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 = (a 1 , a 2 , a 3 ) 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 − a 1 , y − a 2 , z − a 3 ) 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 a 1 (x − a 1 ) + a 2 (y − a 2 ) + a 3 (z − a 3 ) = 0. Expanding the brackets: a 1 x − a 1 2 + a 2 y − a 2 2 + a 3 z − a 3 2 = 0. Move the squared terms to the right-hand side: a 1 x + a 2 y + a 3 z = a 1 2 + a 2 2 + a 3 2 . In vector notation, this becomes A · P = A · A = ‖ A ‖ 2 . ...