Question

The plane $2x–3y+6z–11=0$ makes an angle ${\sin }^{–1}\left(\alpha \right)$ with the x-axis. Then the value of $\alpha$ is

$\left(a\right)\frac{1}{7}\left(b\right)\frac{2}{7}\left(c\right)\frac{3}{7}\left(d\right)\frac{6}{7}$

Answer 1

Let the acute angle made by the plane with the x-axis is $\alpha$ and can be calculated as
$\sin \alpha =\frac{\text{coefficient of}x}{\sqrt{(\text{coefficient of}x{)}^2+(\text{coefficient of}y{)}^2+(\text{coefficient of}z{)}^2}}$
$=\frac{2}{\sqrt{(2^2+(-3{)}^2+6^2)}}=\frac{2}{7}$
$\Rightarrow \alpha ={\sin }^{-1}\frac{2}{7}$
$\therefore \alpha =\frac{2}{7}$
Ans : $\left(b\right)$