Chapter 10, Problem 58CR

To determine

To find: the distance between plane and the point.

Answer to Problem 58CR

The distance between plane and point is $\frac{6\sqrt{14}}{7}$ .

Explanation of Solution

Given information:

The point is $\left(0,0,0\right)$ and plane is $2x+3y+z=12$ .

Formula used: the distance between point and plane is,

  $D=\frac{|P\stackrel{\rightharpoonup }{Q}·n|}{\left|\right|n\left|\right|}$

Calculation: since, the plane is $2x+3y+z=12$ ,

Thus, the normal vector is $n=\langle 2,3,1\rangle$ and to find the point in the plane substitute $y=0,z=0$ in the plane,

  $\begin{array}{l}2x+3y+z=12\\ 2x+3\left(0\right)+\left(0\right)=12\text{ [substitute]}\\ x=6\text{ [simplify]}\end{array}$

Thus, the point in the plane is $P\left(6,0,0\right)$ .

Since, the point (not in the plane) is $Q\left(0,0,0\right)$ ,

The vector from $P\text{ to }Q$ is

  $\begin{array}{l}P\stackrel{\rightharpoonup }{Q}=\langle 6-0,0-0,0-0\rangle \\ P\stackrel{\rightharpoonup }{Q}=\langle 6,0,0\rangle \text{ [subtract]}\end{array}$

Now, use the distance formula and then substitute and simplify,

  $\begin{array}{l}D=\frac{|P\stackrel{\rightharpoonup }{Q}·n|}{\left|\right|n\left|\right|}\\ D=\frac{|\langle 6,0,0\rangle ·\langle 2,3,1\rangle |}{\sqrt{{\left(2\right)}^2+{\left(3\right)}^2+{\left(1\right)}^2}}\text{ [substitute]}\\ D=\frac{|12+0+0|}{\sqrt{4+9+1}}\text{ [dot product and multiply]}\\ D=\frac{12}{\sqrt{14}}\text{ [simplify]}\\ D=\frac{6\sqrt{14}}{7}\text{ [simplify]}\end{array}$

Thus, the distance between point and plane is $\frac{6\sqrt{14}}{7}$ .