Question

The angle between the lines $2x=3y=-z$ and $6x=-y=-4z$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$
• E. $\frac{2\pi }{3}$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

Given lines can be written as
$\frac{x}{\left(\frac{1}{2}\right)}=\frac{y}{\left(\frac{1}{3}\right)}=\frac{z}{-1}$
and $\frac{x}{\left(\frac{1}{6}\right)}=\frac{y}{(-1)}=\frac{z}{(-\frac{1}{4})}$
If $\theta$ is angle between them
$\cos \theta =\frac{\frac{1}{2}\times \frac{1}{6}+\frac{1}{3}(-1)+(-1)\times (-\frac{1}{4})}{\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{3}\right)}^2+{(-1)}^2}\sqrt{{\left(\frac{1}{6}\right)}^2+{(-1)}^2+{(-\frac{1}{4})}^2}}$
$\Rightarrow \cos \theta =0$

Thus $\theta =\frac{\pi }{2}$