Question

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

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

Answer 1

Answer: (D)

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