Question

Show that the planes given by $2x-3y+5z=7$ and $-6x+9y-15z=8$ are parallel, and find the distance between the planes.

Short answer

The distance between parallel to the plane is $\frac{29\sqrt{38}}{152}$

Step-by-step solution

Given information

The planes $2x-3y+5z=7$and $-6x+9y-15z=8$

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:

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

The normal vector of the plane $2x-3y+5z=7$is:

${\mathbf{N}}_1=⟨2,-3,5⟩$

The normal vector of the plane $-6x+9y-15z=8$is:

${\mathbf{N}}_2=⟨-6,9,-15⟩$

The normal vectors ${\mathbf{N}}_1=⟨2,-3,5⟩$and ${\mathbf{N}}_2=⟨-6,9,-15⟩$are scalar multiple of each other. Therefore, the planes are parallel.

Calculation

The point $R_1=\left(0,0,\frac{7}{5}\right)$is lies on the plane $2x-3y+5z=7$and the point $R_2=\left(0,0,-\frac{8}{15}\right)$lies on the plane $-6x+9y-15z=8$

The vector $\overrightarrow{R_1{R}_2}$is given by:

The distance from $P$ to the plane is:

Therefore, the distance between parallel to the plane is $\frac{29\sqrt{38}}{152}$