Find the point symmetric to M(1,2,8) with respect to…

Find the point symmetric to $M(1,2,8)$ with respect to $\tfrac{x-1}2=\tfrac y{-1}=\tfrac z1$.

Explanation of Method

The mirror image of the point M(1, 2, 8) in a line.

Step 1: Solution

\[M^{*}=2M'-M=2(3,-1,1)-(1,2,8)=(5,-4,-6)\]

Check: the midpoint is $(3,-1,1)$; $\overrightarrow{MM^{*}}=\{4,-6,-14\}$ is perpendicular to $\{2,-1,1\}$ ($8+6-14=0$); and $|M'M^{*}|=\sqrt{62}$.

$M(1,2,8)$ $M'(3,-1,1)$ $M^{*}(5,-4,-6)$

$M'$ is the foot of the perpendicular from $M$ to the line (Problem 261); $M^{*}=2M'-M$ is the mirror image. $|MM'|=|M'M^{*}|=\sqrt{62}$.

Result

\[\boxed{\,M^{*}=2M'-M=2(3,-1,1)-(1,2,8)=(5,-4,-6)\,}\]

Remark

In the source, the heading “$\overrightarrow{MM^{*}}\parallel\vec d$” is a slip, which the source itself then corrects. The correct statement is $\overrightarrow{MM^{*}}\perp\vec d$.