Chapter 10.4, Problem 56E

To determine

To find: the distance between plane and the point.

Answer to Problem 56E

The distance between plane and point is $\frac{4\sqrt{6}}{3}$ .

Explanation of Solution

Given information:

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

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 $x-2y+z=3$ ,

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

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

Thus, the point in the plane is $P\left(3,0,0\right)$ .

Since, the point (not in the plane) is $Q\left(1,4,2\right)$ ,

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

  $\begin{array}{l}P\stackrel{\rightharpoonup }{Q}=\langle 1-3,4-0,2-0\rangle \\ P\stackrel{\rightharpoonup }{Q}=\langle -2,4,2\rangle \text{ [subtract]}\end{array}$

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

  $\begin{array}{l}D=\frac{|P\stackrel{\rightharpoonup }{Q}·n|}{\left|\right|n\left|\right|}\\ D=\frac{|\langle -2,4,2\rangle ·\langle 1,-2,1\rangle |}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(1\right)}^2}}\text{ [substitute]}\\ D=\frac{|-2-8+2|}{\sqrt{1+4+1}}\text{ [dot product and multiply]}\\ D=\frac{8}{\sqrt{6}}\text{ [simplify]}\\ D=\frac{4\sqrt{6}}{3}\text{ [simplify]}\end{array}$

Thus, the distance between point and plane is $\frac{4\sqrt{6}}{3}$ .