Question

Find the angles between the lines:
$\frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\text{and}\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$

Answer 1

Let

$L_1:\frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}$

$\Rightarrow \frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3}$

And,

$L_2:\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$

$\Rightarrow \frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4}$

Now,

$\cos \theta =\frac{\left|-2+14-12\right|}{\sqrt{4+49+9}\sqrt{1+4+16}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \cos \theta =\cos {90}^{\circ }$

$\therefore \theta ={90}^{\circ }$

Hence, the line are perpendicular and angle between line is ${90}^{\circ }.$