Chapter 10.4, Problem 57E

To determine

To find: The distance between the point $\left(1,3,4\right)$ and the plane $4x-5y+2z=6$ .

Answer to Problem 57E

The distance between the point $\left(1,3,4\right)$ and the plane $4x-5y+2z=6$ is $\frac{9}{\sqrt{45}}$ .

Explanation of Solution

Given information:

The points and plane is $\left(1,3,4\right)$ and $4x-5y+2z=6$ .

Calculation:

The normal vector will be the coefficients of $x,y$ and $z$ of the plane.

  $n=\langle 4,-5,2\rangle$

To find the point in the plane let $y=0$ and $z=0$ . Then,

  $\begin{array}{c}4x-0+0=6\\ 4x=6\\ x=\frac{3}{2}\end{array}$

The point in the plane will be $P\left(\frac{3}{2},0,0\right)$ .

Find the vector from P and Q .

  $\begin{array}{c}\overrightarrow{PQ}=\langle 1-\frac{3}{2},3-0,4-0\rangle \\ =\langle -\frac{1}{2},3,4\rangle \end{array}$

The formula for the distance between a point and a plane is given by,

  $D=\frac{\left|\overrightarrow{PQ}\cdot n\right|}{\Vert n\Vert }$

Find the distance between a point and a plane as follows.

  $\begin{array}{c}D=\frac{\left|\langle -\frac{1}{2},3,4\rangle \cdot \langle 4,-5,2\rangle \right|}{\sqrt{4^2+{\left(-5\right)}^2+2^2}}\\ =\frac{\left|-2-15+8\right|}{\sqrt{16+25+4}}\\ =\frac{9}{\sqrt{45}}\end{array}$

Therefore, the distance between the point $\left(1,3,4\right)$ and the plane $4x-5y+2z=6$ is $\frac{9}{\sqrt{45}}$ .