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 $a_2=1,b_2=-1,c_2=-2$
$\therefore$ $\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{1\times 1+(-1)\times (-1)+(-2)\times (-2)}{\sqrt{1+1+4}\sqrt{1+1+4}}$
$=\frac{6}{\sqrt{6}\sqrt{6}}=1$
$\Rightarrow$ $\theta =\frac{\pi }{2}$