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Find parametric question for the plane

by bonbon
Tags: parametric, plane
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bonbon
#1
May17-07, 12:35 PM
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1. The problem statement, all variables and given/known data
Give parametric questions for the plane : 2x-3y+z-6=0


3. The attempt at a solution

i know that the normal is (2,-3,1)
how do i find the direction vector of the plane?
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Dick
#2
May17-07, 02:59 PM
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The plane has two linearly independent 'direction vectors'. Find any two independent vectors perpendicular to your normal.
HallsofIvy
#3
May18-07, 07:44 AM
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Dick is assuming that by 'direction vector', you mean two independent vectors in the plane. I've only seen the term used with lines.

I wonder if you are not confusing this with "find the parametric equations for a line". To do that you would find 'direction vector' for the line. Parametric equations for a plane will involve two parameters. Here, since you can write z as a function of x and y, z= 6- 2x+ 3y, you can use x and y themselves as parameters or, if you prefer distinct variables, x= u, y= w, z= 6- 2u+3w.

Dick
#4
May18-07, 07:48 AM
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Find parametric question for the plane

That's a much more direct way. I was fixated on the 'direction vector' (basis) picture.


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