Given a Parallellogram $OACB$, How to Evaluate $\vec{OC}.\vec{OB}$? Question
Given a parallellogram $OACB$. The lengths of the vectors $\vec{OA},\vec{OB}  \ \text{and}\ \vec{AB}$ are $a,b$ and $c$ respectively. The scalar product of the vectors $\vec{OC}$ and $\vec{OB}$ is

$(A)\frac{a^2-3b^2+c^2}{2}\hspace{1cm}(B)\frac{3a^2+b^2-c^2}{2}\hspace{1cm}(C)\frac{3a^2-b^2+c^2}{2}\hspace{1cm}(D)\frac{a^2+3b^2-c^2}{2}$

My Attempt:
Let $O$ be $(0,0,0)$,position vector of $A$ be $\vec{a}$,position vector of $B$ be $\vec{b}$ and position vector of $C$ is $\vec{a}+\vec{b}$. Next, I write
$$\vec{OC}.\vec{OB}=|\vec{OC}||\vec{OB}|\cos\angle BOC$$
but here I am stuck. I don't know the angle $\angle BOC$! 
Any help will be appreciated.
 A: I think there is no need to memorize the cosine law. Put the origin of the coordinate system at O, then we can write
$$\mathop {OB}\limits^ \to  .\mathop {OC}\limits^ \to   = {\bf{B}}.{\bf{C}} = {\bf{B}}.\left( {{\bf{A}} + {\bf{B}}} \right) = {\bf{A}}.{\bf{B}} + {\bf{B}}.{\bf{B}} = {\bf{A}}.{\bf{B}} + {b^2}\tag{1}$$
so, your main task will be to compute ${\bf{A}}.{\bf{B}}$, you can do it as follows
$$\eqalign{
  & \mathop {AB}\limits^ \to  .\mathop {AB}\limits^ \to   = \left( {{\bf{B}} - {\bf{A}}} \right).\left( {{\bf{B}} - {\bf{A}}} \right) = {\bf{A}}.{\bf{A}} + {\bf{B}}.{\bf{B}} - 2{\bf{A}}.{\bf{B}}  \cr 
  & {c^2} = {a^2} + {b^2} - 2{\bf{A}}.{\bf{B}}  \cr 
  & {\bf{A}}.{\bf{B}} = \frac{{{a^2} + {b^2} - {c^2}}}{2} \cr}\tag{2}$$
Now, combine $(1)$ and $(2)$ to get
$$\mathop {OB}\limits^ \to  .\mathop {OC}\limits^ \to   = \frac{{{a^2} + {b^2} - {c^2}}}{2} + {b^2} = \frac{{{a^2} + 3{b^2} - {c^2}}}{2}\tag{3}$$
A: By the law of cosines,
$$\cos\angle{AOB}=\frac{|\vec{OA}|^2+|\vec{OB}|^2-|\vec{AB}|^2}{2|\vec{OA}||\vec{OB}|}=\frac{a^2+b^2-c^2}{2ab}$$
So,
$$\begin{align}\vec{OC}\cdot\vec{OB}&=(\vec{OA}+\vec{OB})\cdot\vec{OB}\\\\&=\vec{OA}\cdot\vec{OB}+|\vec{OB}|^2\\\\&=|\vec{OA}||\vec{OB}|\cos\angle{AOB}+b^2\\\\&=ab\cdot\frac{a^2+b^2-c^2}{2ab}+b^2\\\\&=\frac{a^2+3b^2-c^2}{2}\end{align}$$
