Question

Show that the angle between the line $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ and plane $\overrightarrow{r}.\overrightarrow{n}=d$ is $\frac{\pi }{2}-{\cos }^{-1}\left(\frac{\overrightarrow{b}.\overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}\right)$.

Answer 1

Let $Q$ is the angle between

Line $r=\overrightarrow{a}+\lambda \overrightarrow{b}$ $\xi$ plane $\overrightarrow{r}.\overrightarrow{n}=p$

${90}^{\circ }-Q\to$ angle between line and $n$.

$\Rightarrow$ $Cos({90}^{\circ }-Q)=\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$

$\Rightarrow 90-Q={Cos}^{-1}\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$

$\Rightarrow$ $Q=\frac{\pi }{2}-{Cos}^{-1}\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$

Hence proved.