Theorem
1. The projection vector of b on a is 
and its magnitude is
Proof:
Let OA = a and OB =
b, P be the foot of the perpendicular from B on OA and θ = ∠AOB
Case–1: θ is an acute angle.
By definition,
| The projection of b on a | = | OP |
| = | |OP|  |
|
| = | (OB) cos θ |
|
| = | (|b| cos θ) |
|
| = | (|a|.|b| cos θ)  |
|
| = |  |
Case–2: θ is an obtuse angle.
OP is in the opposite direction of a
and hence the angle (b, OP) is π – θ.
| ∴ The projection of b on a | = | OP |
| = | |OP|  |
|
| = | (OB) cos (π – θ) |
|
| = | – (OB) cos θ |
|
| = | (OB) cos θ |
|
| = | (|a| |b| cos θ) |
|
| = | |
Case–3: θ is a right angle.
Then P coincides with ' O ' so that OP
= O and also a.b = 0
∴ OP =
∴ The projection vector of b on a =
and its magnitude is