Finding the equation of a plane

Formulas for the equation of a plane

The equation of a plane is given by the formula

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

where $\langle{a},b,c\rangle$ are the direction numbers from the normal vector to the plane.

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Given three points in the plane $P(P_1,P_2,P_3)$, $Q(Q_1,Q_2,Q_3)$ and $R(R_1,R_2,R_3)$, we can find the equation of the plane by

using the points to generate two vectors

$\vec{PQ}=\langle(Q_1-P_1),(Q_2-P_2),(Q_3-P_3)\rangle$

$\vec{PR}=\langle(R_1-P_1),(R_2-P_2),(R_3-P_3)\rangle$,

taking the cross product of $\vec{PQ}$ and $\vec{PR}$ to get the normal vector to the plane

$\vec{PQ}\times\vec{PR}=\begin{vmatrix}\bold i&\bold j&\bold k\\PQ_1&PQ_2&PQ_3\\PR_1&PR_2&PR_3\end{vmatrix}=\bold i\begin{vmatrix}PQ_2&PQ_3\\PR_2&PR_3\end{vmatrix}-\bold j\begin{vmatrix}PQ_1&PQ_3\\PR_1&PR_3\end{vmatrix}+\bold k\begin{vmatrix}PQ_1&PQ_2\\PR_1&PR_2\end{vmatrix}$

$=\bold i(PQ_2PR_3-PQ_3PR_2)-\bold j(PQ_1PR_3-PQ_3PR_1)$

$+\bold k(PQ_1PR_2-PQ_2PR_1)$

and then plugging the given points and the normal vector into the formula for the equation of the plane.

Given three points that lie in the plane, we’ll use a specific formula to find th equation of the plane

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Finding a plane from three points that lie in the plane

Example

Find the equation of the plane that passes through the given points.

$P(1,0,2)$

$Q(2,-1,3)$

$R(1,-1,2)$

We’ll start by using the given points $P$, $Q$ and $R$ to find two vectors $\vec{PQ}$ and $\vec{PR}$ that lie in the plane.

$\vec{PQ}=\left\langle(2-1),(-1-0),(3-2)\right\rangle$

$\vec{PQ}=\left\langle1,-1,1\right\rangle$

and

$\vec{PR}=\left\langle(1-1),(-1-0),(2-2)\right\rangle$

$\vec{PR}=\left\langle0,-1,0\right\rangle$

We’ll start by using the given points P, Q and R to find two vectors PQ and PR that lie in the plane.

Taking the cross product of these two vectors, we get

$\vec{PQ}\times\vec{PR}=\begin{vmatrix}\bold i&\bold j&\bold k\\1&-1&1\\0&-1&0\end{vmatrix}$

$\vec{PQ}\times\vec{PR}=\bold i\begin{vmatrix}-1&1\\-1&0\end{vmatrix}-\bold j\begin{vmatrix}1&1\\0&0\end{vmatrix}+\bold k\begin{vmatrix}1&-1\\0&-1\end{vmatrix}$

$\vec{PQ}\times\vec{PR}=\left[(-1)(0)-(1)(-1)\right]\bold i-\left[(1)(0)-(1)(0)\right]\bold j+\left[(1)(-1)-(-1)(0)\right]\bold k$

$\vec{PQ}\times\vec{PR}=(0+1)\bold i-(0-0)\bold j+(-1-0)\bold k$

$\vec{PQ}\times\vec{PR}=1\bold i-0\bold j-1\bold k$

$\vec{PQ}\times\vec{PR}=\langle 1,0,-1\rangle$

Now we’ll plug any of the given points, we’ll use $P$, and the direction numbers from the cross product into the formula for the equation of the plane.

$(1)(x-1)+(0)(y-0)+(-1)(z-2)=0$

$x-1-z+2=0$

$x-z=-1$

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