Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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 $\overrightarrow{\mathrm{n}}=10\hat{\mathrm{i}}+2\hat{\mathrm{j}}-11\hat{\mathrm{k}}$ $$\begin{array}{l}\sin \varphi =\left|\frac{(2\hat{i}+3\hat{j}+6\hat{k})\cdot (10\hat{i}+2\hat{j}-11\hat{k})}{\sqrt{2^2+3^2+6^2}\sqrt{10^2+2^2+11^2}}\right|\\ =\left|\frac{-40}{7\times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21}\text{ or }\varphi ={\sin }^{-1}\left(\frac{8}{21}\right)\end{array}$$