Write the plane perpendicular to xOy that contains the…

Write the plane perpendicular to $xOy$ that contains the perpendicular from $M(1,-1,1)$ to the line $\ell:\ x=0,\ y-z+1=0$.

Explanation of Method

The plane perpendicular to xOy that contains the perpendicular from a point to a line.

Step 1: Foot of the perpendicular

$\ell:\ (0,\ t-1,\ t)$, with $\vec d=\{0,1,1\}$. Then

\[ \overrightarrow{FM}=\{1,\,-t,\,1-t\},\qquad \overrightarrow{FM}\cdot\vec d=1-2t=0\ \Rightarrow\ t=\tfrac12 . \]

\[ F\left(0,-\tfrac12,\tfrac12\right),\qquad \overrightarrow{FM}=\tfrac12\{2,-1,1\}. \]

Step 2: The plane

A plane perpendicular to $xOy$ contains the direction $\vec k$. It also contains $\overrightarrow{FM}$, so its normal is

\[ \vec N=\{2,-1,1\}\times\{0,0,1\}=\{-1,-2,0\}\parallel\{1,2,0\}. \]

\[x+2y+1=0\]

Checks: $M$ gives $1-2+1=0$ and $F$ gives $0-1+1=0$. The normal has no $z$-component, so the plane is vertical. $\checkmark$

$M(1,-1,1)$ $F$ $\ell$

The perpendicular $MF$ (red) and the vertical direction $\vec k$ (teal dashed) span the plane $x+2y+1=0$ (gray), which is therefore perpendicular to $xOy$.

Result

\[\boxed{\,x+2y+1=0\,}\]