Question

Find the angle between the line $\frac{x+1}{2}=\frac{4}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$:

Answer 1

Line:$\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ Plane:$10x+2y-11z=3$

Angle between line and plane is complement of angle between line and its normal .Directions of normal $(10,2,-11)$

$\cos (90-\theta )=\frac{2\times 10+3\times 2+6\times (-11)}{\sqrt{(2^2+3^2+6^2)({10}^2+2^2+{11}^2)}}=\frac{-8}{21}\therefore \theta ={90}^{\circ }-{\cos }^{-1}\left(\frac{-8}{21}\right)or\theta =\frac{\pi }{2}-{\cos }^{-1}\left(\frac{-8}{21}\right)$