Q.If $A(\bar a)$ and $B(\bar b)$ be any two points in the space and $R(\bar r)$ be a point on the line segment AB dividing it internally in the ratio $m : n$ then prove that $\bar r = \dfrac{m\bar b + n\bar a}{m+n}$.

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.

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

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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.

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

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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:

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

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

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