Find the angle between the plane through M_1(0,0,0)…

Find the angle between the plane through $M_1(0,0,0)$, $M_2(2,-2,0)$, $M_3(2,2,2)$ and the $xOy$ plane.

Explanation of Method

The angle between the plane through three points and the xOy plane.

Step 1: Solution

\[ \vec n=\{2,-2,0\}\times\{2,2,2\}=\{-4,-4,8\}\parallel\{-1,-1,2\},\qquad -x-y+2z=0 . \]

With $\vec k=\{0,0,1\}$: $\cos\varphi=\frac{|2|}{\sqrt6\cdot1}=\frac{\sqrt6}{3}$.

\[\varphi=\arccos\frac{\sqrt6}{3}\approx35^\circ16'\]

Check: $M_2$ gives $-2+2=0$ and $M_3$ gives $-2-2+4=0$. Equivalently $\tan\varphi=\tfrac{1}{\sqrt2}$: the steepest line of the plane, direction $(1,1,1)$, rises $1$ over a horizontal run of $\sqrt2$.

$\varphi$ $M_1$ $M_2$ $M_3$

The plane $M_1M_2M_3$ meets $xOy$ along $y=-x$ (dashed); in the cross-section perpendicular to this line the angle is $\varphi=\arccos\tfrac{\sqrt6}{3}\approx35^\circ16'$.

Result

\[\boxed{\,\varphi=\arccos\frac{\sqrt6}{3}\approx35^\circ16'\,}\]