Question

The plane 2x-3y+6z-11=0 makes an angle ${\sin }^{-1}\alpha$ with x-axis. The value of $\alpha$ is equal to
(a) $\frac{\sqrt{3}}{2}$

(b) $\frac{\sqrt{2}}{3}$

(c) $\frac{2}{7}$

(d) $\frac{3}{7}$

Answer 1

Given, plane 2x – 3y + 6z – 11 = 0 makes an angle sin^–1α with x-axis
$$\begin{gathered} \mathrm{Since}\overrightarrow{n}=2\hat{i}-3\hat{j}+6\hat{k}\mathrm{for}\mathrm{given}\mathrm{plane}\mathrm{and}\sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ \mathrm{Here}\theta \mathrm{is}\mathrm{angle}\mathrm{between}\overrightarrow{n}\mathrm{and}\overrightarrow{b}\mathrm{where}\overrightarrow{b}=\hat{i}\left(\because x-\mathrm{axis}\mathrm{is}\mathrm{given}\right) \\ \therefore \sin \left({\sin }^{-1}\alpha \right)=\frac{\hat{i}.\left(2\hat{i}-3\hat{j}+6\hat{k}\right)}{\sqrt{1^2}\sqrt{2^2+3^2+6^2}} \\ \mathrm{i}.\mathrm{e}.\sin \left({\sin }^{-1}\alpha \right)=\frac{2}{1\left(4+36+9\right)} \\ \mathrm{i}.\mathrm{e}.\alpha =\frac{2}{7} \end{gathered}$$
Hence, the correct answer is option C.