Angle Between a Line and a Plane

Angle θ between line with DRs (a₁,b₁,c₁) and plane ax+by+cz+d=0. Note sin (not cos) because θ is measured from the plane, not from the normal. Complementary to the angle between the line and the normal.

Let line $L$ have direction vector $\mathbf{b} = (a_1,b_1,c_1)$ and plane $\Pi$ have normal $\mathbf{n} = (a,b,c)$.

The angle $\phi$ between $\mathbf{b}$ and $\mathbf{n}$ satisfies:

$$\cos\phi = \frac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$$

The angle $\theta$ between the line and the plane is the complement of $\phi$:

$$\theta = 90° - \phi \implies \sin\theta = \cos\phi$$ $$\sin\theta = \frac{|a_1a+b_1b+c_1c|}{\sqrt{a_1^2+b_1^2+c_1^2}\cdot\sqrt{a^2+b^2+c^2}}$$

Line perpendicular to plane: $\mathbf{b} \parallel \mathbf{n}$, i.e. $a_1/a = b_1/b = c_1/c$. Then $\theta = 90°$.

Line parallel to plane: $\mathbf{b} \perp \mathbf{n}$, i.e. $a_1a+b_1b+c_1c=0$. Then $\theta = 0°$ (line lies in the plane or is parallel to it).

Line in the plane: Satisfies both $\mathbf{b}\cdot\mathbf{n}=0$ (direction) and the point lies on the plane.