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. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

Answer 2

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 ${{a}_{2}}=1,{{b}_{2}}=-1,{{c}_{2}}=-2$
$\therefore$ $\sin \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1\times 1+(-1)\times (-1)+(-2)\times (-2)}{\sqrt{1+1+4}\sqrt{1+1+4}}$
$=\frac{6}{\sqrt{6}\sqrt{6}}=1$
$\Rightarrow$ $\theta =\frac{\pi }{2}$