Plane (geometry)

In a Euclidean space of any number of dimensions, a plane is uniquely determined by any of the following:

• Three non-collinear points (points not on a single line).
• A line and a point not on that line.
• Two distinct but intersecting lines.
• Two distinct but parallel lines.

The following statements hold in three-dimensional Euclidean space but not in higher dimensions, though they have higher-dimensional analogues:

• Two distinct planes are either parallel or they intersect in a line.
• A line is either parallel to a plane, intersects it at a single point, or is contained in the plane.
• Two distinct lines perpendicular to the same plane must be parallel to each other.
• Two distinct planes perpendicular to the same line must be parallel to each other.

In a manner analogous to the way lines in a two-dimensional space are described using a point-slope form for their equations, planes in a three dimensional space have a natural description using a point in the plane and a vector orthogonal to it (the normal vector) to indicate its "inclination".

Specifically, let r₀ be the position vector of some point P₀ = (x₀, y₀, z₀), and let n = (a, b, c) be a nonzero vector. The plane determined by the point P₀ and the vector n consists of those points P, with position vector r, such that the vector drawn from P₀ to P is perpendicular to n. Recalling that two vectors are perpendicular if and only if their dot product is zero, it follows that the desired plane can be described as the set of all points r such that

$${\displaystyle {\boldsymbol {n}}\cdot ({\boldsymbol {r}}-{\boldsymbol {r}}_{0})=0.}$$

The dot here means a dot (scalar) product.
Expanded this becomes

$${\displaystyle a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0,}$$

which is the point–normal form of the equation of a plane.[5] This is just a linear equation

$${\displaystyle ax+by+cz+d=0,}$$

where

$${\displaystyle d=-(ax_{0}+by_{0}+cz_{0}),}$$

which is the expanded form of ${\displaystyle -{\boldsymbol {n}}\cdot {\boldsymbol {r}}_{0}.}$

In mathematics it is a common convention to express the normal as a unit vector, but the above argument holds for a normal vector of any non-zero length.

Conversely, it is easily shown that if a, b, c, and d are constants and a, b, and c are not all zero, then the graph of the equation

$${\displaystyle ax+by+cz+d=0,}$$

is a plane having the vector n = (a, b, c) as a normal.[6] This familiar equation for a plane is called the general form of the equation of the plane.[7]

Thus for example a regression equation of the form y = d + ax + cz (with b = −1) establishes a best-fit plane in three-dimensional space when there are two explanatory variables.

Alternatively, a plane may be described parametrically as the set of all points of the form

$${\displaystyle {\boldsymbol {r}}={\boldsymbol {r}}_{0}+s{\boldsymbol {v}}+t{\boldsymbol {w}},}$$

Vector description of a plane

where s and t range over all real numbers, v and w are given linearly independent vectors defining the plane, and r₀ is the vector representing the position of an arbitrary (but fixed) point on the plane. The vectors v and w can be visualized as vectors starting at r₀ and pointing in different directions along the plane. The vectors v and w can be perpendicular, but cannot be parallel.

Let p₁ = (x₁, y₁, z₁), p₂ = (x₂, y₂, z₂), and p₃ = (x₃, y₃, z₃) be non-collinear points.

The plane passing through p₁, p₂, and p₃ can be described as the set of all points (x,y,z) that satisfy the following determinant equations:

$${\displaystyle {\begin{vmatrix}x-x_{1}&y-y_{1}&z-z_{1}\\x_{2}-x_{1}&y_{2}-y_{1}&z_{2}-z_{1}\\x_{3}-x_{1}&y_{3}-y_{1}&z_{3}-z_{1}\end{vmatrix}}={\begin{vmatrix}x-x_{1}&y-y_{1}&z-z_{1}\\x-x_{2}&y-y_{2}&z-z_{2}\\x-x_{3}&y-y_{3}&z-z_{3}\end{vmatrix}}=0.}$$

To describe the plane by an equation of the form ${\displaystyle ax+by+cz+d=0}$, solve the following system of equations:

$${\displaystyle ax_{1}+by_{1}+cz_{1}+d=0}$$

$${\displaystyle ax_{2}+by_{2}+cz_{2}+d=0}$$

$${\displaystyle ax_{3}+by_{3}+cz_{3}+d=0.}$$

This system can be solved using Cramer's rule and basic matrix manipulations. Let

$${\displaystyle D={\begin{vmatrix}x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\\x_{3}&y_{3}&z_{3}\end{vmatrix}}.}$$

If D is non-zero (so for planes not through the origin) the values for a, b and c can be calculated as follows:

$${\displaystyle a={\frac {-d}{D}}{\begin{vmatrix}1&y_{1}&z_{1}\\1&y_{2}&z_{2}\\1&y_{3}&z_{3}\end{vmatrix}}}$$

$${\displaystyle b={\frac {-d}{D}}{\begin{vmatrix}x_{1}&1&z_{1}\\x_{2}&1&z_{2}\\x_{3}&1&z_{3}\end{vmatrix}}}$$

$${\displaystyle c={\frac {-d}{D}}{\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}}.}$$

These equations are parametric in d. Setting d equal to any non-zero number and substituting it into these equations will yield one solution set.

This plane can also be described by the "point and a normal vector" prescription above. A suitable normal vector is given by the cross product

$${\displaystyle {\boldsymbol {n}}=({\boldsymbol {p}}_{2}-{\boldsymbol {p}}_{1})\times ({\boldsymbol {p}}_{3}-{\boldsymbol {p}}_{1}),}$$

and the point r₀ can be taken to be any of the given points p₁, p₂ or p₃[8] (or any other point in the plane).

For a plane ${\displaystyle \Pi :ax+by+cz+d=0}$ and a point ${\displaystyle {\boldsymbol {p}}_{1}=(x_{1},y_{1},z_{1})}$ not necessarily lying on the plane, the shortest distance from ${\displaystyle {\boldsymbol {p}}_{1}}$ to the plane is

${\displaystyle D={\frac {\left|ax_{1}+by_{1}+cz_{1}+d\right|}{\sqrt {a^{2}+b^{2}+c^{2}}}}.}$

It follows that ${\displaystyle {\boldsymbol {p}}_{1}}$ lies in the plane if and only if D = 0.

If ${\displaystyle a^{2}+b^{2}+c^{2}=1}$, meaning that a, b, and c are normalized,[9] then the equation becomes

${\displaystyle D=\left|ax_{1}+by_{1}+cz_{1}+d\right|.}$

Another vector form for the equation of a plane, known as the Hesse normal form relies on the parameter D. This form is:[7]

${\displaystyle {\boldsymbol {n}}\cdot {\boldsymbol {r}}-D_{0}=0,}$

where ${\displaystyle {\boldsymbol {n}}}$ is a unit normal vector to the plane, ${\displaystyle {\boldsymbol {r}}}$ a position vector of a point of the plane and D₀ the distance of the plane from the origin.

The general formula for higher dimensions can be quickly arrived at using vector notation. Let the hyperplane have equation ${\displaystyle {\boldsymbol {n}}\cdot ({\boldsymbol {r}}-{\boldsymbol {r}}_{0})=0}$, where the ${\displaystyle {\boldsymbol {n}}}$ is a normal vector and ${\displaystyle {\boldsymbol {r}}_{0}=(x_{10},x_{20},\dots ,x_{N0})}$ is a position vector to a point in the hyperplane. We desire the perpendicular distance to the point ${\displaystyle {\boldsymbol {r}}_{1}=(x_{11},x_{21},\dots ,x_{N1})}$. The hyperplane may also be represented by the scalar equation ${\textstyle \sum _{i=1}^{N}a_{i}x_{i}=-a_{0}}$, for constants ${\displaystyle \{a_{i}\}}$. Likewise, a corresponding ${\displaystyle {\boldsymbol {n}}}$ may be represented as ${\displaystyle (a_{1},a_{2},\dots ,a_{N})}$. We desire the scalar projection of the vector ${\displaystyle {\boldsymbol {r}}_{1}-{\boldsymbol {r}}_{0}}$ in the direction of ${\displaystyle {\boldsymbol {n}}}$. Noting that ${\displaystyle {\boldsymbol {n}}\cdot {\boldsymbol {r}}_{0}={\boldsymbol {r}}_{0}\cdot {\boldsymbol {n}}=-a_{0}}$ (as ${\displaystyle {\boldsymbol {r}}_{0}}$ satisfies the equation of the hyperplane) we have

${\displaystyle {\begin{aligned}D&={\frac {|({\boldsymbol {r}}_{1}-{\boldsymbol {r}}_{0})\cdot {\boldsymbol {n}}|}{|{\boldsymbol {n}}|}}\\&={\frac {|{\boldsymbol {r}}_{1}\cdot {\boldsymbol {n}}-{\boldsymbol {r}}_{0}\cdot {\boldsymbol {n}}|}{|{\boldsymbol {n}}|}}\\&={\frac {|{\boldsymbol {r}}_{1}\cdot {\boldsymbol {n}}+a_{0}|}{|{\boldsymbol {n}}|}}\\&={\frac {|a_{1}x_{11}+a_{2}x_{21}+\dots +a_{N}x_{N1}+a_{0}|}{\sqrt {a_{1}^{2}+a_{2}^{2}+\dots +a_{N}^{2}}}}.\end{aligned}}}$

In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line.

Two intersecting planes in three-dimensional space

The line of intersection between two planes ${\displaystyle \Pi _{1}:{\boldsymbol {n}}_{1}\cdot {\boldsymbol {r}}=h_{1}}$ and ${\displaystyle \Pi _{2}:{\boldsymbol {n}}_{2}\cdot {\boldsymbol {r}}=h_{2}}$ where ${\displaystyle {\boldsymbol {n}}_{i}}$ are normalized is given by

${\displaystyle {\boldsymbol {r}}=(c_{1}{\boldsymbol {n}}_{1}+c_{2}{\boldsymbol {n}}_{2})+\lambda ({\boldsymbol {n}}_{1}\times {\boldsymbol {n}}_{2})}$

where

${\displaystyle c_{1}={\frac {h_{1}-h_{2}({\boldsymbol {n}}_{1}\cdot {\boldsymbol {n}}_{2})}{1-({\boldsymbol {n}}_{1}\cdot {\boldsymbol {n}}_{2})^{2}}}}$

${\displaystyle c_{2}={\frac {h_{2}-h_{1}({\boldsymbol {n}}_{1}\cdot {\boldsymbol {n}}_{2})}{1-({\boldsymbol {n}}_{1}\cdot {\boldsymbol {n}}_{2})^{2}}}.}$

This is found by noticing that the line must be perpendicular to both plane normals, and so parallel to their cross product ${\displaystyle {\boldsymbol {n}}_{1}\times {\boldsymbol {n}}_{2}}$ (this cross product is zero if and only if the planes are parallel, and are therefore non-intersecting or entirely coincident).

The remainder of the expression is arrived at by finding an arbitrary point on the line. To do so, consider that any point in space may be written as ${\displaystyle {\boldsymbol {r}}=c_{1}{\boldsymbol {n}}_{1}+c_{2}{\boldsymbol {n}}_{2}+\lambda ({\boldsymbol {n}}_{1}\times {\boldsymbol {n}}_{2})}$, since ${\displaystyle \{{\boldsymbol {n}}_{1},{\boldsymbol {n}}_{2},({\boldsymbol {n}}_{1}\times {\boldsymbol {n}}_{2})\}}$ is a basis. We wish to find a point which is on both planes (i.e. on their intersection), so insert this equation into each of the equations of the planes to get two simultaneous equations which can be solved for ${\displaystyle c_{1}}$ and ${\displaystyle c_{2}}$.

If we further assume that ${\displaystyle {\boldsymbol {n}}_{1}}$ and ${\displaystyle {\boldsymbol {n}}_{2}}$ are orthonormal then the closest point on the line of intersection to the origin is ${\displaystyle {\boldsymbol {r}}_{0}=h_{1}{\boldsymbol {n}}_{1}+h_{2}{\boldsymbol {n}}_{2}}$. If that is not the case, then a more complex procedure must be used.[10]

Given two intersecting planes described by ${\displaystyle \Pi _{1}:a_{1}x+b_{1}y+c_{1}z+d_{1}=0}$ and ${\displaystyle \Pi _{2}:a_{2}x+b_{2}y+c_{2}z+d_{2}=0}$, the dihedral angle between them is defined to be the angle ${\displaystyle \alpha }$ between their normal directions:

${\displaystyle \cos \alpha ={\frac {{\hat {n}}_{1}\cdot {\hat {n}}_{2}}{|{\hat {n}}_{1}||{\hat {n}}_{2}|}}={\frac {a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{{\sqrt {a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}{\sqrt {a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}}}.}$