Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Equations of Planes

A plane is a flat surface in 3D that goes on forever. You can pin one down two ways: one point and two directions inside it, or one point and a normal vector sticking straight out of it. Each gives a different equation.

How It Works

Vector and parametric form. Start at a point on the plane and move along two directions that lie in it: $\vec{r} = \vec{r_0} + t\vec{u} + s\vec{v}$ . The two directions must not be parallel, or you only sweep out a line. Split it into components and you have the parametric equations, one each for $x$ , $y$ and $z$ .

Scalar (Cartesian) form. $Ax + By + Cz + D = 0$ , where $\vec{n} = [A, B, C]$ is a normal, a vector perpendicular to the plane. Why it works: for any point on the plane, the vector from a known point to it lies in the plane, so its dot product with $\vec{n}$ is zero. Expand that and you get the scalar equation.

The recipe for the scalar equation. Get a normal, put its components in as $A$ , $B$ and $C$ , then substitute a known point to find $D$ . Every question comes down to finding the normal.

When you are given two directions in the plane, their cross product is perpendicular to both, so it is a normal. This is how you turn a vector equation into a scalar one.

When you are given three points, subtract to get two direction vectors from the same point, then cross them.

When you are given a line and a point, the line's direction is one direction. The vector from a point on the line to the given point is the second.

Parallel planes share a normal, so a plane parallel to a given one keeps $A$ , $B$ and $C$ and only needs a new $D$ .

Always check by substituting each given point back into your answer. If any point fails, the normal or $D$ is off. A normal can be scaled freely, so divide out common factors to keep the numbers small.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the scalar equation of the plane with normal $\vec{n} = [2, -1, 3]$ through $(1, 4, -2)$ .
Show solution
1. The normal gives the coefficients: $2x - y + 3z + D = 0$ .
2. Substitute the point: $2(1) - 4 + 3(-2) + D = 0$ , so $-8 + D = 0$ and $D = 8$ .
Answer: $2x - y + 3z + 8 = 0$
Easy
Find the scalar equation of the plane through $(3, 0, -1)$ parallel to $4x + y - 2z + 7 = 0$ .
Show solution
1. Parallel planes share a normal, so the plane is $4x + y - 2z + D = 0$ .
2. Substitute: $4(3) + 0 - 2(-1) + D = 0$ , so $14 + D = 0$ and $D = -14$ .
Answer: $4x + y - 2z - 14 = 0$
Medium
Find the scalar equation of the plane $\vec{r} = [1, 2, 0] + t[1, 0, 2] + s[0, 1, -1]$ .
Show solution
1. Normal: $[1, 0, 2] \times [0, 1, -1] = [0(-1) - 2(1), ; 2(0) - 1(-1), ; 1(1) - 0(0)] = [-2, 1, 1]$ .
2. So $-2x + y + z + D = 0$ . Substitute $(1, 2, 0)$ : $-2 + 2 + 0 + D = 0$ , so $D = 0$ .
3. Multiply by $-1$ to tidy it.
Answer: $2x - y - z = 0$
Medium
Find vector and scalar equations of the plane through $A(1, 0, 1)$ , $B(2, 1, 0)$ and $C(0, 3, 2)$ .
Show solution
1. Two directions from $A$ : $\vec{AB} = [1, 1, -1]$ and $\vec{AC} = [-1, 3, 1]$ .
2. Vector equation: $\vec{r} = [1, 0, 1] + t[1, 1, -1] + s[-1, 3, 1]$ .
3. Normal: $\vec{AB} \times \vec{AC} = [1(1) - (-1)(3), ; (-1)(-1) - 1(1), ; 1(3) - 1(-1)] = [4, 0, 4]$ . Divide by $4$ : $\vec{n} = [1, 0, 1]$ .
4. So $x + z + D = 0$ . Substitute $A$ : $1 + 1 + D = 0$ , so $D = -2$ .
5. Check $B$ : $2 + 0 - 2 = 0$ . Check $C$ : $0 + 2 - 2 = 0$ .
Answer: $\vec{r} = [1, 0, 1] + t[1, 1, -1] + s[-1, 3, 1]$ and $x + z - 2 = 0$
Test level
Find the scalar equation of the plane that contains the line $\vec{r} = [2, -1, 0] + t[1, 1, 3]$ and the point $P(1, 2, 4)$ .
Show solution
1. One direction is the line's: $\vec{d_1} = [1, 1, 3]$ .
2. The second runs from the line's point $(2, -1, 0)$ to $P$ : $\vec{d_2} = [-1, 3, 4]$ .
3. Normal: $\vec{d_1} \times \vec{d_2} = [1(4) - 3(3), ; 3(-1) - 1(4), ; 1(3) - 1(-1)] = [-5, -7, 4]$ . Use $\vec{n} = [5, 7, -4]$ .
4. So $5x + 7y - 4z + D = 0$ . Substitute $(2, -1, 0)$ : $10 - 7 + D = 0$ , so $D = -3$ .
5. Check $P$ : $5 + 14 - 16 - 3 = 0$ . Check the line lies in the plane: $\vec{n} \cdot \vec{d_1} = 5 + 7 - 12 = 0$ .
Answer: $5x + 7y - 4z - 3 = 0$

Learn It on Martian Lab

Martian Lab teaches this in the Equations of Planes lesson of its MCV4U Lines and Planes unit. Watch it, then take the Lines and Planes practice tests to see if it stuck.

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Free Lines and Planes practice test, with solutions

More Lines and Planes topics: Intersection of Three Planes

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque

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