Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Direction Cosines and Direction Ratios Solutions Exercise 6(a)

I.

Question 1.

A line makes angles 90°, 60° and 30° with positive directions of X, Y, Z – axes respectively. Find the direction cosines.

Solution:

Suppose l, m, n are the direction cosines of the line.

l = cos α = cos 90° = 0

m = cos β = cos 60° = \(\frac{1}{2}\)

n = cos γ = cos 30° = \(\frac{\sqrt{3}}{2}\)

Direction cosines of the line are (0, \(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\))

Question 2.

If a line makes angles α, β, γ with the positive direction of x, y, z axes, what is the value of sin² α + sin² β + sin² γ?

Solution:

We know that cos² α + cos² β + cos² γ = 1

1 – sin² α + 1 – sin² β + 1 – sin² γ = 1

sin² a + sin² P + sin² γ = 3 – 1 = 2.

Question 3.

If P(√3, 1, 2√3) is a point in space, find the direction cosines of \(\overrightarrow{O P}\).

Solution:

Direction ratios of P are (√3, 1, 2√3)

a² + b² + c² = 3 + 1 + 12 = 16

⇒ \(\sqrt{a^{2}+b^{2}+c^{2}}\) = 4

Direction cosines of \(\overrightarrow{O P}\) are

Question 4.

Find the direction cosines of the line joining the points (-4, 1, 7) are (2, -3, 2).

Solution:

A(- 4, 1, 2) and B(2, -3, 2) are the given points

d.rs of PQ are (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

(2 + 4, 1 + 3, 2 – 7) ie., (6, 4, -5)

Dividing with

II.

Question 1.

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2).

Solution:

A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2) are the vertices of ∆ABC

d.rs of AB are (-1 -3, 1 – 5, 2 + 4) (-4, -4, 6)

Dividing with \(\sqrt{16+16+36}=\sqrt{68}=2 \sqrt{17}\)

D.Rs of AB are \(\frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}}\)

ie., \(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\)

D.Rs of BC are (-5 + 1, -5 -1, -2 -2)

i.e., (-4, -6, -4)

Dividing with \(\sqrt{16+16+36}=\sqrt{68}=2 \sqrt{17}\)

d.cs of BC are \(\frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}\)

ie., \(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\)

d.rs of CA are 3 + 5, 5 + 5, -4 + 2

= 8, 10, -2

Dividing with \(\sqrt{64+100+4}=\sqrt{168}=2 \sqrt{42}\)

d.cs of CA are \(\frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{-2}{2 \sqrt{42}}\)

ie., \(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}\)

Question 2.

Show that the lines \(\stackrel{\leftrightarrow}{P Q}\) and \(\stackrel{\leftrightarrow}{R S}\) are parallel where P, Q, R, S are two points (2, 3, 4), (4, 7, 8), (-1, -2, 1) and (1, 2, 5) respectively.

Solution:

P(2, 3, 4), Q(4, 7, 8), R(-1, -2, 1)

and S(1, 2, 5) are the given points.

d.rs of PQ are 4 -2, 7 -3, 8 – 4 i.e., 2, 4, 4

d.rs of RS are 1 + 1, 2 + 2, 5 – 1 i.e., 2, 4, 4

d.rs of PQ are RS are proportional

∴ PQ and RS are parallel.

III.

Question 1.

Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l² + 5m² – 3n² = 0.

Solution:

Given l – 5m + 3n = 0

⇒ l = 5m – 3n ………….. (1)

7l² + 5m² – 3n² = 0 …………. (2)

Substituting the value of l in (2)

7(5m – 3n)² + 5m² – 3n² = 0

7(25m² + 9n² – 30 mn) + 5m² – 3n² = 0

175 m² + 63n² – 210 mn + 5m² – 3n² = 0

180m² – 210mn + 60n² = 0

Dividing with 30.

6m² – 7mn + 2n² = 0

(3m – 2n) (2m – n) = 0

3m = 2n or 2m = n

Case (i): 3m_{1} = 2n_{1} ⇒ \(\frac{m_{1}}{2}=\frac{n_{1}}{3}\)

and m_{1} = \(\frac{2}{3}\) n_{1}

From (1) l_{1} = 5m_{1} – 3n_{1} = \(\frac{10}{3}\)n_{1} – 3n_{1}

d.rs of the first line are (1, 2, 3)

Dividing with \(\sqrt{1+4+9}=\sqrt{14}\)

d.cs of the first line are (\(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\))

Case (ii): 2m_{2} = n_{2}

From (1) l_{2} – 5m_{2} + 3n_{2} = 0

l_{2} – 5m_{2} + 6m_{2} = 0

l_{2} = m_{2}

∴ \(\frac{l_{2}}{-1}=\frac{m_{2}}{1}=\frac{n_{2}}{2}\)

d.rs of the second line are -1, 1, 2

Dividing with \(\sqrt{1+1+4}=\sqrt{6}\)

d.cs of the second line are (\(\frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\))