Plane Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

medium
Three points A(0,0,0)(0,0,0), B(1,0,0)(1,0,0), C(0,1,0)(0,1,0) lie on a plane. Write the equation of that plane.

Solution

  1. 1
    Step 1: Find two vectors in the plane: ABโƒ—=(1,0,0)\vec{AB} = (1,0,0) and ACโƒ—=(0,1,0)\vec{AC} = (0,1,0).
  2. 2
    Step 2: The normal to the plane is nโƒ—=ABโƒ—ร—ACโƒ—=(0โ‹…0โˆ’0โ‹…1,ย 0โ‹…0โˆ’1โ‹…0,ย 1โ‹…1โˆ’0โ‹…0)=(0,0,1)\vec{n} = \vec{AB} \times \vec{AC} = (0\cdot0-0\cdot1,\ 0\cdot0-1\cdot0,\ 1\cdot1-0\cdot0) = (0,0,1).
  3. 3
    Step 3: Plane equation using point A(0,0,0)(0,0,0): 0(xโˆ’0)+0(yโˆ’0)+1(zโˆ’0)=00(x-0) + 0(y-0) + 1(z-0) = 0, giving z=0z = 0.
  4. 4
    Step 4: Verify: all three points satisfy z=0z=0.

Answer

z=0z = 0 (the xy-plane)
Three non-collinear points determine a unique plane. The cross product of two vectors lying in the plane gives the normal vector, which is then used in the plane equation nโƒ—โ‹…(rโƒ—โˆ’rโƒ—0)=0\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0.

About Plane

A perfectly flat surface extending infinitely in all directions with zero thickness; defined by three non-collinear points.

Learn more about Plane โ†’

More Plane Examples