let ABCD is trapezium such that $- - \to A B = 3 - - \to D E$ , E divides line segement AB internally in the ratio 2:1 and F is mid point of DC. if position vector of A,B and C are $\to a, \to b a n d \to c$ respectively then find the vector $- - \to F E$ .

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Solution

Solution :-
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Given $\to A B = 3 \to D C$

$(\to b - \to a) = 3 (\to c - \to d) - (1)$

$\Rightarrow 2 \to A E = E B$

$∴ \to E = (\frac{\to a + 2 \to b}{3})$ (By section formula)

Similarly $\to F = \frac{\to c + \to d}{2}$

$\to F E = \to E - \to F = \frac{\to a + 2 \to b}{3} - \frac{\to c + \to d}{2}$

$= \frac{2 \to a + 4 \to b - 3 \to c - 3 \to d}{6}$

From (1) $3 \to d = - \to b + \to a + 3 \to c$

$∴ \to F E = \frac{2 \to a + 4 \to b - 3 \to c - \to a + \to b - 3 \to c}{6}$

$\to F E = \frac{\to a + 5 \to b - 6 \to c}{6}$ Required vector

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let ABCD is trapezium such that −−→AB=3−−→DE , E divides line segement AB internally in the ratio 2:1 and F is mid point of DC. if position vector of A,B and C are →a,→band→c respectively then find the vector −−→FE.

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