angle between line and plane

The angle between a line $l$ and a plane $\tau$ is defined as the least possible angle $\omega$ between $l$ and a line contained by $\tau$.

It is apparent that $\omega$ satisfies always  $0\leqq \omega \leqq {90}^{\circ }$.

Let the plane $\tau$ be given by the equation (http://planetmath.org/EquationOfPlane)  $Ax+By+Cz+D=0$,  i.e. its normal vector has the components $A,B,C$. Let a direction vector of the line $l$ have the components $a,b,c$. Then the angle $\omega$ between $l$ and $\tau$ is obtained from the equation

$$\sin \omega =\frac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}\sqrt{a^2+b^2+c^2}}.$$

In fact, the right hand side (http://planetmath.org/Equation) is the cosine of the angle $\alpha$ between $l$ and the surface normal of $\tau$ (see angle between two lines), and $\omega$ is the complementary angle of $\alpha$.

Example.  Consider the $xy$-plane and the line $l$ through the origin and the point  $(1,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}1})$. We can use the components $1,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}1}$ for the direction vector of $l$ and the components $0,\mathrm{\hspace{0.17em}0},\mathrm{\hspace{0.17em}1}$ for the normal vector of the plane. We have

$$\omega =\arcsin \frac{1\cdot 0+1\cdot 0+1\cdot 1}{\sqrt{1^2+1^2+1^2}\sqrt{0^2+0^2+1^2}}=\arcsin \frac{1}{\sqrt{3}}\approx {35.26}^{\circ }.$$