How would I show that the vector v=(-7,3,1) is not in the palne X+Y+2z=0?

Question

What vector in the plane would be the closest to v

Answer 1

if the vector is in the plane we at least have:
v.n=0
hence: n=(1,1,2) v=(-7,3,1) so : v.n = -7 + 3 +2 = -2
so the vector is not in the plane!

Answer 2

How would I show that the vector v=(-7,3,1) is not in the plane x+y+2z=0?

The vector is in the plane if you can put both ends of it into the plane. Pick an arbitrary point in the plane. I suggest (0,0,0). If the vector starts there it will end at point (-7,3,1). Now we need to check to see if that point is in the plane or not.

x+y+2z=0
-7+3+2*1 = -2 ≠ 0

So the vector is not in the plane.

Answer 3

-7 + 3 + 2 = -2 , not 0 as the equation says it should.

Answer 4

I don't know what would be considered "closest". (-7,3,2) is in the plane, and looks pretty close.

Answer 5

did you try -b/2a