#### Example 1

Find the equation for the plane through the point $(0,1,-7)$ perpendicular to the vector $(4, -1, 6)$.

**Solution:** Let $\vc{a} = (0,1,-7)$. Let $\vc{n} = (4,-1,6)$. Then, for
$\vc{x} = (x,y,z)$, the equation for the plane is
\begin{align*}
\vc{n} \cdot (\vc{x}-\vc{a})=0.
\end{align*}
This becomes
\begin{align*}
(4,-1,6) \cdot (x-0, y-1, z+7) = 0
\end{align*}
or
\begin{align*}
4x -(y-1) + 6(z+7) = 0.
\end{align*}
Often, we prefer to write this as
\begin{align*}
4x -y +6z +43 =0.
\end{align*}
(Note that you can read the normal vector $\vc{n} = (4,-1,6)$ right
from the equation for the plane; the components of $\vc{n}$ are simply
the coefficients of $x$, $y$, and $z$.)

#### Example 2

Find the equation for the plane through the points $(0,1,-7)$, $(3,1,-9)$, and $(0, -5, -8)$.

**Solution:** Let $\vc{b} = (0, 1, -7) - (3,1,-9) = (-3, 0, 2)$. Let
$\vc{c} = (0, 1, -7) - (0 -5, -8) = (0, 6, 1)$. Then, we can find a
normal vector by taking their cross product
\begin{align*}
\vc{n} &= \vc{b} \times \vc{c}\\
&=
\left|
\begin{array}{rrr}
\vc{i} & \vc{j} & \vc{k}\\
-3 & 0 & 2\\
0 & 6 & 1
\end{array}
\right|
\\
&= \vc{i} (0 -12) - \vc{j} (-3 -0) + \vc{k}(-18 -0)\\
&= (-12, 3, -18).
\end{align*}
We'll pick the first point and let $\vc{a} = (0, 1, -7)$. The
equation for the plane becomes
\begin{align*}
(-12, 3, -18) \cdot (x-0, y-1, z+7) = 0
\end{align*}
which we rewrite as
\begin{align*}
-12x +3(y-1) -18(z+7) = 0
\end{align*}
or
\begin{align*}
-12x +3y -18z - 129 = 0.
\end{align*}

The planes from both examples went through the same point $(0, 1, -7)$. Did you also notice that their normal vectors are parallel to each other? If you multiply the $\vc{n}$ from Example 1 by $-3$, you obtain the $\vc{n}$ from Example 2: \begin{align*} -3(4, -1, 6) = (-12, 3, -18). \end{align*}

What does that mean about the relationship between the two planes? The two planes must be equal. (See the page on forming planes for a discussion of this fact.) In fact, if you divide both sides of the equation of the second plane by $-3$, you get the equation of the first plane.

#### Example 3

Find the equation for the plane through the points $(1,2,3)$, $(2,4,6)$, and $(-3,-6,-9)$.

**Solution:** Let $\vc{b} = (2,4,6) - (1,2,3) = (1,2,3)$ and let $\vc{c} =
(1,2,3) - (-3,-6,-9) = (4, 8, 12)$. Then, we attempt to find a normal
vector as their cross product:
\begin{align*}
\vc{n} &= \vc{b} \times \vc{c}\\
&=
\left|
\begin{array}{rrr}
\vc{i} & \vc{j} & \vc{k}\\
1 & 2 & 3\\
4 & 8 & 12
\end{array}
\right|
\\
&= \vc{i} (24 -24) - \vc{j} (12 -12) + \vc{k}(8 -8)\\
&= (0, 0, 0).
\end{align*}

Something clearly went wrong, as we weren't able to find a normal direction. Do the three points actually determine a plane? If you are still puzzled, check the page on forming planes for any conditions on the ability of three points to determine a plane.