ABCD is a parallelogram and $A_{1}$ and $B_{1}$ are the mid-points of the sides BC and CD respectively.
$I f \to A A_{1} + \to A B_{1} = λ \to A C,$ then $^{'} λ^{'}$ is equal to

$\frac{1}{2}$

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$\frac{3}{2}$

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Solution

The correct option is C
$\frac{3}{2}$

$Let the position vectors of A, B, C, D be \to 0, \to b, \to c a n d \to d$
$Then position vector of C, \to c = \to b + \to d$
$Also, position vector of A_{1} = \to b + \frac{\to d}{2}$

$and position vector of B_{1} = \to d + \frac{\to b}{2} \Rightarrow {\to A A}_{1} + {\to A B}_{1} = \frac{3}{2} (\to a + \to d) = \frac{3}{2} \to A C$

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ABCD is a parallelogram and $A_{1}$ and $B_{1}$ are the mid-points of the sides BC and CD respectively.
$I f \to A A_{1} + \to A B_{1} = λ \to A C,$ then $^{'} λ^{'}$ is equal to

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ABCD is a parallelogram and $A_{1}$ and $B_{1}$ are the mid-points of the sides BC and CD respectively.
$I f \to A A_{1} + \to A B_{1} = λ \to A C,$ then $^{'} λ^{'}$ is equal to