91影视

The planes $y-7z=16$and $2y-14z=5$

The goal is to calculate the distance between a specified point and the plane. The distance between a point and the plane is calculated as follows:

The normal vector of the plane $y-7z=16$is:

${\mathbf{N}}_1=鉄�0,1,-7鉄�$

The normal vector of the plane $2y-14z=5$ is:

${\mathbf{N}}_2=鉄�0,2,-14鉄�$

The normal vectors ${\mathbf{N}}_1=鉄�0,1,-7鉄�$ and ${\mathbf{N}}_2=鉄�0,2,-14鉄�$ are scalar multiple of each other. Therefore, the planes are parallel.

The point $R_1=(0,16,0)$is lies on the plane $y-7z=16$and the point $R_2=\left(0,\frac{5}{2},0\right)$lies on the plane $2y-14z=5$

The vector $\stackrel{鈫�}{R_1{R}_2}$is given by:

$$\begin{gathered} R_1{R}_2=\left(0-0,\frac{5}{2}-16,0-0\right) \\ =\left(0,-\frac{27}{2},0\right) \end{gathered}$$

The distance from $P$to the plane is:

distance $=\frac{\left|\mathbf{N}路\stackrel{鈫�}{R_1{R}_2}\right|}{鈥�\mathbf{N}鈥�}$

$=\frac{鉄�0,1,-7鉄�路\left\{0,-\frac{27}{2},0\right\}}{鈥�(0,1,-7)鈥�}$(Substitution)

$=\frac{\left|0-\frac{27}{2}+0\right|}{\sqrt{0+1+49}}$(Dot product)

$=\frac{-\frac{27}{2}}{\sqrt{50}}$

$=\frac{27}{2\sqrt{50}}($Simplify)

Therefore, the distance between parallel to the plane is $\frac{27}{2\sqrt{50}}$