Let the position vectors of two points A and B be $a + b + c$ and $a - 2 b + 3 c$ respectively. If the points P and Q divide AB in the ratio 1:3 internally and externally respectively, they $3 |AB| =$

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$4 |PQ|$

$3 |PQ|$

$\frac{1}{2} |PQ|$

$|PQ|$

answer is A.

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Detailed Solution

$OP = \frac{a - 2 b + 3 c + 3 a + 3 b + 3 c}{4} = \frac{4 a + b + b c}{4} OQ = \frac{a - 2 b + 3 c - 3 a - 3 b - 3 c}{- 2} = \frac{- 2 a - 5 b}{- 2} 3 |AB| = 3 |- 3 b + 2 c| = \frac{2 a + 5 b}{2} |PQ| = |9 b - 6 c| = \frac{4 a + 10 b}{4}$

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