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 + λ²

⇒ λ² = 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

⇒

⇒ x² - 2 - 2 > 0

⇒ x² > 4

x ∈ (2, -2)

Case II: When angle between and is obtuse:-

⇒ 

⇒ x² - 5 - 4 < 0

⇒ x² < 9

x ∈ (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.

⇒ 3² + x² + (-1)² = 2² + 1² + y²

⇒ x²+10 = y²+5

⇒ x² - y² + 5 = 0    ....(1)

Now, 

⇒ 6 + x - y = 0

⇒ y = x + 6      .....(2)

From eq(1)

x² - (x + 6)² + 5 = 0

⇒ x² - (x² + 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 

Summary

This set focuses on more advanced applications of the dot product.

Key points include:

• Using dot product to solve complex geometric problems
• Applying dot product properties in vector equations
• Proving vector identities involving dot products
• Utilizing dot product in problems related to direction cosines and ratios
• Combining dot product with other vector operations

Practice Questions

1). If a, b, c are unit vectors such that a + b + c = 0, prove that a · b = b · c = c · a = -1/2.

2). Find the value of λ for which the vectors 2i + j - k, i - 2j + 3k, and λi + 2j - k are coplanar.

3). Prove that (a × b) · (b × c) = (a · b)(b · c) - b²(a · c).

4). If a · b = 3, b · c = 4, c · a = 5, |a| = 2, |b| = 3, and |c| = 4, find the value of |a × b + b × c + c × a|.

5). Show that (a - b) · (a + b) = |a|² - |b|² for any two vectors a and b.

6). If a, b, c are non-coplanar vectors and p = λa + μb + νc, express λ, μ, ν in terms of dot products.

7). Prove that the sum of squares of the direction cosines of a line is always 1.

8). If a = 3i - 2j + k, b = i + 2j - 2k, and c = 2i + j + 3k, find the angle between a + b and a - c.

9). Show that (a · b)² + |a × b|² = |a|²|b|² for any two vectors a and b.

10). If a, b, c are unit vectors such that a · b = b · c = c · a = 1/2, find the value of |a + b + c|.