# Intersection of Line and Plane

Angle between Line and a Plane: The angle between a line and a plane is the complement of the angle between the line and normal to the plane.

Thus, if θ is the angle between a lines $$\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$$. And the plane ax + by + cz + d = 0. Then, the angle between the normal to the plane and the line is (90°- θ).

$$\cos \,\left( {{90}^{0}}-\theta \right)=\frac{al\,+\,bm\,+\,cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$$.

∴ $$\sin \theta =\,\frac{al\,+\,bm\,+\,cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$$.

If the line is parallel to the plane,

∴ al + bm + cn = 0.

If the line is perpendicular to the plane,

∴ $$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}$$

Example: Find the angle between the line $$\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-4}{0}$$ and the plane y + z + 2 = 0.

Solution: Let θ be the angle between the line $$\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-4}{0}$$ and the plane y + z + 2 = O.

∴ $$\sin \theta =\,\frac{al\,+\,bm\,+\,cn}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$$.

⇒ $$\sin \theta =\,\frac{1\times 1\,+1\times 0\,+0}{\sqrt{{{1}^{2}}+{{1}^{2}}+0}\sqrt{{{1}^{2}}+{{1}^{2}}+0}}$$.

Then,

⇒ Sin θ = ½

⇒ θ = π/6