= Foot of a perpendicular
{title2=$\mathbf a+\frac{(\mathbf c-\mathbf a)\cdot\mathbf u}{\|\mathbf u\|^2}\mathbf u$}
= Perpendicular foot
{synonym}
The foot from $\mathbf c$ to the <line> $\mathbf a+t\mathbf u$, with $\mathbf u\ne0$, is its <orthogonal projection> onto that <line>. Solving $(\mathbf c-\mathbf a-t\mathbf u)\cdot\mathbf u=0$ gives the displayed formula and proves uniqueness. It belongs to the <line segment> with endpoints $\mathbf a$ and $\mathbf a+\mathbf u$ exactly when $0\leq(\mathbf c-\mathbf a)\cdot\mathbf u\leq\|\mathbf u\|^2$.
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