Q.Find the position vector of point R which divides the line joining the points P and Q whose position vectors are $2\hat i-\hat j+3\hat k$ and $-5\hat i+2\hat j-5\hat k$ in the ratio $3:2$ internally.

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.

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

The external division of segment $BA$ in ratio $m:n$ gives point $\dfrac{m\bar a-n\bar b}{m-n}$. Comparing $\dfrac{5\bar a-3\bar b}{5-3}=\dfrac{5\bar a-3\bar b}{2}$, we get $m=5,n=3$.

So $C$ divides $BA$ externally in the ratio $5:3$.

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.

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

By the section formula, a point dividing $AB$ in ratio $m:n$ (from $A$) has position vector $\dfrac{n\bar a+m\bar b}{m+n}$. Comparing, $n=5,\ m=3$.

So $C$ divides $AB$ internally in the ratio $3:5$.

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:

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

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

$=\frac{(2\hat i+8\hat j-4\hat k)+(\hat i-2\hat j+\hat k)}{3}=\frac{3\hat i+6\hat j-3\hat k}{3}$

$=\hat i+2\hat j-\hat k$