Chapter 16, Problem 9RP

To determine

To find: the distance a point in $3$ dimensional space to the plane represented by the equation.

Answer to Problem 9RP

  $d=0.2307$ .

Explanation of Solution

Given:

The point on the space are $\left(3,4,2\right)$ .

The equation of the plane is:

  $3x-4y+12z=20$ .

Concept used:

The equation of straight line of distance of a point to the plane is

  $d=\frac{\left|A{x}_1+B{y}_1+C{z}_1+D\right|}{\sqrt{A^2+B^2+C^2}}$ .

Calculation:

According to the given:

The point on the space are $\left(3,4,2\right)$ .

The equation of the plane is:

  $3x-4y+12z=20$ .

From the equation of straight line of distance of a point to the plane is:

  $\begin{array}{l}d=\frac{\left|A{x}_1+B{y}_1+C{z}_1+D\right|}{\sqrt{A^2+B^2+C^2}}.\\ d=\frac{\left|3\left(3\right)+\left(-4\right)\left(4\right)+12\left(2\right)+\left(-20\right)\right|}{\sqrt{3^2+{\left(-4\right)}^2+{\left(12\right)}^2}}.\\ d=\mathrm{0.2307.}\end{array}$

Hence, distance from the point to the plane is $d=0.2307$ .