Chapter 16.1, Problem 10PS

To determine

To find: Thedistance from the given point to the graph of the given equation.

Answer to Problem 10PS

The distance from the given point to the graph of the given equation is $2.208\text{units}$ .

Explanation of Solution

Given information:

The given equation is,

  $3x-7y+5z+13=0$

The distance from the given point to the graph of the given equation is calculated as,

  $\begin{array}{c}d=\frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}}\\ =\frac{\left|3\left(5\right)+\left(-7\right)\left(2\right)+5\left(1\right)+13\right|}{\sqrt{{\left(3\right)}^2+{\left(-7\right)}^2+{\left(5\right)}^2}}\\ =\frac{19}{\sqrt{83}}\\ =2.208\text{units}\end{array}$

Thus, the distance from the given point to the graph of the given equation is $2.208\text{units}$ .