# RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

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## RD Sharma Class 12 Ex 24.2 Solutions Chapter 24 Scalar or Dot Product

### (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)

||= 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)+ (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, θbe the angle between the coordinate axes and the

cos θ1 =

= 1/√3

cos θ2 =

= 1/√3

cos θ3 =

= 1/√3

Since, cos θ= cos θ= 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)√(2+ 3+ 62) = (1/7)(√49) = 1

= (1/7)√(3+ (-6)+ 22) = (1/7)(√49) = 1

= (1/7)√(6+ 2+ (-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 + λ+ 9 = 1 + 9 + 25

⇒ λ2 = 1

⇒ λ = 1

### 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

a= 1/√2

Again we take

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

a= 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

Solution:

We have,

= 2– 2 × 8 + 52

⇒ = 4 – 16 + 25

⇒ = 13

= √13

Solution:

We have,

⇒

= 3– 2 × 1 + 42

⇒ = 9 – 2 + 16

⇒ = 23

= √23

Solution:

We have,

⇒

= 2– 2 × 4 + 32

⇒ = 4 – 8 + 9

= 5

⇒ = √5

### (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

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

⇒ λ= 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

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.

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

and is obtuse:-

⇒

⇒ x– 5 – 4 < 0

⇒ x< 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.

⇒ 3+ x+ (-1)2 = 2+ 1+ y2

⇒ x2+10 = y2+5

⇒ x– y+ 5 = 0    ….(1)

Now,

⇒ 6 + x – y = 0

⇒ y = x + 6      …..(2)

From eq(1)

x– (x + 6)+ 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

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