How to prove that point $P$ is collinear with $A$ and $B$? Referred to the origin $O$, $A$ and $B$ are two points which have position vectors a and b respectively.
Prove  that the point $P$ whose position vector p is given by p=$\lambda $a $+ (1-\lambda)$b is collinear with $A$ and $B$.
From the ratio theorem, I know that if a point $P$ divides a line $AB$ in the ratio $\lambda:\mu$, then $\overrightarrow {OP}= \displaystyle\frac{\lambda b +\mu a }{\lambda+\mu}$.
In this case, I know that $P$ lies on $AB$ such that $AP:PB=\lambda:1-\lambda$, but how do I prove it?
 A: Just rewrite the equality 
p=$\lambda $a $+ (1-\lambda)$b 
as    
$p-b = \lambda (a-b) $ 
which is the same as  
$\overrightarrow {BP} = \lambda \overrightarrow {BA} $ 
which gives you that the 3 points are collinear.  
Alternatively (if you're still not convinced): pick the point $M$ such that $M$ lies on $AB$ and divides it in a ratio $\lambda : 1-\lambda$. Then prove that $\overrightarrow {OM} = \overrightarrow{OP}$.  
But the last gives you that $M$ and $P$ coincide, and so $P$ also lies on $AB$ (because of how you chose $M$).  
I mean, this statement is trivial which (ironically) kind of makes it hard to prove if one gets somewhat confused about what is known already and what needs to be proved. 
A: A general point on the line $AB$ has position vector $$\mathbf p = \mathbf b+ \lambda (\mathbf a - \mathbf b) = \lambda \mathbf a + (1 - \lambda) \mathbf b$$
Since to be on the line we can first go to $B$ and then travel any distance in the direction parallel to $AB$.
Or, if you like, take that formula as given and note that $AP$ and $PB$ are both parallel to $AB$.
