Directional derivative question Calc 3 Help!!!

I just cannot figure out this problem. If someone can please show me how to solve it and explain how you reach your answer that would be very helpful.

The function f(x,y) at (1,2) has a directional derivative equal to2 in the direction towards (2,2) and equal to -2 in the direction toward (1,1). What is the directional derivative in the direction toward (2,3)?

Direction vector from (1, 2) to (2, 2) is (1, 0) [and is normalized].
Direction vector from (1, 2) to (1, 1) is (0, -1) [and is also normalized].

Direction vector from (1, 2) to (2, 3) is u = (1, 1) (normalized is (1, 1)/√2).

So, Du f(1, 2)
= ∇f(1, 2) · u
= ∇f(1, 2) · (1, 1)/√2
= ∇f(1, 2) · ((1, 0) - (0, -1))/√2
= (1/√2) [∇f(1, 2) · (1, 0) - ∇f(1, 2) · (0, -1)]
= (1/√2) [2 - (-2)], by the direction derivative information originally given
= 2√2.

I hope this helps!

From (1,2) to (2,2) is (1,0) and the directional derivative is 2.
So ∂f/∂x = 2.
From (1,2) to (1,1) is (0,-1) and the directional derivative is -2.
So ∂f/∂y = 2.
The gradient is grad(f) = (∂f/∂x, ∂f/∂y).
From (1,2) to (2,3) is (1,1). Unit vector is (.7,.7).
The directional derivitive is grad(f) • v = (2,2) • (.7,.7) = 2.8.
What bothers me is that this site
http://en.wikipedia.org/wiki/Gradient
doesn't say v has to be a unit vector.
It does, doesn't it?