>Parametrization>Parameterization of a Plane

Parameterization of a Plane

A plane inside a three-dimentional coordinate system

A plane can be expressed through parametrization or an equation. The equation of a plane through a point $(x_{0}, y_{0}, z_{0})$ with a normal vector $\vec{n} = (a, b, c)$ looks like this:

Formula

The Equation of a Plane

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

This can also be written as

\begin{array}{l} a x + b y + c z + d = 0, \\ d = - a x_{0} - b y_{0} - c z_{0} \end{array}

a x + b y + c z + d = 0, d = - a x_{0} - b y_{0} - c z_{0}

if you solve the equation above. You can see that if you have the equation of a plane, you can use the numbers in front of

x

y

and

z

to find the normal vector.

You can find the parametric equation of a plane by using two vectors and a point. Given that $\vec{u} = (a_{1}, b_{1}, c_{1})$ and $\vec{v} = (a_{2}, b_{2}, c_{2})$ spans the plane and $(x_{0}, y_{0}, z_{0})$ is a point in the plane, the parametric equation can be written like this:

Theory

The Parametric Equation of a Plane

\begin{array}{l} x & = x_{0} + a_{1} s + a_{2} t \\ y & = y_{0} + b_{1} s + b_{2} t \\ z & = z_{0} + c_{1} s + c_{2} t \end{array}

You can find both the equation and the parametric equation of a plane if you know three points in the plane. With three points $A$ , $B$ and $C$ you can find $\vec{A B}$ and $\vec{A C}$ , which you can use to find a parametric equation for the plane. You can also use the cross product of these vectors to find a normal vector to the plane, which then can be used to find the equation of the plane.

Rule

The equation of the $x y$ -plane is $z = 0$

The equation of the $x z$ -plane is $y = 0$

The equation of the $y z$ -plane is $x = 0$

Example 1

If a plane has the equation $2 x + y - 3 z + 8 = 0$ , you can find a normal vector by looking at the numbers in front of the variables, giving you $\vec{n} = (2, 1, - 3)$ .

Example 2

The points $A = (2, 0, 1)$ , $B = (3, 1, 2)$ and $C = (0, 0, 4)$ are in a plane $α$ . Describe the plane with an equation and a parametric equation.

First, you find the vectors $\vec{A B}$ and $\vec{A C}$ : $\begin{array}{l} \vec{A B} & = (3 - 2, 1 - 0, 2 - 1) = (1, 1, 1), \\ \vec{A C} & = (0 - 2, 0 - 0, 4 - 1) = (- 2, 0, 3) \end{array}$

To find the parametric equation, you insert the point $A$ and the vectors $\vec{A B}$ and $\vec{A C}$ into the formula:

x = 2 + s - 2 t \land y = s \land z = 1 + s + 3 t

x = 2 + s - 2 t \land y = s \land z = 1 + s + 3 t

To find the equation of the plane, you need to find the cross product of

\vec{A B}

and

\vec{A C}

to get a normal vector

\vec{n}

\begin{array}{l} \vec{n} & = \vec{A B} \times \vec{A C} = (1, 1, 1) \times (- 2, 0, 3) \\ = (1 \cdot 3 - 1 \cdot 0, 1 \cdot (- 2) - 1 \cdot 3, \\ 1 \cdot 0 - 1 \cdot (- 2)) \\ = (3, - 5, 2) \end{array}

\begin{array}{l} \vec{n} = \vec{A B} \times \vec{A C} & = (1, 1, 1) \times (- 2, 0, 3) \\ = (1 \cdot 3 - 1 \cdot 0, 1 \cdot (- 2) - 1 \cdot 3, 1 \cdot 0 - 1 \cdot (- 2)) \\ = (3, - 5, 2) \end{array}

Then you can just insert

\vec{n}

and

A

into the general equation of a plane:

\begin{array}{l} 3 (x - 2) - 5 (y - 0) + 2 (z - 1) & = 0 \\ 3 x - 6 - 5 y + 2 z - 2 & = 0 \\ 3 x - 5 y + 2 z & = 8 . \end{array}

Example 3

A point $P = (4, - 4, 4)$ is in a plane with a normal vector $\vec{n} = (6, 1, 1)$ . What is the equation and the parametric equation of the plane?

You can find the equation of the plane first. You already have everything you need, so you can just insert the coordinates of the point and the values in the normal vector into the general equation of a plane:

\begin{array}{l} a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) & = 0 \\ 6 (x - 4) + 1 (- y - (- 4)) + 1 (z - 4) & = 0 \\ 6 x - 24 - y + 4 + z - 4 & = 0 \\ 6 x - y + z - 24 = 0 \end{array}

You can find another two points in the plane by using the equation of the plane. If you set two of the coordinates to $0$ , you will find the third coordinate very quickly. For example, you can set $x = y = 0$ to find $z$ : $\begin{array}{l} 6 \cdot 0 - 0 + z - 24 & = 0 \\ z & = 24 . \end{array}$

This means that $A = (0, 0, 24)$ is a point in the plane. You can do the same with $x = z = 0$ to find $y$ : $\begin{array}{l} 6 \cdot 0 - y + 0 - 24 & = 0 \\ y & = - 24 . \end{array}$

Then, $B = (0, - 24, 0)$ is a point in the plane as well. You now have three points in the plane, which can be used to create two vectors! That will give you

\begin{array}{l} \vec{P A} & = (0 - 4, 0 - (- 4), 24 - 4) \\ = (- 4, 4, 20), \\ \vec{P B} & = (0 - 4, - 24 - (- 4), 0 - 4) \\ = (- 4, - 20, - 4) . \end{array}

\begin{array}{l} \vec{P A} & = (0 - 4, 0 - (- 4), 24 - 4) = (- 4, 4, 20), \\ \vec{P B} & = (0 - 4, - 24 - (- 4), 0 - 4) = (- 4, - 20, - 4) . \end{array}

To find the parametric equation, insert the point

P

and the vectors

\vec{P A}

and

\vec{P B}

into the formula, and you get

\begin{array}{l} x & = 4 - 4 s - 4 t, \\ y & = - 4 + 4 s - 20 t, \\ z & = 4 + 20 s - 4 t . \end{array}

Want to know more?Sign UpIt's free!

Parameterization of Lines in Space

Go back