$$
\mathbf{c}=(70.7,70.7) \mathrm{N}, \quad \mathbf{d}=(80.0,0.0) \mathrm{N} \quad\left(\mathrm{N}=\mathrm{Newton}=1 \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}}\right)
$$
It holds that
$$
\begin{aligned}
\mathbf{F}{\text {ges }} &=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(-95.0,71.0) \mathrm{N} \ \left|\mathbf{F}{\text {ges }}\right| &=\sqrt{95.0^{2}+71.0^{2}} \mathrm{~N}=118.6 \mathrm{~N}
\end{aligned}
$$
We remember that
$$
\mathbf{a}+\mathbf{b}=\sum_{i} a_{i} \mathbf{e}{i}+\sum{i} b_{i} \mathbf{e}{i}=\sum{i}\left(a_{i}+b_{i}\right) \mathbf{e}{i}=\left(a{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}\right)
$$
Graphical determination of the force: Representation by means of polygon of forces.
The angle $\beta$ enclosed by $\mathbf{F}$ and the $x$-axis may be calculated easily. One has
a
$$
\mathbf{F}=(-95.0,71.0) \mathrm{N}, \quad \frac{F_{y}}{F_{x}}=\tan \beta=-\frac{71.0}{95.0}
$$
b
from there it follows that $\beta=143^{\circ}$.
PHS1011 COURSE NOTES ：
For the vector product in it then results that $\mathbf{R}=\left(\mathbf{r}{1}-\mathbf{r}{2}\right) \times\left(\mathbf{r}{1}-\mathbf{r}{3}\right)=(-2,3,-1) \mathrm{m}^{2}=\left(R_{x}, R_{y}, R_{z}\right) .$
Inserting these values in one obtains
$$
\begin{aligned}
\lambda &=\frac{1}{2 \mathrm{~N}} R_{x}=-1 \mathrm{~m}^{2} \mathrm{~N}^{-1} \
F_{y} &=\frac{R_{y}}{\lambda}=-3 \mathrm{~N} \
F_{z} &=\frac{R_{z}}{\lambda}=1 \mathrm{~N}
\end{aligned}
$$
Thus, the components of the force $\mathbf{F}$ are
$$
\mathbf{F}=(2,-3,1) \mathrm{N} \text {. }
$$