Question

If the two lines $l_1:\frac{x-2}{3}=\frac{y+1}{-2},z=2$ and $l_2:\frac{x-1}{1}=\frac{2y+3}{\alpha }=\frac{z+5}{2}$ are perpendicular, then an angle between the lines $l_2$ and $l_3:\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is :

Answer 1

$\because l_1\text{and}{l}_2$ are perpendicular, so

$3\times 1+(-2)\left(\frac{\alpha }{2}\right)+0\times 2=0$

$\Rightarrow \alpha =3$

Now angle between $l_2\text{and}{l}_3$,

$\cos \theta =\frac{1(-3)+\frac{\alpha }{2}(-2)+2\left(4\right)}{\sqrt{1+\frac{{\alpha }^2}{4}+4\sqrt{9+4+16}}}$
$\Rightarrow \cos \theta =\frac{2}{\frac{29}{2}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{29}\right)={\sec }^{-1}\left(\frac{29}{4}\right)$