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• RD Sharma Class 12 Solutions for Maths

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

### (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

### (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

### (i) and

Solution:

Let the angle be θ between and

cos 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 θ1

Now,

= (1)(1) + (-1)(0) + (√2)(0)

= 1

= √4 = 2

= √1 = 1

cos θ1

Now, cos θ1 = 1/(2×1)

= 1/2

θ1 = cos-1(1/2) = π/3

Let the angle between  and  be θ2

Now,

=

= (1)(0) + (-1)(1) + (√2)(0)

= -1

= √1 = 1

cos θ2 =

Now, cos θ2 = -1/(2×1)

= -1/2

θ2 = cos-1(-1/2) = 2π/3

Let the angle between  and  be θ3

Now,

=

= (1)(0) + (-1)(0) + (√2)(1)

= √2

= √1 = 1

cos θ3

= 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

### (i) cos θ/2 = 1/2

Solution:

|| = || = 1

||2 =()2

= 1 + 1 + 2

= 2 + 2||cos θ

= 2(1 + (1)(1)cos θ)

= 2(2cos2 θ/2)

||2 = 4cos2 θ/2

= 2 cos θ/2

cos θ/2 = 1/2||

### (ii) tan θ/2 =

Solution:

= 1

=

=

= tan2 θ/2

Therefore, 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(a2 + b2)

⇒

⇒ 602 + 402 = 2(2 + 492)

⇒ 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 θ1, θ2, θ3 be the angle between the coordinate axes and the

cos θ1

= 1/√3

cos θ2

= 1/√3

cos θ3

= 1/√3

Since, cos θ1 = cos θ2 = cos θ3

Therefore, Given vector is equally inclined with coordinate axis.

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

Solution:

Given,

= (1/7)√(22 + 32 + 62) = (1/7)(√49) = 1

= (1/7)√(32 + (-6)2 + 22) = (1/7)(√49) = 1

= (1/7)√(62 + 22 + (-3)2) = (1/7)(√49) = 1

Now, 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.

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 + λ2 + 9 = 1 + 9 + 25

⇒ λ2 = 1

⇒ λ = 1

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