Question 10.5.9: Finding the Distance between Parallel Planes Find the distan...

First, observe that the planes are parallel, since their normal vectors $\langle 2,-3,1\rangle \text { and }\langle 4,-6,2\rangle$ are parallel. Further, since the planes are parallel, the distance from the plane $P_1 \text { to every point in the plane } P_2 \text { is the same. So, pick any point in } P_2 \text {, }$
say (0, 0, 4). (This is certainly convenient.) The distance D from the point (0, 0, 4) to the plane $P_1 \text { is then given by }$ (5.8) to be

$\begin{aligned} \left|\operatorname{comp}_{\mathbf{a}} \vec{P}_1 P_0\right| &=\left|\overrightarrow{P_1 P_0} \cdot \frac{\mathbf{a}}{\|\mathbf{a}\|}\right| \\ &=\left|\left\langle x_0-x_1, y_0-y_1, z_0-z_1\right\rangle \cdot \frac{\langle a, b, c\rangle}{\|\langle a, b, c\rangle\|}\right| \\ &=\frac{\left|a\left(x_0-x_1\right)+b\left(y_0-y_1\right)+c\left(z_0-z_1\right)\right|}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\left|a x_0+b y_0+c z_0-\left(a x_1+b y_1+c z_1\right)\right|}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\left|a x_0+b y_0+c z_0+d\right|}{\sqrt{a^2+b^2+c^2}} \end{aligned}$    (5.8)