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?
Best Answer
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.