## You are here

Homeangle between line and plane

## Primary tabs

# 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 $Ax\!+\!By\!+\!Cz\!+\!D=0$, 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

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

## Mathematics Subject Classification

51N20*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections