- #1
BraedenP
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Homework Statement
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]
Homework Equations
Equation of a plane:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]
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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