Question

Find the angle between the line $\overrightarrow{r}=(3,-2,4)+\lambda (2,2,1);\lambda \in R$ and the plane $2x-2y+z+7=0$.

Answer 1

$\overrightarrow{r}=(3,-2,4)+\lambda (2,2,1);\lambda \in R$
$\Rightarrow \overrightarrow{r}=(3\hat{i}-2\hat{j}+4\hat{k})+\lambda (2\hat{i}+2\hat{j}+\hat{k})$
Direction ratios of the given line are $(2,2,1)$
$\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$

Plane: $2x-2y+z+7=0$.
Direction ratios of the normal of given plane are $(2,-2,1)$
$\overrightarrow{n}=2\hat{i}-2\hat{j}+\hat{k}$

Hence, the required angle is
$\sin \theta =\left|\frac{\overrightarrow{b}\cdot \overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}\right|$
$\Rightarrow \sin \theta =\left|\frac{2\times 2+2\times (-2)+1\times 1}{\sqrt{4+4+1}\cdot \sqrt{4+4+1}}\right|$
$\Rightarrow \sin \theta =\frac{1}{9}$
$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{9}\right)$