Find the projection of M(1,2,8) onto tfracx-12=tfrac…

Find the projection of $M(1,2,8)$ onto $\tfrac{x-1}2=\tfrac y{-1}=\tfrac z1$.

Step 1: Solution

$M'=(1+2t,-t,t)$, with $\overrightarrow{M'M}\cdot\{2,-1,1\}=0$. This gives $-6t+6=0$, so $t=1$.

\[M'(3,-1,1),\qquad d(M,\ell)=|MM'|=\sqrt{62}\]

Check: $\overrightarrow{M'M}=\{-2,3,7\}$ and $\{-2,3,7\}\cdot\{2,-1,1\}=-4-3+7=0$.