# 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.

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 $(-5, -5, -2)$

(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 $(-1, -2, 1), (1, 2, 5).$

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 $\\$