 × 2

3 3√ 3
= 2 î + 2 ĵ
→ → →
OB = OA + AB

( )
3 3√3
= ( − 4î) + î + ĵ
2 2

( ) √
3 3 3
= −4+ î + ĵ
2 2

( ) √
−8 + 3 3 3
= î + ĵ
2 2

−5 3√3
= î +
3̂
2 2
Hence, the girl's displacement from her initial point of departure is
−5 3√ 3
2
î + 2 ĵ

Page : 458 , Block Name : Miscellaneous Exercise

→ →
Q4 If →
a = b+→
c , then is it true that = | b | + | →
c | ? Justify your answer.

→ → → →
ΔABC, let CB = → c (as shown in the following gure)
a, CA = b, and AB = →

→
Now, by the triangle law of vector addition we have →
a = b+→
c

Page 20

¯
→
It is clearly known that | a | , | b | , and | →
c|
Also, it is known that the sum of the lengths of any two sides of a triangle is greater
¯ ¯
→
than the third side. ∴ | a | < | b | + | c |
→
Hence it is not true that | →
a | = |b | + |→ c|

Page : 458 , Block Name : Miscellaneous Exercise

Q5 Find the value of x x for which x(î + ĵ + k̂) is a unit vector .

x(î + ĵ + k̂) is a unit vector if | x(î + ĵ + k̂) | = 1
Now,
| x(î + ĵ + k̂) | = 1
⇒ √x 2 + x 2 + x 2 = 1
⇒ √3x 2 = 1

⇒ √3x = 1
1
⇒x= ±
√3

Page : 458 , Block Name : Miscellaneous Exercise

Q6
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
→ →
a = 2î + 3ĵ − k̂ and b = î − 2ĵ + k̂

→ →
a = 2î + 3ĵ − k̂ and b = î − 2ĵ + k̂
→
Let →
c be the resultant of →
a and b
Then,
→ →
c =→
a + b = (2 + 1)î + (3 − 2)ĵ + ( − 1 + 1)k̂ = 3î + ĵ
→ →
c =→
a + b = (2 + 1)î + (3 − 2)ĵ + ( − 1 + 1)k̂ = 3î + ĵ
∴ |→
c| = √3 2 + 1 2 = √9 + 1 = √10
→
c ( 3iˆ + jˆ )
∴ ĉ = | c | = →
√10
→
Hence, the vector of magnitude 5 units and parallel to the resultant of vectors →
a and b is
1 3√10iˆ √10
± 5 ⋅ ĉ = ± 5 ⋅ (3î + ĵ) = ± 2
± 2 ĵ
√10

Page : 458 , Block Name : Miscellaneous Exercise

Q7
→
If →
a = î + ĵ + k̂, b = 2î − ĵ + 3k̂ and →
c = î − 2ĵ + k̂, find a unit vector parallel to the
→
vector 2→
a − b + 3→
c

we have,
→ →
a = î + ĵ + k̂, b = 2î + 3k̂ and →
c = î − 2ĵ + k̂
→
2→
a − b + 3→
c = 2(î + ĵ + k̂) − (2î − ĵ + 3k̂) + 3(î − 2ĵ + k̂)
= 2î + 2ĵ + 2k̂ − 2î + ĵ − 3k̂ + 3î − 6ĵ + 3k̂
= 3î − 3ĵ + 2k̂
→ 2 2 2 =
→
| 2 a − b + 3c | = √3 + ( − 3) + 2 √9 + 9 + 4 = √22
→
Hence, the unit vector along 2→
a − b + 3→
c is
3iˆ − 3jˆ + 2k̂
→ → →
2a − b + 3 c 3 3 2
→ = = î − ĵ + k̂
| 2a→ − b + 3 →c √22 √22 √22 √22

Page : 458 , Block Name : Miscellaneous Exercise

Q8 Show that the points A (1, -2, -8), B (5, 0, -2) and C (11, 3, 7) are collinear, and nd the ratio in which B divides AC.

Page 21

The given points are A(1, − 2, − 8), B(5, 0, − 2), and C(11, 3, 7) .
→
∴ AB = (5 − 1)î + (0 + 2)ĵ + ( − 2 + 8)k̂ = 4î + 2ĵ + 6k̂
→
BC = (11 − 5)î + (3 − 0)ĵ + (7 + 2)k̂ = 6î + 3ĵ + 9k̂
→
AC = (11 − 1)î + (3 + 2)ĵ + (7 + 8)k̂ = 10î + 5ĵ + 15k̂
→
| AB | = √4 2 + 2 2 + 6 2 = √16 + 4 + 36 = √56 = 2√14
→
| BC | = √6 2 + 3 2 + 9 2 = √36 + 9 + 81 = √126 = 3√14
→
| AC | = √10 2 + 5 2 + 15 2 = √100 + 25 + 225 = √350 = 5√14
Thus, the given points A, B, and C are collinear.
Now, let point B divide AC in the ratio λ : 1 . Then, we have:
→ →
→ λOC + OA
OB = ( λ + 1 )
→ →
→ λOC + OA
OB = ( λ + 1 )
λ ( 11î + 3ĵ + 7k̂ ) + ( î − 2ĵ − 8k̂ )
⇒ 5î − 2k̂ = λ+1
⇒ (λ + 1)(5î − 2k̂) = 11λî + 3λĵ + 7λk̂ + î − 2ĵ − 8k̂
⇒ 5(λ + 1)î − 2(λ + 1)k̂ = (11λ + 1)î + (3λ − 2)ĵ + (7λ − 8)k̂
On equating the corresponding components, we get:
5(λ + 1) = 11λ + 1
⇒ 5λ + 5 = 11λ + 1
⇒ 6λ = 4
4 2
⇒λ= 6 = 3

Hence, point B divides AC in the ratio 2 : 3 .

Page : 458 , Block Name : Miscellaneous Exercise

→ →
Q9 Find the position vector of a point R which divides the line joining two points P and Q Whose position vector are (2→ a − 3 b) externally
a + b) and (→
in the ratio 1:2 Also, show that P is the mid point of the line segment RQ.

¯ →
→ →
It is given that OP = 2→
a + b, OQ = →
a − 3b
It is given that point R divides a line segment joining two points P and Q externally in
the ratio 1 : 2. Then, on using the section formula, we get:
→ → →
2 ( 2a + b ) − ( a − 3b ) → →
4a + 2b − a + 3b → → → →

→
OR = 2−1
= 1
= 3→
a + 5b
→
Therefore, the position vector of point R is 3→
a + 5b
→ →
OQ + OR
Position vector of the mid-point of RQ = 2
→ →
( a→ − 3b ) + ( 3a→ + 5b )
= 2
→
2a→ + b
= OP

Hence, P is the mid-point of the line segment RQ.

Page : 458 , Block Name : Miscellaneous Exercise

Q10
The two adjacent sides of a parallelogram are 2î − 4ĵ + 5k̂ and î − 2ĵ − 3k̂ .
Find the unit vector parallel to its diagonal. Also, find its area.

→
Adjacent sides of a parallelogram are given as: →
a = 2î − 4ĵ + 5k̂ and b = î − 2ĵ − 3k̂
→
Then, the diagonal of a parallelogram is given by →
a+b

Page 22

→ →
a + b = (2 + 1)î + ( − 4 − 2)ĵ + (5 − 3)k̂ = 3î − 6ĵ + 2k̂
Thus, the unit vector parallel to the diagonal is
→
a→ + b 3î − 6ĵ + 2k̂ 3î − 6ĵ + 2k̂ 3î − 6ĵ + 2k̂ 3 6 2
→ = = = 7
= 7 î − 7 ĵ + 7 k̂
√ 32 + ( − 6 )2 + 22 √9 + 36 + 4
→
|a+b|
→
∴ Area of parallelogram ABCD = | →
a × b|

→ →
î
a×b= 2
| 1
ĵ
−4
−2
k̂
5
−3
|
= î(12 + 10) − ĵ( − 6 − 5) + k̂( − 4 + 4)
= 22î + 11ĵ
= 11(2î + ĵ)
→
a × b | = 11 2 2 + 1 2 = 11√5
∴ |→ √
Hence, the area of the parallelogram is 11√5 square units.

Page : 458 , Block Name : Miscellaneous Exercise

1 1 1
Q11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are , ,
√3 √3 √3

Let a vector be equally inclined to axes ox, oy, and oz at angle a .
Then, the direction cosines of the vector are cos a, cosa, a and cos a .
Now,
cos 2α + cos 2α + cos 2α = 1
⇒ 3cos 2α = 1
1
⇒ cosα =
√3
Hence, the direction cosines of the vector which are equally inclined to the axes
1 1 1
, ,
√3 √3 √3

Page : 458 , Block Name : Miscellaneous Exercise

¯
→
Q12 Let →
a = î + 4ĵ + 2k̂, b = 3î − 2ĵ + 7k̂ and →
c = 2î − ĵ + 4k̂ . Find a vector d
→ →
Perpendicular to both →
a and b, and →
c d = 15

¯
Let d = d 1î + d 2ĵ + d 3k̂
¯
→
since d is perpendicular to both →
a and b , we have:
→
d⋅→
a =0
⇒ d 1 + 4d 2 + 2d 3 = 0
And
→ →
d⋅b =0
⇒ 3d 1 − 2d 2 + 7d 3 = 0
Also, it is given that:
→
c ⋅ d = 15
→

⇒ 2d 1 − d 2 + 4d 3 = 15
On solving (i), (ii), and (iii), we get:
160 5 70
d 1 = 3 , d 2 = − 3 and d 3 = − 3
→ 160 5 70 1
∴ d = 3 î − 3 ĵ − 3 k̇ = 3 (160î − 5ĵ − 70k̂)
1 ˆ ˆ
Hence the required vector is 3 3 ( 160i − 5j − 70k̂ )

Page : 458 , Block Name : Miscellaneous Exercise

Q13

Page 23

The scalar product of the vector î + ĵ + k̂ with a unit vector along the sum of vectors
2î + 4ĵ − 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ

(2î + 4ĵ − 5k̂) + (λî + 2ĵ + 3k̂)
= (2 + λ)î + 6ĵ − 2k̂
Therefore, unit vector along
( 2 + λ ) iˆ + 6jˆ − 2jˆ ( 2 + λ ) î + 6ĵ − 2k̂ ( 2 + λ ) î + 6ĵ − 2k̂
2 2 2
= 2
=
√(2+λ) +6 + ( −2) √4 + 4λ + λ + 36 + 4 √λ2 + 4λ + 44
Scalar product of (t ′ + ĵ + k̂) with this unit vector is 1
( 2 + λ ) î + 6ĵ − 2k̂
⇒ (î + ĵ + k̂) ⋅ =1
√λ2 + 4λ + 44
(2+λ) +6−2
⇒ =1
√λ2 + 4λ + 44
⇒ √λ 2 + 4λ + 44 = λ + 6
⇒ λ 2 + 4λ + 44 = λ + 6
⇒ 8 2 + 4λ + 44 = λ 2 + 12λ + 36
⇒λ=1
Hence, the value of λ is 1 .

Page : 458 , Block Name : Miscellaneous Exercise

Q14
→
If →
a, b, →
c are mutually perpendicular vectors of equal magnitudes, show that the vector
→ → →
a+b+→
c is equally inclined to →
a, b and →
c

→
since →
a, b, and →
c are mutually perpendicular vectors, we have
→ →
a⋅b = b⋅→
→
c ⋅→
c =→ a =0
It is given that:
→
|→
a | = |b | = |→
c|
→ →
Let vector →
a+b+→ c be inclined to →
a, b, and →
c at angles θ 1, θ 2, and θ 3 respectively.
Then, we have:
→ →
( a→ + b + →c ) ⋅ a→ a→ ⋅ a→ + b ⋅ a→ + →c ⋅ a→
cosθ 1 = ¯ = →
| a + b + →c | a→ |
→

| a→ + b + →c | | a→ |
|→a|2
= →
|→
a+b+→c |→
a
|→
a
=
|→
a+b+→
c|
→ → → → → →
( a→ + b + →c ) ⋅ b a→ ⋅ b + b ⋅ b + →c ⋅ b
cosθ 2 = → → = → →
→
|u+b+c| |b| →
|u+b+c| ⋅ |b|
→ →

|b|2
→

= → →
| a→ + b + →c | ⋅ | b |
→
|b
= →
| a→ + b + →c |
→ →
(→ c) ⋅ →
a+b+→ c a⋅→
→
c+b⋅→ c⋅→
c+→ c
cosθ 3 = →
= →
|→
a+b+→
c | |→
c| |→
a+b+→
c |→
c|
¯
|c|2
= →
|→
a+b+→c |→
c|
→
|c
= →
|→a+b+→ c|
→
Now, as | →
a | = |b | = |→
c | , cosθ 1 = cosθ 2 = cosθ 3
∴ θ1 = θ2 = θ3
→ →
Hence, the vector (→
a+b+→
c ) is equally inclined to →
a, b, and →
c

Page : 458 , Block Name : Miscellaneous Exercise

Q15

Page 24

¯
→ → →
a + b) ⋅ (→
Prove that (→ a | 2 + | b | 2, if and only if →
a + b) = | → a, b are perpendicular,
→ → →
given →
a ≠ 0, b ≠ 0

→ → →
a + b) ⋅ (→
(→ a|2+ |b|2
a + b) = | →
→ → → → →
⇔→
a⋅→ a⋅b+b⋅→
a+→ a|2+ |b
a + b ⋅ b = |→
→
⇔ 2→
a⋅b =0
→
⇔→
a⋅b =0
→
∴→
a and b are perpendicular.

Page : 459 , Block Name : Miscellaneous Exercise

→ →
Q16 If θ is the angle between two vectors →
a and b, then →
ab ≥ 0
π π
(A)0 < θ < 2 (B)0 ≤ θ ≤ 2

(C) 0 < θ < π(D)0 ≤ θ ≤ π

→
Let θ be the angle between two vectors →
a and b
→
Then, without loss of generality, →
a and b are non-zero vectors so
→
that | →
a | and | b | are positive
¯
→ →
It is known that a ⋅ b = | →
a | b | cosθ
→
∴→
a⋅b ≥0
⇒ cosθ | cosθ ≥ 0
⇒ cosθ ≥ 0
π
⇒0≤θ≤ 2
→
Hence, →
a b ≥ 0 when
The correct answer is B.

Page : 459 , Block Name : Miscellaneous Exercise

Q17
→ →
Let →
a and b be two unit vectors and θ is the angle between them. Then →
a + b is a unit
vector if
π
(A)θ = 4
π
(B)θ = 3
π
(C)θ = 2
2π
(D)θ = 3

→
Let →
a and b be two unit vectors and θ be the angle between them.
→
Then, | →
a| = |b| = 1
→ →
Now, →
a + b is a unit vector if | →
a + b| = 1
→
|→
a + b| = 1
→
a + b) 2 = 1
⇒ (→
→ →
⇒ (→
a + b) ⋅ (→
a + b) = 1
→ → →→
⇒→
a⋅→ a⋅b+b⋅→
a+→ a + bb = 1
→→ →
⇒ |→a | 2 + 2a b + | b | 2 = 1
→ 2 → →
⇒ | a | + 2→ ab + | b | 2 = 1
→
⇒ 12 + 2 | →
a | | b | cosθ + 1 2 = 1
⇒ 1 + 2.1.1cosθ + 1 = 1
1
⇒ cosθ = − 2
2π
⇒θ= 3

Page 25

→ 2π
Hence, →
a + b is a unit vector if θ = 3

The correct answer is D .

Page : 459 , Block Name : Miscellaneous Exercise

Q18 The value of î ⋅ (ĵ × k̂) + ĵ ⋅ (î × k̂) + k̂. (î × ĵ) is is
(A) 0 (B)-1 (C) 1 (D) 3

î ⋅ ( ĵ × k̂) + ĵ ⋅ ( î × k̂) + k̂ ⋅ ( î × ĵ)
= î ⋅ î + ĵ ⋅ ( − ĵ) + k̂ ⋅ k̂
= 1 − ĵ ⋅ ĵ + 1
=1−1+1
=1

Page : 459 , Block Name : Miscellaneous Exercise

Q19
θ is the angle between any two vectors →
a and b, then | = | →
→
a × b when
→
|
π π
(A)0(B) 4 (C) 2 (D)n

→
Let θ be the angle between two vectors →
a and b
→
Then, without loss of generality, →
a and b are non-zero vectors, so
→
that | →
a | and | b | are positive
→ →
a ⋅ b | = |→
|→ a × b|
⇒ cosθ = sinθ
⇒ tanθ = 1
π
⇒θ= 4
¯
→
Hence, | →
ab | = | →
a × b|
The correct answer is B.

Page : 459 , Block Name : Miscellaneous Exercise

Document Details

Board / Org	NCERT
Exam	Class 12
Type	Solution
Pages	25
Updated	22 Jul 2026

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