Question

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

• A. ${\sin }^{-1}\left(\frac{8}{21}\right)$
• B. ${\sin }^{-1}\left(\frac{5}{21}\right)$
• C. ${\sin }^{-1}\left(\frac{7}{21}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{21}\right)$

Answer 1

Answer: (A)

Let $\theta$ be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have
$\vec{r}=\left(-\hat{i}+3\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$ and
$\vec{r}.\left(10\hat{i}+2\hat{j}-11\hat{k}\right)=3$
Here, $\vec{b}=2\hat{i}+3\hat{j}+6\hat{k}$ and $\vec{n}=10\hat{i}+2\hat{j}-11\hat{k}$
$\sin \varphi =\left|\frac{\left(2\hat{i}+3\hat{j}+6\hat{k}\right).\left(10\hat{i}+2\hat{j}-11\hat{k}\right)}{\sqrt{2^2+3^2+6^2}\sqrt{10^2+2^2+11^2}}\right|$
$=\left|\frac{-40}{7\times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21}$ or $\varphi ={\sin }^{-1}\left(\frac{8}{21}\right)$