Three Dimensional Geometry

Class 12 NCERT

NCERT

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

Solution :

Let direction cosines of the line be $l, m,$ and $n.$ $l = \cos 90^o=0\\ m= \cos 135^o =-\dfrac{1}{\sqrt{2}}\\ n = \cos 45^o=\dfrac{1}{\sqrt{2}}$$\\$ Therefore, the direction cosines of the line are $0,-\dfrac{1}{\sqrt{2}} ,$ and $ \dfrac{1}{\sqrt{2}}.$

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

Solution :

Let the direction cosines of the line make an angle $\alpha$ with each of the coordinates axes.$\\$ $\therefore 1 = \cos \alpha , m = \cos \alpha , n = \cos \alpha \\ 1^2+m^2+n^2=1\\ \implies \cos ^2 \alpha + \cos^2\alpha +\cos^2 \alpha =1\\ \implies 3 \cos^2 \alpha =1\\ \cos^2\alpha =\dfrac{1}{3}\\ \cos \alpha =\pm \dfrac{1}{\sqrt{3}}$ $\\$ Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are$\\$ $\pm\dfrac{1}{\sqrt{3}} , \pm\dfrac{1}{\sqrt{3}} $ and $\pm\dfrac{1}{\sqrt{3}} $

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

Solution :

If a line has direction ratios $- 18, 12, -4,$ then its direction cosines are $\\$ $ \dfrac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\dfrac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\dfrac{-4}{\sqrt{(-18)^2+(12)^2+(-4)^2}}$$\\$ $ i. e. , \ \ \ \dfrac{-18}{22},\dfrac{12}{22},\dfrac{-4}{22}\\ \dfrac{-9}{11},\dfrac{6}{11},\dfrac{-2}{11}$$\\$ Thus, the direction cosines are $\dfrac{-9}{11},\dfrac{6}{11},$ and $\dfrac{-2}{11}.$

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

Solution :

The given points are $A (2, 3, 4), B (-1, -2, 1),$ and $C (5, 8, 7).$ It is known that the direction ratios of the line joining the points, $(x _1 , y_ 1 , z_ 1 )$ and $( x_ 2 , y_ 2 , z_ 2 )$, are 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.$$\\$ It can be seen that the direction ratios of $BC$ are $-2$ times that $AB$ i.e., they are proportional.$\\$ Therefore, $AB$ is parallel to $BC.$ Since point $B$ is common to both $AB$ and $BC$, points $A, B,$ and $C$ are collinear

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 $\Delta ABC$ are $A (3, 5, -4), B (-1, 1, 2),$ and $C (-5, -5, -2).$$\\$ The direction ratios of the side $AB$ are $(-1 -3), (1 -5),$ and $(2 -(-4)) i.e., -4, -4, $ and $6.$$\\$ Then,$\sqrt{(-4)^2+(-4)^2+(6)^2}=\sqrt{16+16+36}\\ =\sqrt{68}\\ =2\sqrt{17}$ $\\$ Therefore, the direction cosines of $AB$ are $\\$ $\dfrac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\dfrac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\dfrac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}\\ \dfrac{-4}{2\sqrt{17}},-\dfrac{4}{2\sqrt{17}},\dfrac{6}{2\sqrt{17}}\\ \dfrac{-2}{\sqrt{17}},\dfrac{-2}{\sqrt{17}},\dfrac{3}{\sqrt{17}}$$\\$ The direction ratios of BC are $(-5 - (-1)), (-5 -1),$ and $(-2 -2) i.e., -4, -6,$ and $-4.$$\\$ Therefore, the direction cosines of $BC$ are $\\$ $\dfrac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}} ,\dfrac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\dfrac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}\\ i. e. , \dfrac{-4}{2\sqrt{17}},\dfrac{-6}{2\sqrt{17}},\dfrac{-4}{2\sqrt{17}}$$\\$ The direction ratios of $CA$ are $(-5 -3), (-5 -5),$ and $(-2 - (-4)) i.e., -8, -10,$and $2.$ Therefore, the direction cosines of $AC$ are$\\$ $\dfrac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\dfrac{-10}{\sqrt{(-8)^2+(10)^2+(2)^2}},\dfrac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}},\\ i. e. , \dfrac{-8}{2\sqrt{42}},\dfrac{-10}{2\sqrt{42}}\dfrac{2}{2\sqrt{42}}$

6   Show that the three lines with direction cosines$\\$ $\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13},\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13},\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$are mutually perpendicular.

Solution :

Two lines with direction cosines $l_ 1 , m_ 1 , n_ 1$ and $l_ 2 , m _2 , n_ 2 $, are perpendicular to each other, if $l_ 1 l_ 2 + m _1 m _2 + n_ 1 n_ 2 = 0$$\\$ (i) For the lines with direction cosines,$\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}$ and $ \dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13, }$ we obtain $l_1l_2+m_1m_2+n_1n_2=\dfrac{12}{13}*\dfrac{4}{13}+(\dfrac{-3}{13})*\dfrac{12}{13}+(\dfrac{-4}{13})*\dfrac{3}{13}\\ =\dfrac{48}{169}-\dfrac{36}{169}-\dfrac{12}{169}\\ =0$$\\$Therefore, the lines are perpendicular.

(ii)For the lines with direction cosines,$\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13}$ and $\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ we obtain $\\$ $l_1l_2+m_1m_2+n_1n_2\\ =\dfrac{4}{13}*\dfrac{3}{13}+\dfrac{12}{13}*(\dfrac{-4}{13})+\dfrac{3}{13}+\dfrac{12}{13}\\ =\dfrac{12}{169}-\dfrac{48}{169}+\dfrac{36}{169}\\ =0$$\\$ Therefore, the lines are perpendicular.$\\$

(iii)For the lines with direction cosines,$\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ and $ \dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}\\ =\dfrac{36}{169}+\dfrac{12}{169}-\dfrac{48}{169}\\ =0$$\\$ Therefore, the lines are perpendicular.$\\$ Thus, all the lines are mutually perpendicular.

7   Show that the line through the points $(1, -1, 2) (3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6).$

Solution :

Let $AB$ be the line joining the points, $(1, -1, 2)$ and $(3, 4, -2),$ and $CD$ be the line through the points $(0, 3, 2)$ and $(3, 5, 6).$$\\$ The direction ratios, $a_ 1 , b_ 1 , c_ 1 $, of $AB$ are $(3 -1), (4 – (-1)), $ and $(-2 -2)$ i.e.,$ 2, 5,$ and $-4.$$\\$ The direction ratios, $a_ 2 , b_ 2 , c_ 2$ , of $CD$ are $(3 -0), (5 -3),$ and $(6 -2)$ i.e.,$ 3, 2,$and $4.$$\\$ $AB$ and $CD$ will be perpendicular to each other, if $a _1 a_ 2 + b _1 b _2 + c _1 c_ 2 =0$$\\$ $a_ 1 a_ 2 + b _1 b_ 2 + c _1 c_ 2 = 2 * 3 + 5 * 2 +(- 4 )* 4\\ = 6 + 10 - 16\\ = 0$$\\$ Therefore, $AB$ and $CD$ are perpendicular to each other.

8   Show that the line through the points $(4, 7, 8) (2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1), (1, 2, 5).$

Solution :

Let $AB$ be the line through the points $(4, 7, 8)$ and $(2, 3, 4), CD$ be the line through the points, $(-1, -2, 1)$ and $(1, 2, 5).$$\\$ The directions ratios,$ a_ 1 , b_ 1 , c_ 1 $, of $AB$ are $(2 -4), (3 -7),$ and $(4 -8)$ i.e., $-2, -4,$ and $-4.$$\\$ The direction ratios, $a _2 , b_ 2 , c_ 2 $, of $CD$ are $(1 - (-1)), (2 - (-2)),$ and $(5 -1)$ i.e.,$ 2, 4,$ and $4.AB$ will be parallel to $CD$, if$\\$ $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\\ \dfrac{a_1}{a_2}=\dfrac{-2}{2}=-1\\ \dfrac{b_1}{b_2}=\dfrac{-4}{4}=-1\\ \dfrac{c_1}{c_2}=\dfrac{-4}{4}=-1\\ \therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$\\$ Thus, $AB$ is parallel to $CD.$

9   Find the equation of the line which passes through point $(1, 2, 3)$ and is parallel to the vector $3\hat{i}+2\hat{j}-2\hat{k}$.

Solution :

It is given that the line passes through the point $A (1, 2, 3)$. Therefore, the position vector through $A$ is $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$\\$ $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$$\\$ It is known that the line which passes through point $A$ and parallel to $\vec{b}$ is given by$\\$ $\vec{r}=\vec{a}+\lambda \vec{b},$ where $ \lambda $ is a constant$\\$ $\Rightarrow \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(3\hat{i}+2\hat{j}-2\hat{k})$$\\$ This is the required equation of the line.

10   Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$

Solution :

It is given that the line passes through the point with positive vector $\\$ $\vec{a}=2\hat{i}-\hat{j}+4\hat{k} \ \ .....(1)$$\\$ $\vec{b}=\hat{i}+2\hat{j}-\hat{k} \ \ ......(2)$$\\$ It is known that a line through a point with positive vector $\vec{a}$ and parallel to $\vec{b}$ is given by the equation,$\\$ $\vec{r}=\vec{a}+\lambda \vec{b}$$\\$ $Rightarrow \vec{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda (\hat{i}+2\hat{j}-\hat{k})$$\\$ This is the required equation of the line in vector form. $\\$ $\vec{r}=x\hat{i}-y\hat{j}+z\hat{k}$$\\$ $\Rightarrow x\hat{i}-y\hat{j}+z\hat{k}\\ =(\lambda +2)\hat{i}+(2\lambda-1)\hat{j}+(-\lambda+4)\hat{k}$$\\$ Eliminating $\lambda $, we obtain the Cartesian form equation as $\\$ $\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}$ This is the required of the given line in Cartesian form $\\$