Question

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

• A. $\frac{2\sqrt{3}}{5}$
• B. $\frac{\sqrt{2}}{10}$
• C. $\frac{4}{5\sqrt{2}}$
• D. $\frac{\sqrt{5}}{6}$

Answer 1

The correct option is B $\frac{\sqrt{2}}{10}$

$\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
$\Rightarrow \overrightarrow{b}=3\hat{i}+4\hat{j}+5\hat{k}$

$2x-2y+z=5$
$\Rightarrow \overrightarrow{n}=2\hat{i}-2\hat{j}+\hat{k}$

$\sin \theta =\left|\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}\right|$
$=\left|\frac{3\cdot 2-4\cdot 2+5\cdot 1}{\sqrt{9+16+25}\cdot \sqrt{4+4+1}}\right|$
$=\frac{3}{5\sqrt{2}\times 3}$
$=\frac{\sqrt{2}}{10}$