Work out equation of a plane instantly with clear inputs, formula shown and shareable results.
Two edge vectors AB and AC lie in the plane, so their cross product is a normal vector (a, b, c). Substituting point A then fixes d in ax + by + cz + d = 0. Half the normal's magnitude is the area of triangle ABC.
Normal from three points
n = (B − A) × (C − A)
Plane equation
a(x − x_A) + b(y − y_A) + c(z − z_A) = 0
The normal is (6, 3, 2) and the plane is 6x + 3y + 2z − 6 = 0, matching the intercept form x/1 + y/2 + z/3 = 1.
Collinear points lie on infinitely many planes, so the cross product is the zero vector and no unique answer exists.