Cartesian equation of plane that i perpendicular to plane and contains line

craka

1. The problem statement, all variables and given/known data
Question states "The plane that contains the line r=<-2,4,3>+t<3,2-1> and is perpendicular to the plane r=<5,0,0>+s<2,1,0>+t<-1,0,1> is:"

Answer is y+2z=10


2. Relevant equations

Cross product and dot product of vectors


3. The attempt at a solution

I found a vector normal to the plane r=<5,0,0>=s<2,1,0>+t<-1,0,1>

by doing the cross product of the two direction vectors is <2,1,0> x <-1,0,1>
getting <1,-2,1>

than apply rule of n.v=0 ie <1,-2,1> . <x-(-2), y-4, z-3>
to get x+2-2y+8+z-3=0
and so x-2y+z = -7

Not sure what I have done wrong here, could someone explain please?

HallsofIvy

You found a plane that has normal vector <1, -2, 1>? Then you found a plane that is [b]paralllel[b] to r, not perpendicular to it!

craka

By doing the cross product of the direction vectors of r=<5,0,0>+s<2,1,0>+t<-1,0,1> am I not find a vector perpendicular to the plane?

HallsofIvy

Yes, that is correct. But in your form x-2y+z = -7, the vector of coefficients, <1, -2, 1> is perpendicular that this new plane. Since it was also perpendicular to the original plane, the two planes are parallel.

craka

Should I than from finding vector <1, -2, 1> do the cross product of it and the direction vector of the line <3,2-1> and than try and find the Cartesian equation of plane ?

gabbagabbahey

That should work.