Position Vector of a point given three other points Given a $\Delta ABC$ and a point $P$ such that $AB=a, BC=b, CA=c$ and $PA=\overrightarrow{A}, PB=\overrightarrow{B}, PC=\overrightarrow{C}$, then it's true that $$\vec P=x\vec A+ y\vec B+ z \vec C$$ for $$x+y+z=1$$.
Now, we can show that: $x=\frac{\text {Area}[PBC]}{\text{Area}[ABC]}$ and similarly for $y$ and $z$. (Take signed area or take $P$ inside $\Delta ABC$ for convenience.
Don't set values for any of the provided points and consider the origin as $O$ different from any of the above points.
How to show this (the bold portion)? One way is to brute force it by considering every length from $O$ and do it. I think that's a way but it might not be a good solution, I presume. Is there any other way using the section formula  $\bigg($ $\vec K= t\vec A + (1-t)\vec B$ (for $A,K,B$ collinear) $\bigg)$? Or any other way that's short and precise?

 A: For the proof you can use cross product showing that assuming $\overrightarrow{P}$ as a linear combination of $\overrightarrow{A}$, $\overrightarrow{B}$, $\overrightarrow{C}$:
$$\overrightarrow{P}=\alpha\overrightarrow{A}+\beta\overrightarrow{B}+\gamma \overrightarrow{C}$$ with $$\alpha+\beta+\gamma=1$$
$$\alpha,\beta,\gamma \in \mathbb{R}$$
then, considering signed areas, it can be shown that:
$[ABC]=\frac12 \vec{AB}\times \vec{AC} =\frac12 \vec{A} \times \vec{B}+\frac12 \vec{B} \times \vec{C}+\frac12 \vec{C} \times \vec{A}$
$[PBC]=\frac12 \vec{PB}\times \vec{PC} =\alpha(\frac12 \vec{A} \times \vec{B}+\frac12 \vec{B} \times \vec{C}+\frac12 \vec{C} \times \vec{A}) = \alpha [ABC]$
and similarly:
$[PAC]=\beta [ABC]$
$[PAB]=\gamma [ABC]$
observe that
$[PBC]+[PAC]+[PAB]=\alpha [ABC]+\beta [ABC]+\gamma [ABC]=(\alpha+\beta+\gamma)[ABC]=[ABC]$
NOTE
Every "point" (i.e. vector) in $\mathbb{R^3}$ can be expressed by a linear combination of three distinct"points" (i.e. vectors).
Thus, in general, the combination:
$$\overrightarrow{P}=\alpha\overrightarrow{A}+\beta\overrightarrow{B}+\gamma \overrightarrow{C}$$
will lead to a point away from the plane passing through A,B,C.
Assuming:
$$\alpha+\beta+\gamma=1$$
we are selecting a particular subspace of $\mathbb{R^3}$ that is just the plane passing through A,B,C.
Infact:


*

*this subspace is a plane since it has 2 dimension (i.e. only 2 parameters can fixed among $\alpha$, $\beta$ and $\gamma$);

*it passes through A,B, and C ($P \equiv A$ when $\alpha=1$ and so on).


Linear Algebra guarantees that the definition is well posed as every point P on the plane can be selected by an unique triple of real numers.
What has been shown is that $\alpha,\beta,\gamma$ have a precise geometric meaning as areal ratio.
