# Ncert Solutions For Class 12 Maths.3 Ex 11.1

## Ncert Solutions For Class 12 Maths Chapter 11.3 Ex 11.1

Q1. Find the cosine directions, if a line makes angles 90°, 135°, 45° with x, y and z- axes respectively.

Sol:

Let, the direction cosines of the line be p, q and r

p = cos 90° = 0

q = cos 135° = $$-\frac{1}{\sqrt{2}}$$

r = cos 45° = $$\frac{1}{\sqrt{2}}$$

Therefore, the direction cosines of the lines are 0, $$\frac{-1}{\sqrt{2}}$$ and $$\frac{1}{\sqrt{2}}$$.

Q2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

Sol:

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

p = cos α, q = cos α and r = cos α

Therefore, p2 + q2 + r2 = 1

cos2 α + cos2 α + cos2 α = 1

3cos2 α = 1

cos2 α = $$\frac{1}{3}$$

cos2 α = ± $$\frac{1}{\sqrt{3}}$$

Therefore, the direction cosines of the line, which is equally inclined to the coordinate axes, are $$\pm \frac{1}{\sqrt{3}}, \;\pm \frac{1}{\sqrt{3}}\; and \; \pm \frac{1}{\sqrt{3}}$$.

Q3. If a line has the direction ratios 12, -18, -4 then what are its direction cosines?

Sol:

If a line has direction ratios of 12, -18, -4 then its direction cosines are:

$$\frac{12}{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}$$, $$\frac{-18}{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}$$, $$\frac{ -4 }{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}$$

That is,

$$\frac{12}{22}, \frac{-18}{22}, \frac{-4}{22}$$ $$\frac{6}{11}, \frac{-9}{11}, \frac{-2}{11}\\$$

Thus, the direction cosines are $$\frac{6}{11}, \frac{-9}{11} \; and \; \frac{-2}{11}$$

Q4. Prove that the given points are collinear:

(-1, -2, 1), (2, 3, 4) and (8, 8, 7)

Sol:

The points are as follows:

P (-1, -2, 1)

Q (2, 3, 4)

R (8, 13, -2)

It is known that the direction ratios of line joining the points (x1, y1, z1) and (x1, y2, z2) are given by:

(x2 – y1), (y2 – y1) and (z2 – z1)

The direction ratios are PQ are [2 – (-1)], [3 – (-2)] and (4 – 1) that is 3, 5 and -3.

The direction ratios are QR are (8 – 2), (13 – 3) and (-2 – 4) that is 6, 10 and -6.

It can be seen that the direction ratios of QR are 2 times that of PQ. That is they are proportional.

Therefore, PQ is parallel to QR and Q is the common point to both PQ and QR. P, Q and R are collinear.

Q5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2).

Sol:

The vertices of $$\Delta ABC$$ are X (3, 5, −4), Y (−1, 1, 2) and Z (−5, −5, −2).

The direction ratios of side XY are (−1 − 3), (1 − 5), and [2 − (−4)], that is −4, −4, and 6.

i.e. $$\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}$$ = $$\sqrt{ 16 + 16 + 36}$$ = $$\sqrt{ 68 }$$ = $$2\sqrt{ 17 }$$

Hence, the direction cosines of XY are:

= $$\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}$$, $$\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}$$, $$\frac{6}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}$$

= $$\frac{-4}{2 \sqrt{17}}$$, $$\frac{-4}{2 \sqrt{17}}$$, $$\frac{6}{2 \sqrt{17}}$$

= $$\frac{-2}{ \sqrt{17}}$$, $$\frac{-2}{ \sqrt{17}}$$, $$\frac{3}{ \sqrt{17}}$$

The direction ratios of YZ are [−5 − (−1)], (−5 − 1), and (−2 − 2), that is −4, −6, and −4.

Hence, the direction cosines of XY are:

= $$\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}$$, $$\frac{-6}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}$$, $$\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}$$

= $$\frac{-4}{2 \sqrt{17}}$$, $$\frac{-6}{2 \sqrt{17}}$$, $$\frac{-4}{2 \sqrt{17}}$$

The direction ratios of ZX are (−5 − 3), (−5 − 5), and (−2 − (−4), that is −8, −10, and 2.

Hence, the direction cosines of XZ are:

= $$\frac{-8}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}$$, $$\frac{-5}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}$$, $$\frac{2}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}$$

= $$\frac{-8}{2 \sqrt{42}}$$, $$\frac{-5}{2 \sqrt{42}}$$, $$\frac{2}{2 \sqrt{42}}$$