Calculating Moment of Forces: Cartesian Vector Analysis for Column at Point A

[suspenc3]

I just realized this is in the wrong place..Move it to intro to physics please. Sorry

Homework Statement

Using Cartesian vector analysis determine the moment of the three forces about the base of the column at A. F1 =400i +300j+120k

Homework Equations

The $$x$$ represents "cross"..i don't know the latex for it

$$M_a = r x F$$ (summation)

The Attempt at a Solution

I keep getting the wrong answer and I think it has something to do with the point E. What does this have to do with anything?

What I am doing is summing all the forces crossed with their distances.

$$r_B={12k}N$$
$$r_{F3}={0i-j+8k}N$$

So $$M_A = (r_B x F_1) + (r_B x F_2) + (r_{F3} x F_3)$$

 

[member 553083]

M_A = {12k x 400i} + {12k x 300j} + {0i-j+8k x 120k}M_A = 4800k x -300j + 1440i x 8kM_A = -1440k^2 + 384k^2M_A = -1056k^2