Step-by-step explanation:
Assuming that all forces extend from the origin point, with
and
lying in the xy plane, so
is along the z axis. So, we have:
![\vec{F_1}=F_1\hat{i}+0\hat{j}+0\hat{k}\\\vec{F_2}=F_2cos\theta\hat{i}+F_2sin\theta\hat{j}+0\hat{k}\\\vec{F_3}=0\hat{i}+0\hat{j}+F_3\hat{k}](https://img.qammunity.org/2020/formulas/physics/high-school/rw8qr965zz8sgfry3ea4svml8wvjyi90q2.png)
The net force is:
![\vec{F}=\vec{F_1}+\vec{F_2}+\vec{F_3}\\\vec{F}=(F_1+F_2cos\theta)\hat{i}+F_2sin\theta\hat{j}+F_3\hat{k}](https://img.qammunity.org/2020/formulas/physics/high-school/obaeyhyuoklg9xkbwlwygz56zmhcskzin2.png)
The force (
) that would exactly counterbalance these three forces will be opposite in direction and equal in magnitude to the net force:
![\vec{F_4}=-(F_1+F_2cos\theta)\hat{i}-F_2sin\theta\hat{j}-F_3\hat{k}\\F_4=√((-(F_1+F_2cos\theta))^2+(-F_2sin\theta)^2+(-F_3)^2)](https://img.qammunity.org/2020/formulas/physics/high-school/38rq6yjucecp39u2wmznayq7lq9xdu0fc1.png)