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, $\sin\, \theta=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|$
$\Rightarrow \sin\, \theta=\left|\frac{(3 \hat{\imath}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{\imath}-2 \hat{\jmath}+\hat{k})}{\sqrt{9+16+25} \sqrt{4+4+1}}\right|$
$=\frac{6-8+5}{\sqrt{50} \sqrt{9}}=\frac{1}{5 \sqrt{2}}$
$=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{10}$