Q.If $\bar a, \bar b, \bar c$ are the position vectors of the points $A, B, C$ respectively and $5\bar a - 3\bar b - 2\bar c = \bar 0$, then find the ratio in which the point $C$ divides the line segment $BA$.

Rewrite $5\bar a-3\bar b-2\bar c=0$ as $\bar c=\dfrac{5\bar a-3\bar b}{5-3}$, the external-division form.

$5\bar a-3\bar b-2\bar c=\bar0 \Rightarrow \bar c=\dfrac{5\bar a-3\bar b}{2}$

…

Rewrite $5\bar a+3\bar b-8\bar c=0$ as $\bar c=\dfrac{5\bar a+3\bar b}{5+3}$ and read off the section-formula weights.

$5\bar a+3\bar b-8\bar c=\bar0 \Rightarrow \bar c=\dfrac{5\bar a+3\bar b}{8}$

…

Use the internal section formula $R=\dfrac{m\bar q+n\bar p}{m+n}$ for ratio $m:n=2:1$.

Given $\bar p=\hat i-2\hat j+\hat k$ (P.V. of P) and $\bar q=\hat i+4\hat j-2\hat k$ (P.V. of Q). R divides PQ internally in ratio $2:1$.

By the section formula, the position vector of R is:

$\bar r=\frac{2\bar q+1\cdot\bar p}{2+1}$

$=\frac{2(\hat i+4\hat j-2\hat k)+(\hat i-2\hat j+\hat k)}{3}$

…