It means simply they are perpindicular
proj = (u*v/|v|^2)v
so u*v = 0 u and v are perpindicular
If vector u and vector v are non-zero vectors, but Proj(u onto v) = vector 0, what conclusion can be drawn? If Proj(u onto v)= vector 0, does it follow that Proj(v onto u)=vector 0? Explain
Ok, so I am having trouble trying to explain this question. This is what I came up with so far, hopefully someone can add on or help answer this question better. Thanks for your help in advance.
Part A
when Proj(u onto v)=vector 0, that means that the projection vector is at the orgin, doesn't it?
Part B
I know that they are not equal to each other.. but I am not sure how to prove this.
No, that would be the 0 vector and you are told that u and v are non-zero vectors. Since u and v are non-zero vectors, |u| and |v| are non-zero so you can write the projection of u on v as . In order that that be the 0 vector, must be 0 so the vectors must be perpendicular.
While, in general, the "projection of u on v" is not the same as the "projection of v on u", being perpendicular to one another is "dual": if u is perpendicular to v, then v is perpendicular to u. If the projection of u on v is 0, then, by A, one of three things must be true:Part B
I know that they are not equal to each other.. but I am not sure how to prove this.
a) u= 0
b) v= 0
c) u is perpendicular to v.
In any of those, it follows that the projection of v on u is also 0.