Geometric Bites

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 . ...

Let A, B and X be column vectors in ℝ³: A = a₁ a₂ a₃ , B = b₁ b₂ b₃ , X = x y z . BAᵀX = B(AᵀX). Matrix multiplication is associative, so AᵀX may be evaluated first. Since Aᵀ is a 1 × 3 row matrix and X is a 3 × 1 column matrix, their product is a scalar: AᵀX = a₁x + a₂y + a₃z. Therefore, BAᵀX = b₁ b₂ b₃ (a₁x + a₂y + a₃z). The quantity in parentheses is a scalar, so it multiplies every component of B: BAᵀX = b₁a₁x + b₁a₂y + b₁a₃z b₂a₁x + b₂a₂y + b₂a₃z b₃a₁x + b₃a₂y + b₃a₃z . Collect the coefficients of x, y and z into a matrix multiplying X: BAᵀX = b₁a₁ b₁a₂ b₁a₃ b₂a₁ b₂a₂ b...