Why is the addition of two vectors and the resultant of two forces are taken in two different ways? The sum of two vectors for example
$\vec{P}$ and $\vec{Q}$ are taken as $\vec{P}+\vec{Q}$ but when we take the resultant of two non parallel forces $\vec{P}$ and $\vec{Q}$, we take it as 
$$R^2 =P^2+ Q^2+ 2PQ \cos \theta$$
Why is that? Why can't we take the resultant of the two forces in the same manner we do in Vectors? As Force is a vector quantity? 
A simple explanation(that is without heavy technical terms) would be preferred)

 A: I think you are a bit confused.
The addition of two vectors (any vectors) is always done element by element
$$\mathbf{R} = \mathbf{P} + \mathbf{Q} = \pmatrix{P_x \\ P_y \\ P_z} + \pmatrix{Q_x\\Q_y\\Q_z} = \pmatrix{ P_x+Q_x \\ P_y+Q_y \\ P_z+Q_z}= \pmatrix{R_x \\ R_y \\ R_z} $$
and the magnitude of a vector is always defined by the Euclidean norm
$$ R =  \| \mathbf{R} \| = \| \pmatrix{R_x \\ R_y \\ R_z} \| = \sqrt{R_x^2+R_y^2+R_z^2} $$
The formula that you state is the cosine rule and you can derive it by taking the magnitude of the addition of two vectors. One with magnitude $P$ along the $x$-axis and one with magnitude $Q$ at an angle $\theta$ from the $x$-axis.
$$\begin{aligned} \mathbf{P} & = \pmatrix{P \\ 0} \\
\mathbf{Q} & = \pmatrix{Q \cos\theta \\ Q \sin \theta} \end{aligned}$$
$$ R = \| \mathbf{P} + \mathbf{Q} \| = \sqrt{ (P+Q \cos\theta)^2 + (Q \sin\theta)^2 } = \sqrt{P^2+Q^2+2 P Q \cos\theta} $$
NOTE: Vector quantities are bold (as in $\mathbf{P}$) and scalars are not (as in $P$).
A: you're talking of $\left |R^2 \right |$ so you are talking of $\left \| P+Q \right \|^2$. You know that $\left \| v \right \| is \sqrt{\left \langle v.v\right \rangle}$, so $\left \| P+Q \right \|^2=\left \langle P+Q.P+Q\right \rangle$ but the inner product is bilinear so $\left \| P+Q \right \|^2=\left \| P \right \|^2+\left \|Q \right \|^2+2\left \langle P.Q\right \rangle$ where $2\left \langle P,Q\right \rangle$ is $2PQcos\theta$.
I hope I didn't do any mistake.
