The equation of a plane

A triangular shade sail is stretched between three points: two pegs in the ground and a hook on a pole in the corner of a yard. With the foot of the pole at the origin and the metre as unit, the corners of the sail are $A(3, 0, 0)$, $B(0, 6, 0)$ and $C(0, 0, 3)$. A filming drone hovers at $M(3, 6, 3)$. How far is the drone from the plane of the sail?

To answer, we need an equation that describes every point of the plane $(ABC)$. This lesson gives us that equation and a distance formula that takes only a few calculations.

Normal vectors of a plane

Exploring: a vector perpendicular to the sail

We have $\overrightarrow{AB}=(-3, 6, 0)$ and $\overrightarrow{AC}=(-3, 0, 3)$. Try the vector $\vec{n}=(2, 1, 2)$: $$\vec{n}\cdot\overrightarrow{AB}=-6+6+0=0,\qquad \vec{n}\cdot\overrightarrow{AC}=-6+0+6=0.$$ So $\vec{n}$ is perpendicular to two non-parallel edges of the sail, hence to the whole sail. A point $P(x, y, z)$ lies in the plane $(ABC)$ exactly when $\overrightarrow{AP}\perp\vec{n}$, that is $$2(x-3)+1\cdot(y-0)+2(z-0)=0\iff 2x+y+2z-6=0.$$

When we only know a pair of direction vectors, we need a quick way to find a vector perpendicular to both.

Why is it true? Take the scalar product with $\vec{u}$: $a(bc'-b'c)+b(ca'-c'a)+c(ab'-a'b)=0$, because the terms cancel in pairs. In the same way $\vec{n}\cdot\vec{v}=0$. Since $\vec{u}$, $\vec{v}$ are not parallel, $\vec{n}\neq\vec{0}$.

How to remember it. Write the coordinates of $\vec{u}$ on top and $\vec{v}$ below. For the first coordinate, cover the first column and work out the remaining $2\times2$ determinant; for the second, cover the second column and cross-multiply column 3 with column 1; for the third, cover the third column.

The Cartesian equation of a plane

Why is it true? A point $M(x, y, z)$ lies in the plane exactly when $\overrightarrow{M_0M}=(x-x_0, y-y_0, z-z_0)$ is perpendicular to $\vec{n}$, that is, when their scalar product is $0$. Expanding gives the form $Ax+By+Cz+D=0$ with $D=-(Ax_0+By_0+Cz_0)$.

When we are given three points, we build a pair of direction vectors from them.

Special planes

Consider $(P): Ax+By+Cz+D=0$.

• $D=0$: $(P)$ passes through the origin $O$.
• $A=0$: $\vec{n}=(0, B, C)$ is perpendicular to $\vec{i}$, so $(P)$ is parallel to the axis $Ox$ or contains it. Likewise for $B=0$ (axis $Oy$) and $C=0$ (axis $Oz$).
• $A=B=0$: $(P): z=z_0$ is parallel to the plane $(Oxy)$ or equal to it. The three coordinate planes are $(Oxy): z=0$, $(Oyz): x=0$, $(Ozx): y=0$.
• A plane meeting the three axes at $(a, 0, 0)$, $(0, b, 0)$, $(0, 0, c)$ with $abc\neq0$ has the intercept form $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.$$

For the sail, $a=3$, $b=6$, $c=3$, so $\dfrac{x}{3}+\dfrac{y}{6}+\dfrac{z}{3}=1$; multiplying both sides by $6$ gives $2x+y+2z-6=0$ again.

Parallel and perpendicular planes

Two parallel planes have parallel normal vectors. Two perpendicular planes have perpendicular normal vectors.

The distance from a point to a plane

Why is it true? Let $H$ be the orthogonal projection of $M_0$ on $(P)$. Then $\overrightarrow{HM_0}=t\vec{n}$ for some number $t$, and $M_0H=|t|\cdot|\vec{n}|$. Since $H$ lies in $(P)$, substituting $H=M_0-t\vec{n}$ into the equation gives $Ax_0+By_0+Cz_0+D=t(A^2+B^2+C^2)$. Solving for $t$ and multiplying by $|\vec{n}|=\sqrt{A^2+B^2+C^2}$ gives the formula.

The last example has more steps, so we follow the four stages.

Key terms

Summary

• The plane through $M_0(x_0, y_0, z_0)$ with normal vector $\vec{n}=(A, B, C)$: $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$.
• $Ax+By+Cz+D=0$ has normal vector $(A, B, C)$.
• Direction vectors $\vec{u}=(a, b, c)$, $\vec{v}=(a', b', c')$ give $\vec{n}=\vec{u}\times\vec{v}=(bc'-b'c,\ ca'-c'a,\ ab'-a'b)$.
• Intercept form: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$; $(Oxy): z=0$, $(Oyz): x=0$, $(Ozx): y=0$.
• Parallel: $\vec{n'}=k\vec{n}$, $D'\neq kD$. Perpendicular: $\vec{n}\cdot\vec{n'}=0$.
• $d(M_0, (P))=\dfrac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Mini test