The angle between the line $\vec{r}=(\hat{i}+2\hat{j}-\hat{k}) +\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\vec{r}.(2\hat{i}-\hat{j}+\hat{k})= 4$ is :
Answer & explanation
Correct answer: option 1
$\vec v$ → vector in direction of line
$\vec n$ → normal vector of plane
θ → angle between plane and line
$\vec v= \hat i-\hat j+\hat k$
$\vec n=2\hat i-\hat j+\hat k$
$|\vec v|=\sqrt{3}, |\vec n|=\sqrt{6}$
so $\vec v.\vec n=|\vec v||\vec n|\cos(90-θ)$
so $\sin θ=\frac{\vec v.\vec n}{|\vec v||\vec n|}$
$=\frac{4}{\sqrt{3}×\sqrt{2}×\sqrt{3}}$
$=\frac{2\sqrt{2}}{3}⇒θ=\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$