Tension between two/three ropes using vectors Say there are three points, $a$, $b$, and $c$, with associated vectors $\vec{r_a}$, $\vec{r_b}$, and $\vec{r_c}$. $a$ and $b$ are both attached to firm surfaces, and each are connected to $c$ by ideal ropes. Force $\vec{F}$ acts on point $c$.
I am trying to find the tension in the rope connecting $a$ and $c$, let's call it $\vec{T_{ac}}$, with $\vec{T_{bc}}$ in the other rope.
First I use Newton's 2nd law to get $\vec{F} = \vec{T_{ac}} + \vec{T_{bc}}$.
Now is it wrong that $\vec{T_{ac}}$ is the projection of $\vec{F}$ onto $\vec{r_{ac}}$? Because that seems too easy. I think it involves balancing the components in each direction.
I can handle that in 2D, but then in the 3D case ($a$, $b$, and $c$ connected to $d$ with $\vec{F}$ acting on $d$), I'm having trouble grasping this.
I can't think of what coordinate system to use to get the components of the $\vec{T_{ij}}$s in each direction.
Any help for me and my 2D brain would be much appreciated.
 A: The same conditions hold for equilibrium. The sum of all forces must be $0$:
$$\sum \vec F=0$$
This can be broken into corresponding components:
$$\sum F_x=0$$
$$\sum F_y=0$$
$$\sum F_z=0$$
Below is an image of what I think the problem you are working with is. I have drawn the force vectors, and drawn a plane perpendicular to the applied force $\vec F$ at the common point of where all the forces are applied.

Let's say $\vec F$ points in the negative $z$ direction. As you can see, we can use the three triangles drawn to determine the $z$ components of each of the tension vectors. Their sum is determined by the magnitude of $\vec F$.
We can also use the triangles to figure out the projections of the tension vectors onto the specified plane (it can be the $x-y$ plane if you would like). These are the primed vectors specified in the diagram. From there the problem reduces into a 2D equilibrium problem in that plane, which it seems like you are able to grasp. 
I will leave the finer details to you.
A: The vector sum of the 3 forces must be zero. This means that :


*

*The 3 forces must lie in the same plane, and

*If you choose any 2 orthogonal directions in that plane, the sum of components of the forces in each of those 2 directions will be zero.

$F$ is the vector sum of $-T_a$ and $-T_b$. However $-T_a$ and $-T_b$ are not the orthogonal projections of $F$ onto ac and bc respectively. 
