
#1
Dec410, 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=2t[/itex] and is perpendicular to the plane [itex]2x4y+2z=9[/itex] 2. Relevant equations Equation of a plane: [tex]a(xx_o)+b(yy_o)+c(zz_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)y1(1)z = 9x+2y+z = 0[/tex] It seems too easy for me. What am I doing wrong? 



#2
Dec410, 12:13 PM

Mentor
P: 20,941





#3
Dec410, 01:23 PM

P: 96





#4
Dec410, 02:01 PM

Mentor
P: 20,941

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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