Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Exercise

The number of straight lines that are equally inclined to three coordinate axes is

The vector equation of the line passing through the point $(-1,5,4)$ and perpendicular to the plane $z=0$ is 

If a vector makes angles $\alpha , \beta , \gamma$ with the co-ordinate axes, then show that, ${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$.

If $P (1, y, z)$ lies on the line through $(3,2,-1)$ and $(-4, 6, 3)$, find $y \& z$.

Find the direction cosines of the line $\frac{x-2}{2}=\frac{2y-5}{-3}=\frac{z+1}{0}$

Find the shortest distance between the two given straight lines $\overrightarrow{r}=\left(2\hat{i}+3\hat{j}+4\hat{k}\right)+t\left(-2\hat{i}+\hat{j}-2\hat{k}\right)$ and $\frac{x-3}{2}=\frac{y}{-1}=\frac{{\displaystyle z+2}}{2}.$

Show that the shortest distance between the lines $\mathrm{x} + \mathrm{a}=2\mathrm{y}=-12\mathrm{z}$ and $\mathrm{x}+\mathrm{y}+2\mathrm{a}=6\mathrm{z}-6\mathrm{a}$, is $2a$.

Find the shortest distance between the lines $\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1} \mathrm{and} \frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}$. Find also the equation of the line of the shortest distance.