Question

Prove that the lines $x=py+q,z=ry+s$ and $x=p^{{}^{\prime }}y+q^{{}^{\prime }},z=r^{{}^{\prime }}y+s^{{}^{\prime }}$ are perpendicular if $p{p}^{{}^{\prime }}+r{r}^{{}^{\prime }}+1=0$

Answer 1

Given: $L_1:x=py+q,z=ry+s$

$\Rightarrow y=\frac{x-q}{p}=\frac{z-s}{r}$

$\Rightarrow \frac{x-q}{p}=\frac{y-0}{1}=\frac{z-s}{r}$

So, the vector parallel to the line $L_1$ is : $p\hat{i}+\hat{j}+r\hat{k}=\overrightarrow{a}$ (say)

Given: $L_2:x=p^{{}^{\prime }}y+q^{{}^{\prime }},z=r^{{}^{\prime }}y+s^{{}^{\prime }}$

$\Rightarrow y=\frac{x-q^{{}^{\prime }}}{p^{{}^{\prime }}}=\frac{z-s^{{}^{\prime }}}{r^{{}^{\prime }}}$

$\Rightarrow \frac{x-q^{{}^{\prime }}}{p^{{}^{\prime }}}=\frac{y-0}{1}=\frac{z-s^{{}^{\prime }}}{r^{{}^{\prime }}}$

So, the vector parallel to the line $L_2$ is: $p^{{}^{\prime }}\hat{i}+\hat{j}+r^{{}^{\prime }}\hat{k}=\overrightarrow{b}$( say )

Angle between two lines is as same as the angle between the parallel vectors to the lines. The lines $L_1$ and $L_2$ is perpendicular, if the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, so we get $\overrightarrow{a}.....\overrightarrow{b}=0$