= Solution
{c}
Write $\mathbf u=\mathbf b-\mathbf a$. Since the <triangle> is nondegenerate, $\mathbf u\ne0$. Every <point> on the <line> $AB$ has a unique <position vector> $\mathbf p=\mathbf a+t\mathbf u$. The condition that $CP$ is <orthogonal> to $AB$ is the <dot product> <equation>
$$
(\mathbf c-\mathbf a-t\mathbf u)\cdot\mathbf u=0.
$$
Its <coefficient> of $t$ is $-\|\mathbf u\|^2\ne0$, so it has exactly one solution. The <foot of a perpendicular> is therefore
$$
\boxed{\mathbf p=\mathbf a+\frac{(\mathbf c-\mathbf a)\cdot(\mathbf b-\mathbf a)}{\|\mathbf b-\mathbf a\|^2}(\mathbf b-\mathbf a).}
$$
This is the <orthogonal projection> of $C$ onto the <line> $AB$, expressed using the <Euclidean norm>.
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