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With $t$ as in i, the foot of a perpendicular lies strictly between the endpoints precisely when $0<t<1$. Since the denominator is positive, this givesEquivalently, both dot productsare positive: the angles of the triangle at $A$ and $B$ are both less than $\pi /2$. If “between” includes the endpoints, replace both strict inequalities by non-strict ones; equality gives $P=A$ or $P=B$.

Write $\mathbf{u}=\mathbf{b}-\mathbf{a}$. Since the triangle is nondegenerate, $\mathbf{u}\ne 0$. 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 equationIts coefficient of $t$ is $-\Vert \mathbf{u}{\Vert }^2\ne 0$, so it has exactly one solution. The foot of a perpendicular is thereforeThis is the orthogonal projection of $C$ onto the line $AB$, expressed using the Euclidean norm.