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(B) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $ABCD$ is a parallelogram,$\vec{d} =

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$\frac{3}{2} AC$

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(B) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $ABCD$ is a parallelogram,$\vec{d} = \vec{a} + \vec{c} - \vec{b}$.$L$ is the mid-point of $BC$,so $\vec{l} = \frac{\vec{b} + \vec{c}}{2}$.$M$ is the mid-point of $CD$,so $\vec{m} = \frac{\vec{c} + \vec{d}}{2} = \frac{\vec{c} + (\vec{a} + \vec{c} - \vec{b})}{2} = \frac{\vec{a} + 2\vec{c} - \vec{b}}{2}$.Now,$\vec{AL} = \vec{l} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.And $\vec{AM} = \vec{m} - \vec{a} = \frac{\vec{a} + 2\vec{c} - \vec{b}}{2} - \vec{a} = \frac{2\vec{c} - \vec{b} - \vec{a}}{2}$.Adding these,$\vec{AL} + \vec{AM} = \frac{\vec{b} + \vec{c} - 2\vec{a} + 2\vec{c} - \vec{b} - \vec{a}}{2} = \frac{3\vec{c} - 3\vec{a}}{2} = \frac{3}{2}(\vec{c} - \vec{a}) = \frac{3}{2} \vec{AC}$.Thus,$AL + AM = \frac{3}{2} AC$.

For a parallelogram $ABCD$,if $L$ and $M$ are mid-points of $BC$ and $CD$ respectively,then $AL + AM =$

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