Chapter 10.4, Problem 55E

To determine

To find: the distance between plane and the point.

Answer to Problem 55E

The distance between plane and point is $\frac{3\sqrt{5}}{5}$ .

Explanation of Solution

Given information:

The point is $\left(1,3,4\right)$ and plane is $4x-5y+2z=6$ .

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 $4x-5y+2z=6$ ,

Thus, the normal vector is $n=\langle 4,-5,2\rangle$ and to find the point in the plane substitute $y=0,z=0$ in the plane,

  $\begin{array}{l}4x-5y+2z=6\\ 4x-5\left(0\right)+2\left(0\right)=6\text{ [substitute]}\\ x=\frac{3}{2}\text{ [simplify]}\end{array}$

Thus, the point in the plane is $P\left(\frac{3}{2},0,0\right)$ .

Since, the point (not in the plane) is $Q\left(1,3,4\right)$ ,

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

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

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

  $\begin{array}{l}D=\frac{|P\stackrel{\rightharpoonup }{Q}·n|}{\left|\right|n\left|\right|}\\ D=\frac{|\langle -\frac{1}{2},3,4\rangle ·\langle 4,-5,2\rangle |}{\sqrt{{\left(4\right)}^2+{\left(-5\right)}^2+{\left(2\right)}^2}}\text{ [substitute]}\\ D=\frac{|-2-15+8|}{\sqrt{16+25+4}}\text{ [dot product and multiply]}\\ D=\frac{3}{\sqrt{15}}\text{ [simplify]}\\ D=\frac{3\sqrt{5}}{5}\text{ [simplify]}\end{array}$

Thus, the distance between point and plane is $\frac{3\sqrt{5}}{5}$ .