Recipe

Write the equation of a plane in space

Use when: you're given three points on a plane, or a point plus a normal vector, and need the plane's scalar equation ax+by+cz=d

Goal: the coefficients a, b, c, d of the plane's scalar equation

Taught in Planes in space

1. 1

   A plane's equation always starts from one point on it and one vector perpendicular to it, but a problem can hand those over in two different shapes.

   Example

   P_1=(1,1,1), P_2=(2,3,0), P_3=(0,2,3): \overrightarrow{P_1P_2}=\langle1,2,-1\rangle and \overrightarrow{P_1P_3}=\langle-1,1,2\rangle. Crossing them gives \mathbf n=\langle5,-1,3\rangle.

   • a point P_0 and normal \mathbf n are given directly — use them as they are, and skip to Step 2.
   • three points P_1,P_2,P_3 are given instead, not all on one line — build two in-plane vectors, \overrightarrow{P_1P_2} and \overrightarrow{P_1P_3}, then cross them to get \mathbf n.

   Why: a plane's equation needs one point on it and one vector perpendicular to it, and three points determine both once crossed, the same cross product from the previous section.

2. 2

   Use \mathbf n=\langle a,b,c\rangle directly as the equation's left-side coefficients.

   Example

   \mathbf n=\langle5,-1,3\rangle gives 5x-y+3z=d.

   Why: the coefficients in a plane's scalar equation literally are a normal vector's components, so this step is a direct read-off, not a computation.

3. 3

   Substitute P_0's coordinates into the equation's left side.

   Example

   Using P_0=P_1=(1,1,1): d=5(1)-1(1)+3(1)=5-1+3=7.

   Why: d is whatever value makes the known point satisfy the equation, since P_0 has to lie on the plane by construction.

4. 4

   Substitute another point that should lie on the plane and confirm the equation still balances.

   Example

   P_2=(2,3,0): 5(2)-1(3)+3(0)=10-3+0=7. P_3=(0,2,3): 5(0)-1(2)+3(3)=-2+9=7.

   Why: this catches a sign error from Step 1's cross product or an arithmetic slip in Step 3 before the equation is used anywhere else.