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Find an Equation of a Plane

by BraedenP
Tags: lines, planes
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BraedenP
#1
Dec4-10, 11:38 AM
P: 96
1. The problem statement, all variables and given/known data
Find the equation of the plane that contains the line [itex]x=-1+3t, y=5+2t, z=2-t[/itex] and is perpendicular to the plane [itex]2x-4y+2z=9[/itex]


2. Relevant equations

Equation of a plane:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]

3. The attempt at a solution
I was thinking that I could simply find an orthogonal normal to the normal of the specified plane. That would give me the normal of my new plane.

I found such a normal to be (3, 1, -1).

Then I'd simply take a the direction of the line (3, 2, -1), and plop it into the plane equation. This is where I get stuck; I can't figure out how to get the point specification into that equation. Is it simply:

[tex]3(3)x+1(2)y-1(-1)z = 9x+2y+z = 0[/tex]

It seems too easy for me. What am I doing wrong?
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Mark44
#2
Dec4-10, 12:13 PM
Mentor
P: 21,313
Quote Quote by BraedenP View Post
1. The problem statement, all variables and given/known data
Find the equation of the plane that contains the line [itex]x=-1+3t, y=5+2t, z=2-t[/itex] and is perpendicular to the plane [itex]2x-4y+2z=9[/itex]


2. Relevant equations

Equation of a plane:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]

3. The attempt at a solution
I was thinking that I could simply find an orthogonal normal to the normal of the specified plane. That would give me the normal of my new plane.

I found such a normal to be (3, 1, -1).
This is one vector that is perpendicular to the normal of the plane you are to find. The problem is that there are an infinite number of vectors that are perpendicular to that plane.
Quote Quote by BraedenP View Post

Then I'd simply take a the direction of the line (3, 2, -1), and plop it into the plane equation. This is where I get stuck; I can't figure out how to get the point specification into that equation. Is it simply:

[tex]3(3)x+1(2)y-1(-1)z = 9x+2y+z = 0[/tex]

It seems too easy for me. What am I doing wrong?
BraedenP
#3
Dec4-10, 01:23 PM
P: 96
Quote Quote by Mark44 View Post
This is one vector that is perpendicular to the normal of the plane you are to find. The problem is that there are an infinite number of vectors that are perpendicular to that plane.
There are, but all the rest of them would simply be scalar multiples of that one, right? Therefore, it doesn't matter which multiple is used in the equation; it'll work either way.

Mark44
#4
Dec4-10, 02:01 PM
Mentor
P: 21,313
Find an Equation of a Plane

No, not at all. All of the vectors that are perpendicular to <2, -4, 2> would lie in the same plane, but they point in all different directions.


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