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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 2

### Question 17. If and , then express  in the form of where is parallel to and is perpendicular to .

Solution:

Given,

According to question

also  = 0

Now,

⇒

⇒

⇒

Now,

⇒

⇒ 3(2-3λ)+4(1-4λ)-5(4+5λ) = 0

⇒ 6-9λ+4-16λ-20-25λ = 0

⇒ -10 -50λ = 0

⇒ λ = -1/5

### Question 18. If either or , then . But, The converse need not be true. Justify your answer with an example.

Solution:

Given,

or  then

Suppose

But,

= √(2)2+(1)2+(1)2

= √4+1+1

= √6 ≠ 0

= √(1)2+(1)2+(1)2

= √3 ≠ 0

Hence Proved

### Question 19. Show that the vectors  form a right-angled triangle.

Solution:

Given,

To prove given vectors form a right angle triangle

= √(32+(-2)2+12) = √14

= √(12+(-3)2+52) = √35

= √(22+12+(-4)2) = √21

= 14 + 21 = 35

Since, (Pythagoras Theorem)

Hence, and  form a right angled triangle.

### Question 20. If , and are such that is perpendicular to , then find the value of λ.

Solution:

Given:

Now,

⇒

⇒

⇒ (2 – λ)3 + (2 + 2λ) + 0 = 0

⇒ 6 – 3λ + 2 + 2λ =0

⇒ λ = 8

### Question 21.  Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).

Solution:

Given that angle of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4) and C(5, 7, 1).

= √98 = 7√2

Now,

= (3 × 2 + 2 × 6 – 6 × 3) = 0

Thus, we can say AB is perpendicular to BC.

Hence, AB = BC = 7, ∠A =∠C and ∠B = 90°

∠A + ∠B + ∠C = 180°

2∠A = 180° – 90°

∠A = 45°

∠C = 45°

∠B = 90°

Solution:

We know

⇒ 1/ 2 =

⇒ 1/2 = (1/2)

⇒

or

⇒

### Question 23. Show that the points whose position vector are  form a right triangle.

Solution:

Given that positions vectors

Now,

⇒

⇒

⇒

Now,

= 2 – 3 – 20 = -21

= -3 – 6 – 5 = -14

= -6 + 2 + 4 = 0

So, AB is perpendicular to CA or the given position vectors form a right-angled triangle.

### Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?

Solution:

Given the vertices of △ABC are A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2)

Now,

Or,

We know that

(2 × 1) + (2 × 1) + (3 × 2)

= 2 + 2 + 6 = 10

Now,  = √17

= √6

Therefore,

cos θ =

cos θ = 10/ √(17×6)

θ = cos-1(10/√102)

### Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.

Solution:

Given, position vectors A(0, 1, 1), B(3, 1, 5), C(0, 3, 3)

Now,

= 2 × 2 – 2 × 2 = 0

Thus,  and  are perpendicular hence △ABC is right-angled at C

### Question 26. Find the projection of on , where and.

Solution:

Given:

To find the projection of  on

Now, Projection of

= 6/9 × 3

= 2

### Question 27. If and , then show that the vectors and  are orthogonal.

Solution:

Given:

To prove

Taking LHS

= √35 – √35

= 0

Thus, the given vectors and are orthogonal.

### Question 28. A unit vector makes angle π/2 and π/3 with and respectively and an acute angle θ with . Find the angle θ and components of .

Solution:

Let us assume

We know that

a12+ a22 + a32 = 1  ….(1)

So,

(1)(1)(1/√2) = a1

a1 = 1/√2

Again we take

(1)(1)(1/2) = a2

a2 = 1/2

Put all these values in eq(1) to find the value of a3

(1/√2)2 + (1/2)2 + a32 = 1  ….(1)

a32 = 1/4

a3 = 1/2

Now we find the value of θ

(1)(1)cosθ = 1/2

cosθ = 1/2

cosθ = π/3

and components of

### Question 29. If two vectors and are such that = 2, = 1, and =1. Find the value of

Solution:

Given,

=

=

= 6(2)2 + 11(1) – 35(1)2

= 24 + 11 – 35

= 35 – 35 = 0

Solution:

Given,

⇒

⇒

⇒

Solution:

Given,

⇒

⇒

⇒

⇒

⇒ =√13

Solution:

Given,  = 12

⇒

⇒  = 12

⇒  = 12

⇒ = 12

⇒ = 2

So,

= 4

Solution:

Given, = 8

⇒

⇒

⇒

⇒ = √(8/63)

So,

= 8√(8/63)

Solution:

Given,

⇒

⇒

⇒

⇒ 3= 3

⇒ = 1

So,

= 2

### (i) and

Solution:

We have,

= 22 – 2 × 8 + 52

⇒ = 4 – 16 + 25

⇒ = 13

= √13

### (ii) = 3, = 4 and  = 1

Solution:

We have,

⇒

= 32 – 2 × 1 + 42

⇒ = 9 – 2 + 16

⇒ = 23

= √23

### (iii)  and = 4

Solution:

We have,

⇒

= 22 – 2 × 4 + 32

⇒ = 4 – 8 + 9

= 5

⇒ = √5

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