Question

Show that the planes given by $y-7z=16$ and $2y-14z=5$ are parallel, and find the distance between the planes.

Short answer

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

Step-by-step solution

Given information

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

Calculation

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:

$\text{distance}=\frac{\left|\mathbf{N}·\overline{R_1{R}_2}\right|}{\Vert \mathbf{N}\Vert }$

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

${\mathbf{N}}_1=⟨0,1,-7⟩$

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

${\mathbf{N}}_2=⟨0,2,-14⟩$

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

Calculation

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 $\overrightarrow{R_1{R}_2}$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 $=\frac{\left|\mathbf{N}·\overrightarrow{R_1{R}_2}\right|}{\Vert \mathbf{N}\Vert }$

$=\frac{⟨0,1,-7⟩·\left\{0,-\frac{27}{2},0\right\}}{\Vert (0,1,-7)\Vert }$(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 $\frac{27}{2\sqrt{50}}$