Question

Find the angle between the line$\hat{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda (\hat{i}-\hat{j}+k)$ and the plane $\overline{r}.(2\hat{i}-\hat{j}+\hat{k})=4.$

Answer 1

$\begin{array}{c}\overrightarrow{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)\\ \overrightarrow{r}.\left(2\hat{i}-\hat{j}+\hat{k}\right)=4\\ \sin \theta =\frac{2+1+1}{\sqrt{6}\sqrt{3}}\\ =\frac{4}{\sqrt{3}\sqrt{3}\sqrt{2}}\\ =\frac{4}{3\sqrt{2}}\\ =\frac{2\sqrt{2}}{3}\\ \therefore \theta ={\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right).\end{array}$