Finding the direction of steepest descent from (1,1)

[dcl]

Heya's
I need to find the direction in the xy plane in which one should travel, starting from point (1,1), to obtain the most rapid rate of decrease of
$$f(x,y,z) = (x + y - 2)^2 + (3x - y - 6)^2$$

now, $$\nabla f = (2(x+y-2), 2(3x-y-6))$$

so I'm thinking now I have to find the the unti vector 'u' which would be the direction in question. Unfortunately I do not know how to go on from here.
Somehow maximise $$\nabla f \cdot u$$ (where 'u' is a unit vector, don't know how to do vectors properly in latex )

 

[Icarus]

\vec{u} (I looked it up myself just a few hours ago).

First of all, $$\nabla f = (2(x + y - 2) + 6(3x - y -6), 2(x + y - 2) - 2(3x - y - 6))$$
Second, you are only concerned with what is going on at (1,1), where $$\nabla f$$ = (-24, 8).

Now, the question is to minimize (-24, 8)$$\cdot\vec{u}$$. Since $$\vec{v}\cdot\vec{w} = vw\cos \theta$$, where $$\theta$$ is the angle between them, we see that the minimum occurs when $$\theta = \pi$$. I.e., when the vectors are pointing in opposite directions.

So $$\vec{u}$$ must be the unital vector in the direction of (24, -8).

 

[dcl]

Silly me, I was straining all that time trying to figure out why my method is wrong and all the time I had been working with the wrong grad f. Thanks.