- 339

- 0

is a vector in 3d space and it gets projected to 2d space. So [1 2 3]--->[1 2 0]

Within the null space is [0 0 3], which is perpendicular to every vector in the x-y plane,

not to mention the inner product of [0 0 3] and (column)[1 2 0] is 0, which shows

that [0 0 3] is perpendicular to [1 2 0]. What also gets me is that I have seen

a picture of two vectors and one vector projected (orthogonally) onto the

other vector, but both were on the x-y plane, yet orthogonality is shown

by taking the inner product between two vectors and getting 0. I can't see

how 0 can be obtained when both vectors are on the same plane (note that they

were in the positive x positive y quadrant).