The angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y -11z=3$ is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\sin^{-1}(\frac{8}{21})$
$\vec v$(parallel to line) = $2\hat i+3\hat j+6\hat k$
$\vec n$ (perpendicular to plane) = $10\hat i+2\hat j-11\hat k$
θ → angle between line and plane
$90 - θ$ → angle between normal and plane
$|\vec v.\vec n|=|\vec v||\vec n|\cos(90 - θ)$
$=|-40|=7×15\sin θ⇒\sin θ=\frac{40}{7×15}$
$θ=\sin^{-1}(\frac{8}{21})$