Three non-collinear points fix a plane: build two vectors, cross them for a normal, then write the equation.
Key formulas
Vectors in the plane
From A, B, C form AB and AC. Their cross product is perpendicular to every line in the plane.
Example outline
A(1,0,0), B(0,1,0), C(0,0,1). AB=(-1,1,0), AC=(-1,0,1). n = (1,1,1). Equation: x+y+z=1.
Collinear trap
If AB and AC are parallel, the cross product is zero and no unique plane exists — the points lie on a line.
Worked example
Worked example
Problem. Find the equation of the plane through A(1,0,0), B(0,1,0) and C(0,0,1).
Solution. AB = (−1,1,0), AC = (−1,0,1). n = AB × AC = (1,1,1). Using A: 1(x−1)+1(y−0)+1(z−0)=0 → x + y + z = 1.
Check
Substitute A, B and C back into x+y+z=1.
Review quiz
Plane through (2,0,0), (0,2,0), (0,0,2) is:
Practice questions
- Points (2,0,0), (0,2,0), (0,0,2). Find the plane equation.
- How do you detect collinear points quickly?
Study notes
Can four points fail to be coplanar?
Yes. The fourth may not satisfy the plane of the first three.
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