Chapter 11.4, Problem 65E

To determine

To find: The distance between a point and the given plane.

Answer to Problem 65E

  $D=\frac{4}{\sqrt{6}}\text{ units}$ .

Explanation of Solution

Given information:

An equation of the plane is given as $2x-y+z=4$ .

A point outside the plane is given as $\left(4,-2,-2\right)$ .

Concepts used:

The distance between a plane and a point (outside the plane) is given as −

  $D=\Vert {\text{proj}}_{\text{n}}\overrightarrow{PQ}\Vert =\frac{\left|\overrightarrow{PQ}\cdot \text{n}\right|}{\Vert \text{n}\Vert }$ .

Calculation:

Equation of a plane is given as,

  $2x-y+z=4$

So, normal vector to this plane will be given as −

  ${\text{n}}_1=\langle 2,-1,1\rangle$

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

  $\begin{array}{l}\Rightarrow 2x-0+0=4\\ \Rightarrow x=2\end{array}$

So, point $P\left(2,0,0\right)$ lies in the plane.

A point outside the plane is given as $Q\left(4,-2,-2\right)$ .

  $\begin{array}{l}\overrightarrow{PQ}=\langle 4-2,-2-0,-2-0\rangle \\ \Rightarrow \overrightarrow{PQ}=\langle 2,-2,-2\rangle \end{array}$

Now distance between point $Q\left(4,-2,-2\right)$ and the plane will be given as −

  $\begin{array}{l}D=\frac{\left|\overrightarrow{PQ}\cdot \text{n}\right|}{\Vert \text{n}\Vert }\\ \Rightarrow D=\frac{\left|\langle 2,-2,-2\rangle \cdot \langle 2,-1,1\rangle \right|}{\sqrt{2^2+{\left(-1\right)}^2+1^2}}\\ \Rightarrow D=\frac{\left|4+2-2\right|}{\sqrt{6}}\\ \Rightarrow D=\frac{4}{\sqrt{6}}\text{ units}\end{array}$

Hence, the distance between given point and the given plane is $D=\frac{4}{\sqrt{6}}\text{ units}$ .