Inner products and projection

Draw u=(2,1) and the line through b=(1,1) and the origin. Predict the nearest point on that line to u. A right-angle drop gives p=(1.5,1.5), leaving residual (0.5,−0.5). It is not enough merely to land on the line: the error must be perpendicular to it. The MIT projection excerpt begins with this perpendicular error at 0:26. [1]

For vectors a and b in real Euclidean space, the dot product a·b sums their coordinate products; a·a is the square of its Euclidean length. Orthogonality means a·b=0; it is a different question from linear independence. To project a vector v onto the line spanned by a nonzero direction u, write the candidate as cu and require the residual v−cu to be perpendicular to u. Solving u·(v−cu)=0 gives the orthogonal-projection theorem:

The symbol projᵤ(v) is the point on the line nearest to v in Euclidean distance; u·v is a scalar, and u·u is positive because u≠0. The residual is perpendicular because u·(v−projᵤ(v))=0. This perpendicularity proves nearest distance: for any other line point cu, the squared distance is the squared residual length plus the squared distance along the line.

Use the projection lab’s Initial vectors u=(2,1), b=(1,1) and calculate p=(1.5,1.5). Scale b by 2 and compare p; try Zero direction · invalid, then Nonorthogonal example. Read the residual dot product and distance in the numeric result.