NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.1 - Three Dimensional Geometry

Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:

If the line is making equal angle with the coordinate axes. Then,

Let the common angle made is $\alpha$ with each coordinate axes.

Therefore, we can write;

$l = \cos \alpha,\ m= \cos \alpha,and\ n= \cos \alpha$

And as we know the relation; $l^2+m^2+n^2 = 1$

$\Rightarrow \cos^2 \alpha +\cos^2 \alpha+\cos^2 \alpha = 1$

$\Rightarrow \cos^2 \alpha = \frac{1}{3}$

or $\cos \alpha =\pm \frac{1}{\sqrt3}$

Thus the direction cosines of the line are $\pm \frac{1}{\sqrt3},\ \pm \frac{1}{\sqrt3},and\ \pm \frac{1}{\sqrt3}$

Question 3: If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

Answer:

GIven a line has direction ratios of -18, 12, – 4 then its direction cosines are;

Line having direction ratio -18 has direction cosine:

$\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{-18}{22} = \frac{-9}{11}$

Line having direction ratio 12 has direction cosine:

$\frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{12}{22} =\frac{6}{11}$

Line having direction ratio -4 has direction cosine:

$\frac{12}{\sqrt{(-4)^2+(12)^2+(-4)^2}} = \frac{-4}{22} = \frac{-2}{11}$

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

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

Answer:

We have the points, A (2, 3, 4),B (– 1, – 2, 1),C (5, 8, 7);

And as we can find the direction ratios of the line joining the points $(x_{1},y_{1},z_{1}) \ and\ (x_{2},y_{2},z_{2})$ is given by $x_{2}-x_{1}, y_{2}-y_{1}, \ and\ z_{2}-z_{1}.$

The direction ratios of AB are $(-1-2), (-2-3),\ and\ (1-4)$ i.e., $-3,\ -5,\ and\ -3$

The direction ratios of BC are $(5-(-1)), (8-(-2)),\ and\ (7-1)$ i.e., $6,\ 10,\ and\ 6$ .

We can see that the direction ratios of AB and BC are proportional to each other and is -2 times.

$\therefore$ AB is parallel to BC. and as point B is common to both AB and BC,

Hence the points 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).

Answer:

Given vertices of the triangle $\triangle ABC$ (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

$\Rightarrow$ Direction ratios of side AB are $(-1-3), (1-5),\ and\ (2-(-4))$ i.e.,

$-4,-4,\ and\ 6.$

Therefore its direction cosines values are;

$\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

SImilarly for side BC;

$\Rightarrow$ Direction ratios of side BC are $(-5-(-1)), (-5-1),\ and\ (-2-2)$ i.e.,

$-4,-6,\ and\ -4.$

Therefore its direction cosines values are;

$\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

$\Rightarrow$ Direction ratios of side CA are $(-5-3), (-5-5),\ and\ (-2-(-4))$ i.e.,

$-8,-10,\ and\ 2.$

Therefore its direction cosines values are;

$\frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}}$ $or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$