(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the equation of a plane with distance 3 units from the origin and perpendicular to the line through P(1,2,3) and Q(-2,4,1).

2. Relevant equations

[tex]\vec{n}[/tex] = [tex]\frac{\vec{PQ}}\left|{\vec{PQ}}\left|[/tex]

Plane Equation: a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0})

3. The attempt at a solution

Ok so I think I can solve this all the way up until the end.

We have [tex]\vec{PQ}[/tex] = <-3, -2, 2>

so [tex]\vec{n}[/tex] = [tex]\frac{1}{\sqrt{17}}[/tex]<-3,-2,2>

Now if I scale that vector by 3 or -3 I can get a point on the plane that I am looking for, I need to put this into the form of a plane so I use the equation above and end up getting:

[tex]\frac{-3}{\sqrt{17}}[/tex]x + [tex]\frac{2}{\sqrt{17}}[/tex]y - [tex]\frac{2}{\sqrt{}17}[/tex]z = [tex]\frac{-27}{17}[/tex] + [tex]\frac{18}{17}[/tex] + [tex]\frac{18}{17}[/tex]

This can be simplified further but this is where mine and my professors work differs. He gets the following on the right side of the equation = [tex]\frac{27}{17}[/tex] + [tex]\frac{12}{17}[/tex] + [tex]\frac{12}{17}[/tex]

Any ideas?

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# Homework Help: Equation of a plane

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