Question

If the plane $2x-3y+6z-11=0$ makes an angle $si{n}^{-1}\left(k\right)$ with x-axis, then k is equal to

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{7}$
• C. $\frac{\sqrt{2}}{7}$
• D. $1$

Answer 1

The correct option is B $\frac{2}{7}$

Given plane is $2x-3y+6z-11=0$

We know $sin\theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}$

Here $\overrightarrow{n}$ is the normal vector to the plane

$\overrightarrow{n}=2\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k}$

and $\overrightarrow{b}$ is along x axis.

$\therefore sin\theta =\frac{(2\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k}).\overrightarrow{i}}{\sqrt{2^2+(-3{)}^2+6^2.}\sqrt{1}2}$

$\therefore \frac{2}{\sqrt{4}9}=\frac{2}{7}$.