Question

The angle between the line $ \frac{3x-1}{3}=\frac{y+3}{-1} $ $ =\frac{5-2z}{4} $ and the plane $ 3x-3y-6z=10 $ is equal to

• 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 and planes are
$ \frac{3x-1}{3}=\frac{y+3}{-1}=\frac{5-2z}{4} $ Or
$ \frac{x-\frac{1}{3}}{1}=\frac{y+3}{-1}=\frac{\left( z-\frac{5}{2} \right)}{-2} $ and $ 3x-3y-6z=0 $
$ \Rightarrow $ $ x-y-2z=0 $
Here, $ {{a}_{1}}=1,{{b}_{1}}=-1,{{c}_{1}}=-2 $
and $ {{a}_{2}}=1,{{b}_{2}}=-1,{{c}_{2}}=-2 $
$ \therefore $ $ \sin \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} $
$=\frac{1\times 1+(-1)\times (-1)+(-2)\times (-2)}{\sqrt{1+1+4}\sqrt{1+1+4}} $
$=\frac{6}{\sqrt{6}\sqrt{6}}=1 $
$ \Rightarrow $ $ \theta =\frac{\pi }{2} $