Question

Show that the planes given by $2x-4y-3z=5$ and $-4x+8y+6z=1$ are parallel, and find the distance between the planes.

Short answer

The distance between parallel to the plane is $\frac{11\sqrt{29}}{58}$

Step-by-step solution

Given information

The planes $2x-4y-3z=5$and $-4x+8y+6z=1$

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 $2x-4y-3z=5$ is:

${\mathbf{N}}_1=⟨2,-4,-3⟩$

The normal vector of the plane $-4x+8y+6z=1$ is:

${\mathbf{N}}_2=⟨-4,8,6⟩$

The normal vectors ${\mathbf{N}}_1=⟨2,-4,-3⟩$ and ${\mathbf{N}}_2=⟨-4,8,6⟩$ are scalar multiple of each other. Therefore, the planes are parallel.

Calculation

The point $R_1=\left(0,0,-\frac{5}{3}\right)$is lies on the plane $2x-4y-3z=5$and the point $R_2=\left(0,0,\frac{1}{6}\right)$lies on the plane $-4x+8y+6z=1$

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 $\frac{11\sqrt{29}}{58}$