Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 11 Three Dimensional Geometry Ex 11.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1

Question 1.

If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.

Solution.

Let direction cosines of the line be l, m and n.

I = cos 90° = 0, m = cos 135° = – \(\frac{1}{\sqrt{2}}\), n = cos 45° = \(\frac{1}{\sqrt{2}}\)

Therefore, the direction cosines of the line are 0, \(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\).

Question 2.

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution.

Let the direction cosines of the line make an angle a with each of the coordinate axes.

∴ l = cos α, m = cos α, n = cos α

We know that l^{2} + m^{2} + n^{2} = 1

⇒ cos^{2} α + cos^{2} α + cos^{2} α = 1

⇒ 3 cos^{2} α = 1

⇒ cos^{2} α = \(\frac{1}{3}\)

⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are ± \(\frac{1}{\sqrt{3}}\), ± \(\frac{1}{\sqrt{3}}\) and ± \(\frac{1}{\sqrt{3}}\).

Question 3.

If a line has the direction ratios -18, 12, – 4, then what are its direction cosines?

Solution.

If a line has direction ratios of – 18,12 and – 4, then its direction cosines are

Question 4.

Show that the points (2, 3, 4), (- 1, – 2, 1), (5, 8, 7) are collinear.

Solution.

Let the given points are A (2, 3, 4), B (- 1, – 2, 1), C(5, 8, 7)

Direction ratios of AB are x_{2} – x_{1}; y_{2} – y_{1}, z_{2} – z_{1}

i.e., (- 1 – 2), (- 2 – 3), (1 – 4) or – 3, – 5, – 3

Direction ratios of BC are

5 – (-1), 8 – (- 2), (7 – 1) or 6, 10, 6

which are – 2 times the direction ratios of AB.

∴ AB and BC have the same direction ratios.

∴ AB || BC, but B is a common point of AB and BC.

Hence A,B,C are collinear.

Question 5.

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

Solution.

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

The direction ratios of side AB are (- 1, – 3), (1, – 5) and (2 – (- 4)) i.e„ – 4, – 4 and 6.

Then, \(\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}\)

= \(\sqrt{16+16+36}\)

= \(\sqrt{68}=2 \sqrt{17}\)