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Q.

In a $∆$ ABC,P,Q,R are points on BC,CA and AB respectively, dividing them in the ratio 1: 4,3: 2 and 3: 7 . The point S divides AB is the ratio 1:3 . Then $|\frac{\bar{AP} + \bar{BQ} + \bar{CR}}{CS}| =$

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a

7/10

b

1/5

c

5/2

d

2/5

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{OP} = \frac{\bar{c} + 4 \bar{b}}{5}; \bar{OQ} = \frac{3 \bar{a} + 2 \bar{c}}{5}, \bar{OR} = \frac{3 \bar{b} + 7 \bar{a}}{10}, \bar{OS} = \frac{\bar{b} + 3 \bar{a}}{4} \\ \bar{AP} = \frac{\bar{c} + 4 \bar{b} - 5 \bar{a}}{5}; \bar{BQ} = \frac{3 \bar{a} - 5 \bar{b} + 2 \bar{c}}{5}, \bar{CR} = \frac{3 \bar{b} + 7 \bar{a} - 10 \bar{c}}{10}, \bar{CS} = \frac{\bar{b} + 3 \bar{a} - 4 \bar{c}}{4} \\ |\frac{\bar{AP} + \bar{BQ} + \bar{CR}}{\bar{CS}}| = \frac{4}{10} = \frac{2}{5} \end{array}$

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In a ∆ABC,P,Q,R are points on BC,CA and AB respectively, dividing them in the ratio 1: 4,3: 2 and 3: 7 . The point S divides AB is the ratio 1:3 . Then AP¯+BQ¯+CR¯CS=