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

$\text{Therefore, the distance from the point to the plane is}\frac{11\sqrt{26}}{13}\text{units.}$

Step-by-step solution

Step 1:Given information

The planes given by y − 7z = 16 and 2y − 14z = 5

Step 2:Calculation 

$\text{Consider the point}P(1,2,-3)\text{and the line with the equation}\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩\text{and the}$

$\text{plane with the equation}x-3y+4z=5\text{.}$

$\text{The objective is to find the distance from the point}P(1,2,-3)\text{to the plane with the equation}$

$x-3y+4z=5$

$\text{The normal vector of the plane}x-3y+4z=5\text{is:}$

$\mathbf{N}=⟨1,-3,4⟩$

$\text{The point}R=(5,0,0)\text{is on the plane}x-3y+4z=5\text{.}$

$\text{The vector}\overrightarrow{RP}\text{is given by:}$

$\overrightarrow{RP}=⟨1-5,2-0,-3-0⟩$

$=⟨-4,2,-3⟩$

$\text{The distance from}P\text{to the plane is:}$

$\left|{comp}_{\mathbf{N}}\overrightarrow{RP}\right|=\frac{|\mathbf{N}·RP|}{\Vert \mathbf{N}\Vert }$

$=\frac{|⟨1,-3,4⟩·⟨-4,2,-3⟩|}{\Vert ⟨1,-3,4⟩\Vert }$

$=\frac{|-4-6-12|}{\sqrt{1+9+16}}$

$=\frac{|-22|}{\sqrt{26}}$

$=\frac{22}{\sqrt{26}}$

$=\frac{11\sqrt{26}}{13}$

$\text{Therefore, the distance from the point to the plane is}\frac{11\sqrt{26}}{13}\text{units.}$