Chapter 11, Problem 17CLT

To determine

To Calculate:

Distance between given point and plane.

Answer to Problem 17CLT

The perpendicular distance between the given point and the plane is $\frac{\sqrt{30}}{2}$ .

Explanation of Solution

Given Information:

Point: $\left(0,0,25\right)$

Plane: $2x-5y+z=10$

Consider the following points and plane:

Equation of the plane: $2x-5y+z=10$

Point: $\left(0,0,25\right)$

To get the perpendicular distance between then procced as follows:

Get a point on the plane by equating $y$ and $z$ equal to zero.

  $\begin{array}{l}2x-5y+z=10\\ 2x-5\left(0\right)+\left(0\right)=10\\ 2x=10\\ x=5\end{array}$

Hence a point on the plane is $\left(4,0,0\right)$

Now get the displacement vector between the given point and the point on the plane.

  $\begin{array}{l}\text{d}=\langle 4,0,0\rangle -\langle 0,0,25\rangle \\ \text{d}=\langle 5,0,-25\rangle \end{array}$

Now get the normal vector of the plane using the standard result:

Normal vector of the plane whose equation is $ax+by+cz=d$ is $\langle a,b,c\rangle$

Hence the normal vector of the plane is $n=\langle 2,-5,1\rangle$

Now find the dot product of normal vector and displacement vector.

  $\begin{array}{l}\text{d}.n=\langle 2,-5,1\rangle .\langle 5,0,-25\rangle \\ =10+0-25\\ =-15\end{array}$

Now, get the modulus of the normal vector

Modulus of the normal vector is

  $\begin{array}{l}\langle 2,-5,1\rangle =\sqrt{2^2+{\left(-5\right)}^2+1^2}\\ =\sqrt{4+25+1}\\ =\sqrt{30}\end{array}$

Now, the value of $\frac{\text{d}.n}{\langle 2,-5,1\rangle }$

Therefore,

  $\begin{array}{l}\frac{\text{d}.n}{\langle 2,-5,1\rangle }=\frac{15}{\sqrt{30}}\\ =\frac{15}{\sqrt{30}}\left(\frac{\sqrt{30}}{\sqrt{30}}\right)\\ =\frac{15\sqrt{30}}{30}\\ =\frac{\sqrt{30}}{2}\end{array}$

Hence, the perpendicular distance between the point and the plane is $\frac{\sqrt{30}}{2}$ .