Q77 / 125 · BITSAT 2014 · Mathematics
Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$
and the plane $10 \mathrm{x}+2 \mathrm{y}-11 \mathrm{z}=3$
- Correct answerA$\sin ^{-1}\left(\frac{8}{21}\right)$
- B$\sin ^{-1}\left(\frac{5}{21}\right)$
- C$\sin ^{-1}\left(\frac{7}{21}\right)$
- D$\sin ^{-1}\left(\frac{1}{21}\right)$
Solution
Let $\theta$ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have $\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$ and
$$
\overrightarrow{\mathrm{r}} \cdot(10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-11 \hat{\mathrm{k}})=3
$$
Here, $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{n}}=10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-11 \hat{\mathrm{k}}$
$$
\begin{array}{l}
\sin \phi=\left|\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k}) \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+11^{2}}}\right| \\
=\left|\frac{-40}{7 \times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21} \text { or } \phi=\sin ^{-1}\left(\frac{8}{21}\right)
\end{array}
$$
Hence, the correct answer is option: 'A'.