# Thread: Vector Proof with projections

1. ## Vector Proof with projections

I'm given a plane in R^3 that passes through the origin with normal vector a. I'm also given that P is spanned by u and v.
The question asked is to show that for every x within P, x= proju(x) + projv(x).
All i can get from this is that the constants s and t for the parametric equation of the plane are equal to
proju(x) and projv(x). other than this I really have no idea where to start.

2. Unless I misunderstand the question, it turns out to be false unless u and v are perpendicular. Consider the following counter example:

Let the plane P be the XY plane (just to keep things simple) and let u=(1,2) and v=(1,0). These vectors span P (since they're linearly independent). Let's take x=(1,1).

Then $proj_u(x)=\frac{(1,2) \bullet (1,1)}{|(1,1)|^2}(1,2)=\frac{3}{2}(1,2)$

And $proj_v(x)=\frac{(1,0) \bullet (1,1)}{|(1,0)|^2}(1,0)=(1,0)$

So, $proj_u(x)+proj_v(x)=(\frac{3}{2}+1, 3+0)=(5/2, 3) \ne x$

So unless I'm misunderstanding or there's something missing, it's not true.