Chapter 10, Problem 67RE

To determine

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

Answer to Problem 67RE

The distance between the point and the plane is $\frac{2}{\sqrt{440}}$ .

Explanation of Solution

Given information:

The given point is $\left(2,3,10\right)$ and the equation of the plane is $2x-20y+6z=6$ .

Formula used: The formula used to calculate the distance between the point P and the plane Q is:

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

Here, $n$ is normal to the plane.

Calculation:

The general equation of the plane is,

  $2x-20y+6z=6$

Substitute 0 for $y$ and 0 for $z$ in the equation of the plane.

  $\begin{array}{c}2x-20\left(0\right)+6\left(0\right)=6\\ x=3\end{array}$

So, the point on the plane be $P\left(3,0,0\right)$ .The vector from $P\left(3,0,0\right)$ to $Q\left(2,3,10\right)$ is:

  $\begin{array}{c}\overrightarrow{PQ}=\langle 2-3,3-0,10-0\rangle \\ =\langle -1,3,10\rangle \end{array}$

The distance between a plane and the point can be calculated as:

  $\begin{array}{c}D=\frac{\left|\overrightarrow{PQ}\cdot n\right|}{\Vert n\Vert }\\ =\frac{\left|\overrightarrow{PQ}\cdot n\right|}{\sqrt{a^2+b^2+c^2}}\end{array}$ …… (1)

Compare the given equation with the general form of the equation of plane $ax+by+cz+d=0$ to get:

  $\begin{array}{l}a=2\\ b=-20\\ c=6\end{array}$

Substitute all the values in the equation (1).

  $\begin{array}{c}D=\frac{\left|\langle -1,3,10\rangle \cdot \langle 2,-20,6\rangle \right|}{\sqrt{4+400+36}}\\ =\frac{\left|-2-60+60\right|}{\sqrt{440}}\\ =\frac{2}{\sqrt{440}}\end{array}$

Therefore, the distance between the point and the plane is $\frac{2}{\sqrt{440}}$ .