# RD Sharma Solutions Class 12 Algebra Of Vectors Exercise 23.9

RD Sharma Solutions for Class 12 Maths Chapter 23 Algebra of Vectors Exercise 23.9 are given here to improve the problem-solving skills of the students. These solutions are developed by subject experts according to the latest CBSE guidelines so that students can achieve high marks in their exams. The RD Sharma Class 12 Solutions is the right resource to kick start preparations for the board exams. Students can now download the PDF containing the solutions for this exercise effortlessly, from the link provided below.

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Exercise 23.9

1. Solution:

We know that, if l, m and n are the direction cosines of a vector and Î±, Î², Î³ be the direction angle

Then,

l = cos Î±

m = cos Î²

n = cos Î³

And, l2 + m2 + n2 = 1 â€¦ (i)

Here,

l = 1/âˆš2

m = Â½

n = -Â½

Putting the value of l, m and n in (i), we get

(1/âˆš2)2 + (1/2)2 + (-1/2)2 = 1

Â½ + Â¼ + Â¼ = 1

(2 + 1 + 1)/4 = 1

1 = 1

L.H.S = R.H.S

Therefore, a vector can have direction angles as 45o, 60o and 120o.

2. Solution:

Given,

l = 1, m = 1 and n = 1

We know that,

l2 + m2 + n2 = 1

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

1 + 1 + 1 = 1

3 â‰  1

L.H.S â‰  R.H.S

Therefore, 1,1,1 cannot be the direction cosines of a straight line.

3. Solution:

We have,

Î± = Ï€/4 and Î² = Ï€/4

So,

l = cos Î± = cos Ï€/4 = 1/ âˆš2

m = cos Î² = cos Ï€/4 = 1/ âˆš2

And,

n = cos Î³

We know that,

l2 + m2 + n2 = 1

(1/âˆš2)2 + (1/âˆš2)2 + cos2 Î³ = 1

Â½ + Â½ + cos2 Î³ = 1

1 + cos2 Î³ = 1

cos2 Î³ = 0

Taking square root on both sides, we get

cos Î³ = 0

Î³ = cos-1 0

Î³ = Ï€/2

Therefore, the angle made by the vector with the z-axis is Ï€/2.

4. Solution:

We have,

Î± = Î² = Î³

So,

cos Î± = cos Î² = cos Î³

Also,

l = m = n = k (say)

We know that,

l2 + m2 + n2 = 1

k2 + k2 + k2 = 1

3k2 = 1

k2 = 1/3

Taking square root on both sides, we get

k = Â± 1/âˆš3

So,

l = Â± 1/âˆš3, m = Â± 1/âˆš3 and n = Â± 1/âˆš3

Hence, the direction cosines of vector r are Â± 1/âˆš3, Â± 1/âˆš3, Â± 1/âˆš3

Now,

5. Solution:

We have,

Î± = 45o, Î² = 60o and Î³ = Î¸ (say)

So,

l = cos Î± = cos 45o = 1/âˆš2

m = cos Î² = cos 60o = Â½

And,

n = cos Î³ = cos Î¸

We know that,

l2 + m2 + n2 = 1

Putting values of l, m and n

(1/âˆš2)2 + (Â½)2 + cos2 Î¸ = 1

Â½ + Â¼ + cos2 Î¸ = 1

Â¾ + cos2 Î¸ = 1

cos2 Î¸ = 1 â€“ Â¾ = (4 – 3)/4

cos2 Î¸ = Â¼

Taking square root on both the sides we get,

cos Î¸ = Â± 1/2

So, the direction cosines of the vector r are Â± Â½, Â± Â½, Â± Â½

Thus, the required vector is given by