# Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 1

### Question 1. Find when

### (i) and

**Solution:**

=

= (1)(4) + (-2)(-4) + (1)(7)

= 4 + 8 + 7

= 19

**(ii)** and

**Solution:**

=

= (0)(2) + (1)(0) + (2)(1)

= 2

**(iii) **and

**Solution:**

=

= (0)(2) + (1)(3) + (-1)(-2)

= 0 + 3 + 2

= 5

### Question 2. For what value of λ are the vector and perpendicular to each other? where:

### (i) and

**Solution:**

and are perpendicular to each other

So

⇒

⇒ λ(4) + (2)(-9) + (1)(2) = 0

⇒ 4λ – 18 + 2 = 0

⇒ 4λ = 16

⇒ λ = 4

### (ii) and

**Solution:**

and are perpendicular to each other

so= 0

⇒

⇒ λ(5) + (2)(-9) + (1)(2) = 0

⇒ 5λ – 18 + 2 = 0

⇒ 5λ = 16

⇒ λ = 16/5

### (iii) and

**Solution:**

and are perpendicular to each other

so = 0

⇒ =0

⇒ (2)(3) + (3)(2) – (4)λ = 0

⇒ 6 + 6 – 4λ = 0

⇒ 4λ = 12

⇒ λ = 3

### (iv) and

**Solution:**

and are perpendicular to each other

so

⇒

⇒ λ(1) + (3)(-1) + (2)(3) = 0

⇒ λ – 3 + 6 = 0

⇒ λ = 3

### Question 3. If and are two vectors such that ||=4, || = 3 and = 6. Find the angle between and

**Solution:**

Let the angle be θ

cos θ =

= 6 /(4×3) = 1/2

Therefore, θ = cos

^{-1}(1/2)= π/3

### Question 4. If and , find .

**Solution:**

=

=

=

=

=

=

Now,

=

= (1)(1) + (1)(-2) + (-4)(2)

= 1 – 2 – 8

= -9

Therefore, = -9

### Question 5. Find the angle between the vectors and where :

### (i) and

**Solution:**

Let the angle be θ between andcos θ =

Now,

=

= (1)(0) + (-1)(1) + (0)(1)

= 0 – 1 + 0 = -1

||= ||

=

= √2

= ||

=

= √2

Now, cos θ = -1/(√2×√2)

= -1/2

θ = cos

^{-1}(-1/2)= 2π/3

### (ii) and

**Solution:**

Let the angle be θ between and

Now,

=

=(3)(4) + (-2)(-1) + (-6)(8)

= 12 + 2 – 48

= -34

|| = ||

=

= √49 = 7

=

= √81 = 9

cos θ =

Now, cos θ = -34/(7×9)

= -34/63

θ = cos

^{-1}(-34/63)

### (iii) and

**Solution:**

Let the angle be θ between and

Now,

=

= (2)(4) + (-1)(4) + (2)(-2)

= 8 – 4 – 4 = 0

|| = ||

=

= √9 = 3

|| = ||

=

= √36 = 6

Now, cos θ =

cos θ = 0/(3×6) = 0

θ = cos

^{-1}(0)θ = π/2

### (iv) and

**Solution:**

Let the angle be θ between and

Now,

=

= (2)(1) + (-3)(1) + (1)(-2)

= 2 – 3 – 2

= -3

|| =

=

= √14

|| =||

=

= √6

cos θ =

Now, cos θ = -3/(√14×√6)

= -3/√84

θ = cos

^{-1}(-3/√84)

### (v) and

**Solution:**

Let the angle be θ between and

Now,

=

= (1)(1) + (2)(-1) + (-1)(1)

= 1 – 2 – 1

= -2

|| = ||

=

= √6

|| = ||

=

= √3

cos θ =

Now, cos θ = -2/(√6×√3)

= -2/√18

= -2/3√2

θ = cos

^{-1}(-√2 /3)

### Question 6. Find the angles which the vectors makes with the coordinate axes.

**Solution:**

Components along x, y and z axis are and respectively.

Let the angle between and be θ

_{1}Now,

=

= (1)(1) + (-1)(0) + (√2)(0)

= 1

=

= √4 = 2

= √1 = 1

cos θ

_{1}=Now, cos θ

_{1}= 1/(2×1)= 1/2

θ

_{1}= cos^{-1}(1/2) = π/3Let the angle between and be θ

_{2}Now,

=

= (1)(0) + (-1)(1) + (√2)(0)

= -1

= √1 = 1

cos θ

_{2}=Now, cos θ

_{2}= -1/(2×1)= -1/2

θ

_{2}= cos^{-1}(-1/2) = 2π/3Let the angle between and be θ

_{3}Now,

=

= (1)(0) + (-1)(0) + (√2)(1)

= √2

= √1 = 1

cos θ

_{3}== 1/(√2)

= cos

^{-1}(1/√2) = π/4

### Question 7(i). Dot product of a vector with and are 0, 5 and 8respectively. Find the vector.

**Solution:**

Let and be three given vectors.

Let be a vector such that its dot products with , and are 0, 5 and 8 respectively. Then,

⇒ = 0

⇒ x + y – 3z = 0 ….(1)

⇒ = 5

⇒ x + 3y – 2z = 5 …..(2)

⇒ = 8

⇒ 2x + y + 4z = 8 …..(3)

Solving 1,2 and 3 we get x = 1, y = 2 and z = 1,

Hence, the required vector is

### Question 8. If and are unit vectors inclined at an angle θ then prove that

### (i) cos θ/2 = 1/2

**Solution:**

|| = || = 1

||

^{2}=()^{2}=

= 1 + 1 + 2

= 2 + 2||cos θ

= 2(1 + (1)(1)cos θ)

= 2(2cos

^{2}θ/2)||

^{2 }= 4cos^{2}θ/2= 2 cos θ/2

cos θ/2 = 1/2||

### (ii) tan θ/2 =

**Solution:**

= 1

=

=

=

=

=

= tan

^{2}θ/2Therefore, tan θ/2 =

### Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is √3.

**Solution:**

Let and be two unit vectors

Then,

According to question:

Taking square on both sides

⇒

⇒

⇒ (1)

^{2}+(1)^{2}+ = 1⇒ 2+ 2 = 1

⇒ 2= -1

⇒ \hat{a}.\hat{b} =-1/2

Now,

=

= (1)

^{2}+ (1)^{2 }– 2 (-1/2)= 2 + 1 = 3

Therefore, = 3

=√3

### Question 10. If are three mutually perpendicular unit vectors, then prove that || =√3.

**Solution:**

Given are mutually perpendicular so,

Now,

=

=

= (1)

^{2}+ (1)^{2}+(1)^{2}+ 0= 3

= √3

### Question 11. If = 60, = 40 and = 46, find

**Solution:**

Given =60, = 40 and = 46

We know that,

(a + b)

^{2 }+ (a – b)^{2}= 2(a^{2}+ b^{2})⇒

⇒ 60

^{2}+ 40^{2}= 2(^{2}+ 49^{2})⇒ 3600 + 1600 = 2+ 2401

⇒ = 968

⇒ = √484 =22

### Question 12. Show that the vector is equally inclined with the coordinate axes.

**Solution:**

Let

√(1+1+1) = √3

Let θ

_{1}, θ_{2}, θ_{3 }be the angle between the coordinate axes and thecos θ

_{1}== 1/√3

cos θ

_{2}== 1/√3

cos θ

_{3}== 1/√3

Since, cos θ

_{1 }= cos θ_{2 }= cos θ_{3}Therefore, Given vector is equally inclined with coordinate axis.

### Question 13. Show that the vectors are mutually perpendicular unit vectors.

**Solution: **

Given,

= (1/7)√(2

^{2 }+ 3^{2 }+ 6^{2}) = (1/7)(√49) = 1= (1/7)√(3

^{2 }+ (-6)^{2 }+ 2^{2}) = (1/7)(√49) = 1= (1/7)√(6

^{2 }+ 2^{2 }+ (-3)^{2}) = (1/7)(√49) = 1Now, 1/49[3 × 2 – 3 × 6 + 6 × 2]

= 1/49[6 – 18 + 12] = 0

1/49[3 × 6 – 6 × 2 – 2 × 3]

= 1/49[18 – 12 – 6] = 0

Since, they are mutually perpendicular unit vectors.

### Question 14. For any two vectors and , Show that

**Solution:**

To prove

⇒

⇒

⇒

Hence Proved

### Question 15. If , and , find such that is perpendicular to .

**Solution:**

Given:

According to question

⇒

⇒

⇒ 2(λ+1) – (λ+3) -2λ-1 = 0

⇒ 2λ + 2 -λ – 3 – 2λ – 1 = 0

⇒ -λ = 2

⇒ λ = -2

### Question 16. If and , then find the value of λ so that and are perpendicular vectors.

**Solution:**

Given,

According to question

⇒

⇒

⇒

⇒ 25 + λ

^{2 }+ 9 = 1 + 9 + 25⇒ λ

^{2}= 1⇒ λ = 1

## Please

Loginto comment...