Question

The angle between the line $\overrightarrow{r}=(5\hat{i}-\hat{j}-4\hat{k})+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}·(3\hat{i}-4\hat{j}-\hat{k})+5=0$ is ____________.

Answer 1

Since the angle between the line

$$\begin{gathered} \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\mathrm{and}\mathrm{plane}\overrightarrow{r}.\overrightarrow{n}=d\mathrm{is}\mathrm{given}\mathrm{by} \\ \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ \mathrm{Given}\mathrm{line}\overrightarrow{r}=\left(5\hat{i}-\hat{j}-4\hat{k}\right)+\lambda \left(2\hat{i}-\hat{j}+\hat{k}\right) \\ \overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k}\mathrm{and}\mathrm{the}\mathrm{plane}\overrightarrow{r}.\left(3\hat{i}-4\hat{j}-\hat{k}\right)+5=0 \\ \overrightarrow{n}=\left(3\hat{i}-4\hat{j}-\hat{k}\right) \end{gathered}$$
$\begin{array}{rcl}\mathrm{i}.\mathrm{e}.\sin \theta & =& \frac{\left(2\hat{i}-\hat{j}+\hat{k}\right).\left(3\hat{i}-4\hat{j}-\hat{k}\right)}{\left|2\hat{i}-\hat{j}+\hat{k}\right|\left|3\hat{i}-4\hat{j}-\hat{k}\right|}\\ & =& \frac{6+4-1}{\sqrt{4+1+1}\sqrt{9+16+1}}\\ & =& \frac{9}{\sqrt{6}\sqrt{26}}\\ \sin \theta & =& \frac{9}{\sqrt{156}}\end{array}$
i.e. angle between given line and plane is ${\sin }^{-1}\left(\frac{9}{\sqrt{156}}\right)$