- #1

Specter

- 120

- 8

## Homework Statement

Find the shortest distance from P(-4,2,6) to the plane 2x-3y+z-8=0.

## Homework Equations

##|proj_\vec n \vec PQ|=|(\frac {\vec PQ \cdot \vec n} {\vec n \cdot \vec n})| |\vec n|##

## The Attempt at a Solution

I kind of had to guess some steps because it was done differently in my lesson and I couldn't figure it out that way. Am I on the right track?

The normal is ##\vec n =(2,-3,1)##

Find another point on the plane, let x=0 and y=0

2x-3y+z-8=0

2(0)-3(0)+z-8=0

z=0

Another point on the plane is Q(0,0,8).

I have to find the direction of ##\vec {PQ}##

##\vec {PQ}=\vec Q - \vec P##

=(0,0,8)-(-4,2,6)

=(4,-2,2)

To find the shortest distance, I think I would project ##\vec {PQ}## onto ##\vec n##.

##|proj_\vec n \vec PQ|=|(\frac {\vec PQ \cdot \vec n} {\vec n \cdot \vec n})| |\vec n|##

=##|\frac {(4,-2,2) \cdot (2,-3,1)} {(2,-3,1) \cdot (2,-3,1)}| |(2,-3,2)|##

=##|\frac {(4)(2)+(-2)(-3)+(2)(1)} {(2)(2)+(-3)(-3)+(1)(1)}| |(2,-3,1)|##

=##|\frac {16} {14}| |(2,-3,1)|##

=##|\sqrt (\frac {16} {7}), (-\frac {24} {7}), (\frac {8} {7})|##

=##\sqrt (\frac {16} {7})^2 + (-\frac {24} {7})^2 + (\frac {8} {7})^2##

=##\sqrt \frac {256} {49} + \frac {576} {49} + \frac {64} {49}##

=##\sqrt \frac {896}{49}##

The shortest distance is ##\sqrt \frac {896}{49}##