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 $Ax\!+\!By\!+\!Cz\!+\!D = 0$ , 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 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 $(1,\,1,\,1)$ . We can use the components $1,\,1,\,1$ for the direction vector of $l$ and the components $0,\,0,\,1$ for the normal vector of the plane. We have $$\omega = \arcsin\frac{1\!\cdot\!0\!+\!1\!\cdot\!0\!+\!1\!\cdot\!1}{\sqrt{1^2\!+\!1^2\!+\!1^2}\sqrt{0^2\!+\!0^2\!+\!1^2}} = \arcsin\frac{1}{\sqrt{3}} \approx 35.26^\circ.$$