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Section Formula

Let position ...

Question

Let position vector of the points A,B and C are $\to a, \to b$ and $\to c$ respectively. Point D divides line segment BC internally in the ratio $2 : 1$ . Find vector AD

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Solution

P.V of $\to D = \frac{2 (\to (c) + 1 (\to b)}{2 + 1} = \frac{2 \to c + \to b}{3}$

we have to find $\to A D$

we know by triangular law of addition

$\to B A + \to A D = \to B D$
$\to A D = \to B D - \to B A$

P.V of $\to B D =$ P.V of $D -$ P.V of $B$

$= \frac{2 \to c + \to b}{3} - \to b = \frac{2 \to c - 2 \to b}{3}$
P.V of $\to B A = (\to a - \to b)$

$∴ \to A D = \frac{2 \to c - 2 \to b}{3} - \to a + \to b$
$= \frac{2}{3} \to c - \to a + \frac{1}{3} \to b$

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