Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1

Question 1.

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

Solution:

Direction angles are 90°, 135° and 45°.

∴ Direction cosines are

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

Hence, direction cosines are 0, – \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\).

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Question 2.

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

Solution:

Let direction angle be α each

∴ Direction cosines are cos α, cos α, cos α.

But l^{2} + m^{2} + n^{2} = 1, where l, m, n are the direction cosines of the line.

∴ cos^{2}α + cos^{2}α + cos^{2}α = 1 .

⇒ 3cos^{2}α = 1

∴ cos α = ± \(\frac{1}{\sqrt{3}}\).

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

Question 3.

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

Solution:

Direction ratios of a line are – 18, 12 and – 4.

Now, \(\sqrt{(-18)^{2}+(12)^{2}+(-4)^{2}}\) = \(\sqrt{484}\) = \(\sqrt{2×2×11×11}\)

= 2 × 11 = 22

∴ Direction cosines are \(\frac{- 18}{22}\), \(\frac{12}{22}\), \(\frac{- 4}{22}\) or \(\frac{- 9}{11}\), \(\frac{6}{11}\), \(\frac{- 2}{11}\).

Question 4.

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

Solution:

The given points are

A(2, 3, 4), B(- 1, – 2, 1) and 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 and 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 triangle ABC are

A (3, 5, – 4), B(- 1, 1, 2) and C(- 5, – 5, – 2).

(i) Direction ratios of AB are x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1}

or – 1 – 3,1 – 5 and 2 – (- 4), i.e., – 4, – 4 and 6.

(ii) Points B and C are (- 1, 1, 2) and (- 5, – 5, – 2) respectively.

∴ Direction ratios of BC are

– 5 – (- 1), – 5 – 1, – 2 – 2 or – 4, – 6 – 4

or – 2, – 3, – 2.

∴ \(\sqrt{(a)^{2}+(b)^{2}+(c)^{2}}\) = \(\sqrt{4+9+4}\) = \(\sqrt{17}\).

⇒ Direction cosines of BC are

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

(iii) Points C and A are (- 5, – 5, – 2) and (3, 5, – 4) respectively.

∴ Direction ratios of CA are

3 – (- 5), 5 – (- 5), – 4 – (- 2) or 8, 10, – 2 or 4, 5, – 1.

∴ \(\sqrt{(a)^{2}+(b)^{2}+(c)^{2}}\) = \(\sqrt{(4)^{2}+(5)^{2}+(1)^{2}}\) = \(\sqrt{16+15+1}\) = \(\sqrt{42}\).

∴ Direction cosines of CA are

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

Thus, the direction cosines of AB, BC and CA are respectively.

\(\frac{-2}{\sqrt{17}}\), \(\frac{-2}{\sqrt{17}}\), \(\frac{3}{\sqrt{17}}\); \(\frac{-2}{\sqrt{17}}\), \(\frac{-3}{\sqrt{17}}\), \(\frac{-2}{\sqrt{17}}\) and \(\frac{4}{\sqrt{42}}\), \(\frac{5}{\sqrt{42}}\), \(\frac{-1}{\sqrt{42}}\).