Answer
$P = \left( { - \frac{{18}}{{17}},3, - \frac{{13}}{{17}}} \right)$
Work Step by Step
By Theorem 1, the plane $x-4z=2$ has normal vector ${\bf{n}} = \left( {1,0, - 4} \right)$.
By Eq. (7) of Exercise 71, the point $P$ on the plane closest to $Q$ is determined by the equation:
$\overrightarrow {OP} = \overrightarrow {OQ} + \left( {\frac{{d - \overrightarrow {OQ} \cdot{\bf{n}}}}{{{\bf{n}}\cdot{\bf{n}}}}} \right){\bf{n}}$
We have $\overrightarrow {OQ} = Q - O = \left( { - 1,3, - 1} \right)$. Substituting the corresponding values in Eq. (7) gives
$\overrightarrow {OP} = \left( { - 1,3, - 1} \right) + \left( {\frac{{2 - \left( { - 1,3, - 1} \right)\cdot\left( {1,0, - 4} \right)}}{{\left( {1,0, - 4} \right)\cdot\left( {1,0, - 4} \right)}}} \right)\left( {1,0, - 4} \right)$
$\overrightarrow {OP} = \left( { - 1,3, - 1} \right) + \left( {\frac{{2 - 3}}{{17}}} \right)\left( {1,0, - 4} \right)$
$\overrightarrow {OP} = \left( { - \frac{{18}}{{17}},3, - \frac{{13}}{{17}}} \right)$
Thus, the point $P = \left( { - \frac{{18}}{{17}},3, - \frac{{13}}{{17}}} \right)$ is nearest to $Q = \left( { - 1,3, - 1} \right)$ on the plane $x-4z=2$.
