Write the equation of the plane through M_1(0,-1,3) normal…

Write the equation of the plane through $M_1(0,-1,3)$ normal to $\overrightarrow{M_1M_2}$, where $M_2(1,3,5)$.

Explanation of Method

The plane through M₁ normal to the vector from M₁ to M₂.

Step 1: Solution

\[ \vec n=\overrightarrow{M_1M_2}=\{1,4,2\},\qquad 1(x-0)+4(y+1)+2(z-3)=0 . \]

\[x+4y+2z-2=0\]

Check: $M_1$ gives $0-4+6-2=0$. The point $M_2$ is at distance $|\overrightarrow{M_1M_2}|=\sqrt{21}$ from the plane, since $\tfrac{|1+12+10-2|}{\sqrt{21}}=\sqrt{21}$.

$M_1$ $M_2$ $\vec n$

The plane $x+4y+2z-2=0$ through $M_1$ with normal $\overrightarrow{M_1M_2}=\{1,4,2\}$.

Result

\[\boxed{\,x+4y+2z-2=0\,}\]