# 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,

â‡’

â‡’

â‡’

Now,

â‡’

â‡’ 3(2-3Î»)+4(1-4Î»)-5(4+5Î») = 0

â‡’ 6-9Î»+4-16Î»-20-25Î» = 0

â‡’ -10 -50Î» = 0

â‡’ Î» = -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,

= âˆš(2)2+(1)2+(1)2

= âˆš4+1+1

= âˆš6 â‰  0

= âˆš(1)2+(1)2+(1)2

= âˆš3 â‰  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

= âˆš(32+(-2)2+12) = âˆš14

= âˆš(12+(-3)2+52) = âˆš35

= âˆš(22+12+(-4)2) = âˆš21

= 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 Î».

Solution:

Given:

Now,

â‡’

â‡’

â‡’ (2 – Î»)3 + (2 + 2Î») + 0 = 0

â‡’ 6 – 3Î» + 2 + 2Î» =0

â‡’ Î» = 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).

= âˆš98 = 7âˆš2

Now,

= (3 Ã— 2 + 2 Ã— 6 – 6 Ã— 3) = 0

Thus, we can say AB is perpendicular to BC.

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

âˆ A + âˆ B + âˆ C = 180Â°

2âˆ A = 180Â° – 90Â°

âˆ A = 45Â°

âˆ C = 45Â°

âˆ B = 90Â°

Solution:

We know

â‡’ 1/ 2 =

â‡’ 1/2 = (1/2)

â‡’

or

â‡’

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

Solution:

Given that positions vectors

Now,

â‡’

â‡’

â‡’

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 â–³ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of âˆ ABC?

Solution:

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

Now,

Or,

We know that

(2 Ã— 1) + (2 Ã— 1) + (3 Ã— 2)

= 2 + 2 + 6 = 10

Now,  = âˆš17

= âˆš6

Therefore,

cos Î¸ =

cos Î¸ = 10/ âˆš(17Ã—6)

Î¸ = cos-1(10/âˆš102)

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

Solution:

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

Now,

= 2 Ã— 2 – 2 Ã— 2 = 0

Thus,  and  are perpendicular hence â–³ABC 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 Ã— 3

= 2

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

Solution:

Given:

To prove

Taking LHS

= âˆš35 – âˆš35

= 0

Thus, the given vectors and are orthogonal.

### Question 28. A unit vector makes angle Ï€/2 and Ï€/3 with and respectively and an acute angle Î¸ with . Find the angle Î¸ and components of .

Solution:

Let us assume

We know that

a12+ a22 + a32 = 1  ….(1)

So,

(1)(1)(1/âˆš2) = a1

a1 = 1/âˆš2

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/âˆš2)2 + (1/2)2 + a32 = 1  ….(1)

a32 = 1/4

a3 = 1/2

Now we find the value of Î¸

(1)(1)cosÎ¸ = 1/2

cosÎ¸ = 1/2

cosÎ¸ = Ï€/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,

â‡’

â‡’

â‡’

â‡’

â‡’

Solution:

Given,

â‡’

â‡’

â‡’

â‡’

â‡’ =âˆš13

Solution:

Given,  = 12

â‡’

â‡’  = 12

â‡’  = 12

â‡’ = 12

â‡’ = 2

So,

= 4

Solution:

Given, = 8

â‡’

â‡’

â‡’

â‡’

â‡’ = âˆš(8/63)

So,

= 8âˆš(8/63)

Solution:

Given,

â‡’

â‡’

â‡’

â‡’ 3= 3

â‡’ = 1

So,

= 2

### (i) and

Solution:

We have,

â‡’

â‡’= 22 – 2 Ã— 8 + 52

â‡’ = 4 – 16 + 25

â‡’ = 13

â‡’= âˆš13

### (ii) = 3, = 4 and  = 1

Solution:

We have,

â‡’

â‡’ = 32 – 2 Ã— 1 + 42

â‡’ = 9 – 2 + 16

â‡’ = 23

â‡’ = âˆš23

### (iii)  and = 4

Solution:

We have,

â‡’

â‡’= 22 – 2 Ã— 4 + 32

â‡’ = 4 – 8 + 9

â‡’ = 5

â‡’ = âˆš5

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