Split a vector into parallel and perpendicular parts

Try

Project b=(3,1) onto the line spanned by u=(1,1). The closest point is p=(2,2), with residual r=(1,−1). Their dot product u·r is zero, so the error points perpendicular to the allowed line.

Predict

Predict why the numerator alone cannot be the projection coefficient. The dot product x·y is Σxᵢyᵢ, and ||x||=√(x·x). If u is not a unit vector, its squared norm must divide the coefficient. Orthogonal means dot product zero; orthonormal means pairwise orthogonal with each norm one.

Experiment

Compute b·u=4 and u·u=2, giving coefficient 2. Compare candidate points (1,1),(2,2),(3,3): their squared distances from b are 4,2,4. Then verify the residual dot product, which provides a general nearest-point certificate rather than a three-point search.

Understand

Worked derivation: a point on the line is p=tu. Requiring (b−tu)·u=0 yields t=(b·u)/(u·u), valid for u≠0. For any other point q on the line, b−q=(b−p)+(p−q), a sum of orthogonal vectors. Squaring norms gives ||b−q||²=||b−p||²+||p−q||², proving p uniquely minimizes distance.

The same argument works for a subspace when the residual is orthogonal to every direction in that subspace. An orthonormal basis qᵢ gives p=Σ(b·qᵢ)qᵢ. The zero subspace has projection zero; it requires no invalid division by a zero direction.

Invalid shortcut: subtract (b·u)u for any u. That formula assumes ||u||=1.

Apply

Separate the projection vector, its scalar coefficient and the residual. Check orthogonality and reconstruction b=p+r. For exact irrational norms, keep a symbolic expression in a written response; the rational checker does not accept square-root expressions.

Continue

Depth: the Cauchy–Schwarz inequality |x·y|≤||x||||y|| follows from nonnegative squared norms and supports angle definitions. General inner products can weight coordinates differently; this lesson uses the ordinary Euclidean dot product.