# Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 3

### Question 33. Find the angle between the two vectors and , if

### (i) =√3, = 2 and = √6

**Solution:**

We know,

⇒ √6 = 2√3 cos θ

⇒ cos θ = 1/√2

⇒ θ = cos

^{-1}(1/√2)⇒ θ = π/4

### (ii) = 3, = 3 and = 1

**Solution:**

We know,

⇒ 1 = 3×3 cos θ

⇒ cos θ = 1/9

⇒ θ = cos

^{-1}(1/9)

### Question 34. Express the vector as the sum of two vectors such that one is parallel to the vector and other is perpendicular to

**Solution: **

Given,

Let the two vectors be

Now, ….(1)

Assuming is parallel to

Then, ……(2)

is perpendicular to

Then, ……(3)

From eq(1)

⇒

⇒

⇒

From eq(3)

⇒

⇒ (5-3λ)3+(5-λ)=0

⇒ 15-9λ+5-λ=0

⇒ -10λ = -20

⇒ λ=2

From eq(2)

**Question 35. If ****and ****are two vectors of the same magnitude inclined at an angle of 30° such that ****= 3, find **

**Solution:**

Given that two vectors of the same magnitude inclined at an angle of 30°, and

To find

We know,

⇒ 3 =

⇒ 3 =

⇒ 3 = (√3/2)

⇒= 6/√3

⇒

### Question 36. Express as the sum of a vector parallel and a vector perpendicular to

**Solution:**

Assuming

Let the two vectors be

Now,

or ….(1)

Assumingis parallel to

then, …(2)

is perpendicular to

then,……(3)

Putting eq(2) in eq(1), we get

⇒

⇒

⇒

From eq(3)

⇒

⇒ (2 – 2λ)2 – (1 + 4λ)4 – (3 + 2λ)2 = 0

⇒ 4 – 4λ – 4 – 16λ – 6 – 4λ = 0

⇒ 24λ = -6

⇒ λ = -6/24

From eq(2)

### Question 37. Decompose the vector into vectors which are parallel and perpendicular to the vector

**Solution:**

Let and

Let be a vector parallel to

Therefore,

to be decomposed into two vectors

⇒

⇒

Now, is perpendicular to

or

⇒

⇒ 6 – λ – 3 – λ – 6 – λ = 0

⇒ λ = -1

Therefore, the required vectors are and

### Question 38. Let and . Find λ such that is orthogonal to

**Solution:**

Given,

According to question

⇒

⇒

⇒

⇒ 25 + 1 + 49 = 1 + 1 + λ

^{2}⇒ λ

^{2 }= 73⇒ λ = √73

### Question 39. If and , what can you conclude about the vector ?

**Solution:**

Given, ,

Now,

We conclude that or or θ = 90°

Thus, can be any arbitrary vector.

### Question 40. If is perpendicular to both and , then prove that it is perpendicular to both and

**Solution:**

Given is perpendicular to both and

….(1)

….(2)

To prove and

Now,

⇒ [From eq(1) and (2)]

Again,

⇒ [From eq(1) and (2)]

Hence Proved

### Question 41. If and , prove that

**Solution:**

Given, and

To prove

Taking LHS

=

=

=

Taking RHS

=

=

LHS = RHS

Hence Proved

### Question 42. If are three non- coplanar vectors such that then show that is the null vector.

**Solution:**

Given that

So either or

Similarly,

Either or

Also,

So or

But can’t be perpendicular to and because are non-coplanar.

So = 0 oris a null vector

### Question 43. If a vector is perpendicular to two non- collinear vectors and , then is perpendicular to every vector in the plane of and

**Solution:**

Given that is perpendicular to and

Let be any vector in the plane of and and is the linear combination of and

[x, y are scalars]

Now

⇒

⇒

⇒

⇒

Therefore, is perpendicular to i.e. is perpendicular to every vector.

### Question 44. If , how that the angle θ between the vectors and is given by cos θ =

**Solution:**

Given that

⇒

⇒

⇒

⇒

⇒

⇒

⇒ cos θ =

### Question 45. Let and be vector such . = 3, = 4 and = 5, then find

**Solution:**

Given that and are vectors such that . = 3, = 4 and =5,

To find

Taking

Squaring on both side, we get

⇒

⇒

⇒

⇒

⇒

Therefore,

### Question 46. Let and be three vectors. Find the values of x for which the angle between and is acute and the angle between and is obtuse.

**Solution:**

Given

Case I: When angle between and is acute:-

>0

⇒

⇒ x

^{2 }– 2 – 2 > 0⇒ x

^{2 }> 4x ∈ (2, -2)

Case II: When angle between and is obtuse:-

⇒

⇒ x

^{2 }– 5 – 4 < 0⇒ x

^{2 }< 9x ∈ (3, -3)

Therefore, x ∈ (-3, -2)∪(2, 3)

### Question 47. Find the value of x and y if the vectorsand are mutually perpendicular vectors of equal magnitude.

**Solution:**

Given are mutually perpendicular vectors of equal magnitude.

⇒ 3

^{2 }+ x^{2 }+ (-1)^{2}= 2^{2 }+ 1^{2 }+ y^{2}⇒ x

^{2}+10 = y^{2}+5⇒ x

^{2 }– y^{2 }+ 5 = 0 ….(1)Now,

⇒ 6 + x – y = 0

⇒ y = x + 6 …..(2)

From eq(1)

x

^{2 }– (x + 6)^{2 }+ 5 = 0⇒ x

^{2}– (x^{2}+ 36 – 12x) + 5 = 0⇒ -12x – 31 = 0

⇒ x = -31/12

Now, y = -31/12 + 6

y = 41/12

### Question 48. If and are two non-coplanar unit vectors such that , find

**Solution:**

Given that and are two non-coplanar unit vectors such that

To find

Now,

Now,

=

= 6 – 13(1/2) – 5

= 1 – 13/2

= -11/2

### Question 49. If are two vectors such that || = , then prove that is perpendicular to

**Solution:**

To prove

Now,

Squaring on both side, we get

⇒

⇒

⇒

⇒

Therefore, is perpendicular to

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