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. None of these

Answer 1

The correct option is D None of these

Given line can be rewritten as
$\frac{x-\frac{1}{3}}{1}=\frac{y+3}{-1}=\frac{(z-\frac{5}{2})}{-2}$

and equation of plane is rewritten as
$x-y-2z=\frac{10}{3}$

Here, direction ratios of line and plane are
$a_1=1$, $b_1=-1$ and $c_1=-2$
and $a_2=1$, $b_2=-1$ and $c_2=-2$

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

$\Rightarrow \sin \theta =1$

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