- #1
craka
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Homework Statement
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
Homework Equations
Cross product and dot product of vectors
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?