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Vector Addition

Let ABCD be...

Question

Let $A B C D$ be a trapezium and let $P, Q$ be the midpoints of the nonparallel sides $A D, B C$ respectively. Then $\to P Q$

\to A B + \to D C

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\frac{1}{4} (\to A B + \to B C)

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\frac{1}{2} (\to A B + \to D C)

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\frac{1}{2} (\to A B + \to B C)

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Solution

The correct option is D $\frac{1}{2} (\to A B + \to D C)$
$A P = P D$ ( $P$ is mid point)
$B Q = Q C$ ( $Q$ is mid point)
Let $\to a, \to b, \to c, \to d$ are position vector of $A B C D$ respectively
position vector of $P = \frac{\to a + \to d}{2}$
position vector of $Q = \frac{\to b + \to c}{2}$
$\to P Q = \frac{1}{2} (\to b - \to a - \to d + \to c)$
$= \frac{1}{2} (\to A B - \to C D)$
$= \frac{1}{2} (\to A B + \to D C)$

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