Coulomb's law, balancing a ball with point charges

[theneedtoknow]

3 equal charges q are at the vertices of an equilateral triangle as shown, with the z-axis running through the midpoint of the triangle (such that the distance from each charge to the midpoint is d)

http://img111.imageshack.us/img111/7159/graphieew7.th.jpg

a bead of charge Qb (of equal sign as the 3 charges q) is supposed to be levitated on the positive z-azis (coming out of the midpoint of the triangle). Derive an expression for the coloumb force exerted on the bead as a function of its position on the positve z-axis

Homework Equations

F = (K * |q| * |Qb| ) / r^2 is the force in the radial direction (straight line connecting the 2 charges)

The Attempt at a Solution

OK I'm kind of having trouble here...

The force from each charge would simply be F = (K * |q| * |Qb| ) / (d^2 + z^2)
(by pythagorean theorem, the square of the line connecting each charge to any point on the positive z-azis would be d^2 + z^2)
So far so good
because of the way the charges are positioned positioned , the x and y components of the vectors will cancel each other in between the 3 charges (I think) so we're left toworry only about the z-component of each radial vector...but how do i do calculate for this z-component?

If i break down the radial force into 2 components, with one in the x-y plane and the other in the z-direction, then the angle etween the xy-plane component and the radial component is tan^-1 (z/d)...but now hat i have the hypotenuse ((K * |q| * |Qb| ) / (d^2 + z^2) ) and the angle tan^-1 (z/d), how do i isolate for just the z-component of the radial vector?
z-component of the vector would be the hypotenuse times cos of the angle...but i have the angle expressed as inverse tan of (z/d) so how can i take the cos of something i only have expressed as that? I am totally lost as to how to derive this expression...

 

[rl.bhat]

Radial force can be written as F = i*Fx + j*Fy + k*Fz
Direction of the F is given by direction cosines. z component of the F = Fz = F*cos(gamma)= F*z/r. This is due to one charge.

 

[theneedtoknow]

So how does cos(gamma) come out to z/r?

 

[theneedtoknow]

OH never midn i udnerstand where it comes from...If i do it by similar triangles than
Fz / Ftot = z / r
Fz = Ftot * z/r

So i guess Fz = kqQb * z / (Z^2+D^2)^2

thanks for the help :)