# equation of plane

The position of a plane $\tau $ can be fixed by giving the position vector $\overrightarrow{OQ}$ of the projection^{} point $Q$ of the origin on the plane.

Let the length of the position vector be $r$ and the angles formed by the vector with the positive coordinate^{} axes $\alpha $, $\beta $, $\gamma $. Let $P=(x,y,z)$ be an arbitrary point. Then $P$ is in the plane $\tau $ iff its projection on the line $OQ$ coincides with $Q$, i.e. iff (http://planetmath.org/Iff) the projection of the coordinate way of $P$ is $r$. This may be expressed as the equation $x\mathrm{cos}\alpha +y\mathrm{cos}\beta +z\mathrm{cos}\gamma =r$ or

$x\mathrm{cos}\alpha +y\mathrm{cos}\beta +z\mathrm{cos}\gamma -r=0,$ | (1) |

which thus is the equation of the plane.

Conversely, we may show that a first-degree equation

$Ax+By+Cz+D=0$ | (2) |

between the variables $x$, $y$, $z$ represents always a plane. In fact, we may without hurting generality suppose that $D\leqq 0$. Now $R:=\sqrt{{A}^{2}+{B}^{2}+{C}^{2}}>0$. Thus the length of the radius vector (http://planetmath.org/PositionVector) of the point $(A,B,C)$ is $R$. Let the angles formed by the radius vector with the positive coordinate axes be $\alpha $, $\beta $, $\gamma $. Then we can write

$$A=R\mathrm{cos}\alpha ,B=R\mathrm{cos}\beta ,C=R\mathrm{cos}\gamma $$ |

(cf. direction cosines^{}). Dividing (2) termwise by $R$ gives us

$$x\mathrm{cos}\alpha +y\mathrm{cos}\beta +z\mathrm{cos}\gamma +\frac{D}{R}=0,$$ |

where $\frac{D}{R}\leqq 0$. The last equation represents a plane whose distance from the origin is $-\frac{D}{R}$ and whose normal line forms the angles $\alpha $, $\beta $, $\gamma $ with the coordinate axes.

Since the coefficients $A,B,C$ are proportional to the direction cosines of the normal vector^{} of this plane, they are direction numbers of the normal line of the plane.

Examples. The equations of the coordinate planes are

$x=0$ ($yz$-plane), $y=0$ ($zx$-plane), $z=0$ ($xy$-plane);

the equation of the plane through the points $(1,\mathrm{\hspace{0.17em}0},\mathrm{\hspace{0.17em}0})$, $(0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}0})$ and $(0,\mathrm{\hspace{0.17em}0},\mathrm{\hspace{0.17em}1})$ is

$x+y+z=1$.

The plane can be represented also in a vectoral form, by using the position vector ${\overrightarrow{r}}_{0}$ of a point of the plane and two linearly independent^{} vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ parallel^{} to the plane:

$\overrightarrow{r}={\overrightarrow{r}}_{0}+s\overrightarrow{u}+t\overrightarrow{v}.$ | (3) |

Here, $\overrightarrow{r}$ means the position vector of arbitrary point of the plane, $s$ and $t$ are real parameters. In the coordinate form, (3) may be e.g.

$\{\begin{array}{cc}x={x}_{0}+sa+td,\hfill & \\ y={y}_{0}+sb+te,\hfill & \\ z={z}_{0}+sc+tf.\hfill & \end{array}$ |

Title | equation of plane |

Canonical name | EquationOfPlane |

Date of creation | 2013-03-22 17:28:48 |

Last modified on | 2013-03-22 17:28:48 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 13 |

Author | pahio (2872) |

Entry type | Topic |

Classification | msc 51N20 |

Related topic | DirectionCosines |

Related topic | SurfaceNormal |

Related topic | RuledSurface |

Related topic | AnalyticGeometry |

Related topic | AngleBetweenLineAndPlane |

Related topic | IntersectionOfQuadraticSurfaceAndPlane |