Question

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

• A. $\frac{3}{\sqrt{30}}$
• B. $\frac{3}{50}$
• C. $\frac{4}{5\sqrt{2}}$
• D. $\frac{\sqrt{2}}{10}$

Answer 1

Answer: (D)

Given line is $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
and plane is $2x-2y+z=5$
So, $\sin \theta =\left|\frac{\vec{b}\cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|$
$\Rightarrow \sin \theta =\left|\frac{(3\hat{ı}+4\hat{j}+5\hat{k})\cdot (2\hat{ı}-2\hat{ȷ}+\hat{k})}{\sqrt{9+16+25}\sqrt{4+4+1}}\right|$
$=\frac{6-8+5}{\sqrt{50}\sqrt{9}}=\frac{1}{5\sqrt{2}}$
$=\frac{1}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{10}$