Question

Find angle between lines $x-3y-4=0,4y-z+5=0$, and $x+3y-11=0,2y-z+6=0$

Answer 1

Vector along the line of intersection of planes

$x-3y-4=0$ and $4y-z+5=0$ is

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 0\\ 0& 4& -1\end{array}\right|=3i+j+4\hat{k}$

Vector along the line of intersection of planes

$x+3y-11=0$ and $2y-z+6=0$ is

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& 0\\ 0& 2& -1\end{array}\right|=-3i+j+2\hat{k}$

$\therefore$ Angle between lines be $\theta$

$\therefore \cos \theta =\frac{-9+1+8}{\sqrt{9+1+16}\sqrt{9+1+4}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta ={90}^o$