= Solution
Put $\mathbf v=\mathbf r_0-\mathbf a$. The squared distance to a point on the line is $|\mathbf v-\lambda\widehat{\mathbf t}|^2$. Since $|\widehat{\mathbf t}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda=\widehat{\mathbf t}\cdot\mathbf v$, with value $|\mathbf v|^2-(\widehat{\mathbf t}\cdot\mathbf v)^2$. The <cross product> identity $|\widehat{\mathbf t}\times\mathbf v|^2=|\mathbf v|^2-(\widehat{\mathbf t}\cdot\mathbf v)^2$ therefore proves
$$
\boxed{\ell=|\widehat{\mathbf t}\times(\mathbf r_0-\mathbf a)|.}
$$
For the given line, $\widehat{\mathbf t}=(1,1,0)/\sqrt2$, so its squared distance to $(x,y,z)$ is $(x-y)^2/2+z^2$. Substituting the <parabola> gives
$$
\ell^2(\mu)=6\mu^2+(\mu^2-4)^2=(\mu^2-1)^2+15.
$$
The global minimum occurs at $\mu=\pm1$, hence
$$
\boxed{\ell_{\min}=\sqrt{15}.}
$$
The corresponding points on the <parabola> are $\pm(1,1+2\sqrt3,0)+(0,0,-3)$; their nearest points on the line are $\pm(1+\sqrt3,1+\sqrt3,0)$. The connecting <vectors> are <orthogonal> to the line, as the projection calculation requires.
Back to article page
