Deriving the Equation of a Plane Through A Perpendicular to the Vector A

Let A be a non-zero vector in ℝ³, and also regard its endpoint as the point through which the plane passes:

A = (a1, a2, a3)T ≠ (0, 0, 0)T.

Let P be an arbitrary point on the plane:

P = (x, y, z)T.

The displacement from A to P is

P − A = (x − a1, y − a2, z − a3)T.
Because the plane is perpendicular to A, the vector A is its normal vector. Every displacement P − A lying in the plane must therefore be perpendicular to A.

Perpendicular vectors have a dot product of zero. Hence

A · (P − A) = 0.

Writing this condition in coordinates gives

a1(x − a1) + a2(y − a2) + a3(z − a3) = 0.

Expanding the brackets:

a1x − a12 + a2y − a22 + a3z − a32 = 0.

Move the squared terms to the right-hand side:

a1x + a2y + a3z = a12 + a22 + a32.

In vector notation, this becomes

A · P = A · A = ‖A‖2.

Therefore, the plane passes through the point A and is perpendicular to the vector A. Equivalently, A is the plane’s normal vector.

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