# Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 1

### (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

â‡’

â‡’ Î»(4) + (2)(-9) + (1)(2) = 0

â‡’ 4Î» – 18 + 2 = 0

â‡’ 4Î» = 16

â‡’ Î» = 4

### (ii) and

Solution:

and are perpendicular to each other

so= 0

â‡’

â‡’ Î»(5) + (2)(-9) + (1)(2) = 0

â‡’ 5Î» – 18 + 2 = 0

â‡’ 5Î» = 16

â‡’ Î» = 16/5

### (iii) and

Solution:

and are perpendicular to each other

so = 0

â‡’ =0

â‡’ (2)(3) + (3)(2) – (4)Î» = 0

â‡’ 6 + 6 – 4Î» = 0

â‡’ 4Î» = 12

â‡’ Î» = 3

### (iv) and

Solution:

and are perpendicular to each other

so

â‡’

â‡’ Î»(1) + (3)(-1) + (2)(3) = 0

â‡’ Î» – 3 + 6 = 0

â‡’ Î» = 3

### Question 3. If and are two vectors such that ||=4, || = 3 and  = 6. Find the angle between  and

Solution:

Let the angle be Î¸

cos Î¸ =

= 6 /(4Ã—3) = 1/2

Therefore, Î¸ = cos-1(1/2)

= Ï€/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 Î¸ between and

cos Î¸ =

Now,

= (1)(0) + (-1)(1) + (0)(1)

= 0 – 1 + 0 = -1

||= ||

= âˆš2

= ||

= âˆš2

Now, cos Î¸ = -1/(âˆš2Ã—âˆš2)

= -1/2

Î¸ = cos-1(-1/2)

= 2Ï€/3

### (ii)  and

Solution:

Let the angle be Î¸ between  and

Now,

=

=(3)(4) + (-2)(-1) + (-6)(8)

= 12 + 2 – 48

= -34

|| = ||

= âˆš49 = 7

= âˆš81 = 9

cos Î¸ =

Now, cos Î¸ = -34/(7Ã—9)

= -34/63

Î¸ = cos-1(-34/63)

### (iii)  and

Solution:

Let the angle be Î¸ between  and

Now,

=

= (2)(4) + (-1)(4) + (2)(-2)

= 8 – 4 – 4 = 0

|| = ||

= âˆš9 = 3

|| = ||

=

= âˆš36 = 6

Now, cos Î¸ =

cos Î¸ = 0/(3Ã—6) = 0

Î¸ = cos-1(0)

Î¸ = Ï€/2

### (iv)  and

Solution:

Let the angle be Î¸ between  and

Now,

=

= (2)(1) + (-3)(1) + (1)(-2)

= 2 – 3 – 2

= -3

|| =

=

= âˆš14

|| =||

=

= âˆš6

cos Î¸ =

Now, cos Î¸ = -3/(âˆš14Ã—âˆš6)

= -3/âˆš84

Î¸ = cos-1(-3/âˆš84)

### (v)  and

Solution:

Let the angle be Î¸ between  and

Now,

=

= (1)(1) + (2)(-1) + (-1)(1)

= 1 – 2 – 1

= -2

|| = ||

=

= âˆš6

|| = ||

=

= âˆš3

cos Î¸ =

Now, cos Î¸ = -2/(âˆš6Ã—âˆš3)

= -2/âˆš18

= -2/3âˆš2

Î¸ = cos-1(-âˆš2 /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 Î¸1

Now,

= (1)(1) + (-1)(0) + (âˆš2)(0)

= 1

= âˆš4 = 2

= âˆš1 = 1

cos Î¸1

Now, cos Î¸1 = 1/(2Ã—1)

= 1/2

Î¸1 = cos-1(1/2) = Ï€/3

Let the angle between  and  be Î¸2

Now,

=

= (1)(0) + (-1)(1) + (âˆš2)(0)

= -1

= âˆš1 = 1

cos Î¸2 =

Now, cos Î¸2 = -1/(2Ã—1)

= -1/2

Î¸2 = cos-1(-1/2) = 2Ï€/3

Let the angle between  and  be Î¸3

Now,

=

= (1)(0) + (-1)(0) + (âˆš2)(1)

= âˆš2

= âˆš1 = 1

cos Î¸3

= 1/(âˆš2)

= cos-1(1/âˆš2) = Ï€/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,

â‡’  = 0

â‡’ x + y – 3z = 0        ….(1)

â‡’ = 5

â‡’ x + 3y – 2z = 5     …..(2)

â‡’  = 8

â‡’ 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 Î¸/2 = 1/2

Solution:

|| = || = 1

||2 =()2

= 1 + 1 + 2

= 2 + 2||cos Î¸

= 2(1 + (1)(1)cos Î¸)

= 2(2cos2 Î¸/2)

||2 = 4cos2 Î¸/2

= 2 cos Î¸/2

cos Î¸/2 = 1/2||

### (ii) tan Î¸/2 =

Solution:

= 1

=

=

= tan2 Î¸/2

Therefore, tan Î¸/2 =

### Question 9. If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is âˆš3.

Solution:

Let  and  be two unit vectors

Then,

According to question:

Taking square on both sides

â‡’

â‡’

â‡’ (1)2+(1)2+ = 1

â‡’ 2+ 2 = 1

â‡’ 2= -1

â‡’ \hat{a}.\hat{b} =-1/2

Now,

= (1)2 + (1)2  – 2 (-1/2)

= 2 + 1 = 3

Therefore,  = 3

=âˆš3

### Question 10. If  are three mutually perpendicular unit vectors, then prove that || =âˆš3.

Solution:

Given  are mutually perpendicular so,

Now,

=

=

= (1)2 + (1)2 +(1)2 + 0

= 3

= âˆš3

### Question 11. If  = 60,  = 40 and = 46, find

Solution:

Given =60,  = 40 and = 46

We know that,

(a + b)2 + (a – b)2 = 2(a2 + b2)

â‡’

â‡’ 602 + 402 = 2(2 + 492)

â‡’ 3600 + 1600 = 2+ 2401

â‡’ = 968

â‡’ = âˆš484 =22

### Question 12. Show that the vector  is equally inclined with the coordinate axes.

Solution:

Let

âˆš(1+1+1) = âˆš3

Let Î¸1, Î¸2, Î¸3 be the angle between the coordinate axes and the

cos Î¸1

= 1/âˆš3

cos Î¸2

= 1/âˆš3

cos Î¸3

= 1/âˆš3

Since, cos Î¸1 = cos Î¸2 = cos Î¸3

Therefore, Given vector is equally inclined with coordinate axis.

### Question 13. Show that the vectors  are mutually perpendicular unit vectors.

Solution:

Given,

= (1/7)âˆš(22 + 32 + 62) = (1/7)(âˆš49) = 1

= (1/7)âˆš(32 + (-6)2 + 22) = (1/7)(âˆš49) = 1

= (1/7)âˆš(62 + 22 + (-3)2) = (1/7)(âˆš49) = 1

Now, 1/49[3 Ã— 2 – 3 Ã— 6 + 6 Ã— 2]

= 1/49[6 – 18 + 12] = 0

1/49[3 Ã— 6 – 6 Ã— 2 – 2 Ã— 3]

= 1/49[18 – 12 – 6] = 0

Since,  they are mutually perpendicular unit vectors.

Solution:

To prove

â‡’

â‡’

â‡’

Hence Proved

### Question 15. If , and , find such that is perpendicular to .

Solution:

Given:

According to question

â‡’

â‡’

â‡’ 2(Î»+1) – (Î»+3) -2Î»-1 = 0

â‡’ 2Î» + 2 -Î» – 3 – 2Î» – 1 = 0

â‡’ -Î» = 2

â‡’ Î» = -2

### Question 16. If and , then find the value of Î» so that  and  are perpendicular vectors.

Solution:

Given,

According to question

â‡’

â‡’

â‡’

â‡’ 25 + Î»2 + 9 = 1 + 9 + 25

â‡’ Î»2 = 1

â‡’ Î» = 1

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