Find the equation of the following plane in both Cartesian and (vector) parametric form.(adsbygoogle = window.adsbygoogle || []).push({});

the plane through the point (6,5,-2) and parallel to the plane x+y-z+1=0

Cartesian form is just -x-y+z=d as both planes share the same normal. solving for d using (6,5,-2) gives -x-y+z=-13.

as for the vector form..

r = (6,5,-2) + s(x,y,z) + t(a,b,c) (s,t are real numbers)

How do i work out the (x,y,z) and (a,b,c)? Can i just take the dot product of each of them with the normal vector and find values that give 0. meaning they are at right angles (i hope thats right). and then just make sure that (x,y,z) and (a,b,c) aren't parallel?

e.g.

(-1,-1,1) . (x,y,z) = 0

-x-y+z=0

so (1,0,1) would work? then do the same with the (a,b,c) and say take (0,1,1)??

is this the right thing to do? is there an easier way?

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# Homework Help: Equation of a plane

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