Q:

if $\|AB\| / \|AC\| = \alpha$ show that $OB = \alpha OC + (1 - \alpha) OA$

Link to the diagram.  Note: this is for self study. 
http://i.imgur.com/nrlSoQO.jpg

How do i get started to solve this?.  
$ AB = OB - OA $ -> eq(1)
$ AC = OC - OA $ -> eq(2)
Taking norm on both sides of the eq(1), eq(2)
$ \|AB\| = \|OB\| - \|OA\| \  (triangle \ inequality \ \|A - B \| \geq \|A\| - \|B\| )$ -> eq(3)
$\|AC\| = \|OC\| - \|OA\| $ -> eq(4)
$\|AC\| = \frac{1}{\alpha}(\|OB\| - \|OA\|)$  [since $\frac{\|AB\|}{\|AC\|} = \alpha $] 
From eq(3)
$\alpha \|AC\| = \|OB\| - \|OA\|$  
$\alpha (\|OC\| - \|OA\|) = \|OB\| - \|OA\|$  
$\|OB\| = \alpha \|OC\| + (1 - \alpha) \|OA\|$ [since norm is positive]
$OB = \alpha OC + (1 - \alpha) OA$

A:

It is easier to solve the second equation for $\alpha$:
\begin{align}
OB 
&= \alpha OC + (1-\alpha) OA \\ 
&= \alpha (OC - OA) + OA \iff \\
OB - OA &= \alpha \, (OC-OA) \iff \\
AB &= \alpha AC \quad (*)
\end{align}
Now we take the vector norm of each side:
\begin{align}
(**) \quad
\lVert AB \rVert &= \lVert \alpha AC\rVert \iff \\
\lVert AB \rVert &= \lvert \alpha \rvert \, \lVert AC\rVert \iff \\
\lvert \alpha \rvert &= \frac{\lVert AB \rVert}{\lVert AC\rVert} \ge 0
\end{align}
Alas we have to prove backwards.  Thus two remarks:
If we define $\alpha$ by the right hand side, it must be non-negative, as a norm is non-negative, so we can drop the bars. 
To make the step from $(**)$ to $(*)$ we need to know that $AB$ and $AC$ are multiples, this can be inferred from the diagram.