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

##### 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   Maximise $Z = 3x + 4y$ Subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0$

##### Solution :

The feasible region determined by the constraints, $x + y \leq 4, x \geq 0, y \geq 0$, is as follows.$\\$ The corner points of the feasible region are $O ( 0, 0 ), A ( 4, 0 )$ , and $B ( 0, 4 )$ .$\\$ The values of $Z$ at these points are as follows.$\\$ table $\\$ Therefore, the maximum value of $Z$ is $16$ at the point $B ( 0, 4 )$

The feasible region determined by the constraints, $x + y \leq 4, x \geq 0, y \geq 0$, is as follows.$\\$ The corner points of the feasible region are $O ( 0, 0 ), A ( 4, 0 )$ , and $B ( 0, 4 )$ .$\\$ The values of $Z$ at these points are as follows.$\\$ table $\\$ Therefore, the maximum value of $Z$ is $16$ at the point $B ( 0, 4 )$