Question

Find the distance from the point $(2,0,-3)$ to the plane $3x-4y+5z=1$

Short answer

The distance from the point to the plane is $\sqrt{2}$

Step-by-step solution

Given information

The planes $3x-4y+5z=1$and the point $P=(2,0,-3)$

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:

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

The normal vector of the plane $3x-4y+5z=1$is:

$\mathbf{N}=⟨3,-4,5⟩$

The point $R=\left(0,0,\frac{1}{5}\right)$is on the plane $3x-4y+5z=1$

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

Calculation

The distance from $P$to the plane is:

Therefore, the distance from the point to the plane is $\sqrt{2}$