I am having trouble with this problem in my Linear Algebra review:
Find an equation for the plane parallel to $2x-y+2z=4 $ such that the point $(3,2,-1) $ is equidistant from both planes.
The answer is $2x-y+2=0$ . How would you go about finding the $0$ ?
On
In order for the point $(3,2,−1)$ to be equidistant from two distinct parallel planes, it must be midway between them. Furthermore, one plane is a reflection of the other plane through the point $(3,2,−1).$
The point $P_2 = (x_2,y_2,z_2)$ is in the second (unknown) plane if and only if there is a corresponding point $P_1 = (x_1,y_1,z_1)$ in the plane $2x−y+2z=4$ such that the point $(3,2,−1)$ is the midpoint of the line segment $\overline{P_1P_2}.$ That is,
\begin{align} \frac{x_1 + x_2}{2} &= 3, \\[3pt] \frac{y_1 + y_2}{2} &= 2, \\[3pt] \frac{z_1 + z_2}{2} &= -1. \end{align}
Solve these equations for the coordinates of $P_1$:
\begin{align} x_1 &= 6 - x_2, \\ y_1 &= 4 - y_2, \\ z_1 &= -2 - z_2. \end{align}
We know that $(x_1,y_1,z_1)$ lies in the given plane, so $$2x_1 − y_1 + 2z_1 = 4.$$
Use the equations above to substitute for $x_1,$ $y_1,$ and $z_1$:
$$ 2(6 - x_2) − (4 - y_2) + 2(-2 - z_2) = 4. $$
That's an equation of a plane, and it is the desired plane. But to put it in the form that is probably expected, we can simplify:
\begin{align} 12 - 2x_2 − 4 + y_2 - 4 - 2z_2 &= 4, \\ - 2x_2 + y_2 - 2z_2 &= 0, && \text{collect all the constant terms}\\ 2x_2 - y_2 + 2z_2 &= 0. && \text{multiply all terms by $-1$}\\ \end{align}
Done!
Since both planes are parallel, the normal vector to both planes is $(2, -1, 2)$. Thus the points $(2k + 3, -k + 2, 2k-1)$ and $(-2k + 3, k + 2, -2k-1)$ are equidistant to $(3,2,-1)$.
Now if the first point lies on the original plane, $2(2k + 3) - (-k + 2) + 2(-2k + 1) = 4$ $ \implies k = -2$. So the original plane is moved $-2$ units to meet the point $(3,2,-1)$, then another $-2$ units for it to be equidistant gives $2x - y + 2z = 4 + (-2) + (-2) = 0$.