What is the equation of the plane determined by the three points \((0,4,0)\text{,}\) \((-3, -6, 2)\) and \((-7, 0,
1)\text{.}\)
First, I need to calculate the local directions. I’ll set \((0,4,0)\) as my local origin. Then I calculate the differences of the two other points from this point.
\begin{gather*}
(-3, -6. 2) - (0,4,0) = (-3, -10, 2) \\
(-7, 0. 1) - (0,4,0) = (-7, -4, 1)
\end{gather*}
These are the two local directions. Then I proceed as with the previous case. I start by taking the cross product of the two local directions to get the normal to the plane.
\begin{align*}
(-3, -10, 2) \times (-7, -4, 1) \amp = (-10 - (-8),
(-14), - (-3), 12 - 70)\\
\amp = (-2, -11, -58)
\end{align*}
Then I again proceed as before, going back to the very first case. I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\) I use the first point, but I could use any of the three points.
\begin{equation*}
(-2, -11, -58) \cdot (0,4,) = 0 - 44 + 0 = -44
\end{equation*}
So \(d = -44\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
-2x - 11y - 58 z = -44
\end{equation*}
I could multiply through by \(-1\) to make everything positive, if I wanted.
\begin{equation*}
2x + 11y + 58z = 44
\end{equation*}
This equation also describes the same plane.