Let ABCD be a quadrilateral. If E and F are the midpoints of the diagonals AC and BD respectively and ( A B - B C ) + ( A D - D C) = k ? F E , then k is equal to [2023]

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Solution

Q.
Let ABCD be a quadrilateral. If E and F are the midpoints of the diagonals AC and BD respectively and $(\vec{A B} - \vec{B C}) + (\vec{A D} - \vec{D C}) = k \vec{F E},$ then $k$ is equal to [2023]

1 - 4

2 2

3 - 2

4 4

Ans.
(1)

$Let the position vectors of A, B, C, D of quadrilateral A B C D be \vec{a}, \vec{b}, \vec{c}, \vec{d} respectively.$

$Given, E is the midpoint of A C$

$∴$ $Position vector of E = \frac{1}{2} (\vec{a} + \vec{c})$

$and F is the midpoint of B D$

$∴$ $Position vector of F = \frac{1}{2} (\vec{b} + \vec{d})$

$\vec{F E} = Position vector of E - Position vector of F$

$= \frac{1}{2} (\vec{a} + \vec{c}) - \frac{1}{2} (\vec{b} + \vec{d}) = \frac{1}{2} (\vec{a} + \vec{c} - \vec{b} - \vec{d})$

$Now,$ $\vec{A B} - \vec{B C} + \vec{A D} - \vec{D C} = (\vec{b} - \vec{a}) - (\vec{c} - \vec{b}) + (\vec{d} - \vec{a}) - (\vec{c} - \vec{d})$

$= \vec{b} - \vec{a} - \vec{c} + \vec{b} + \vec{d} - \vec{a} - \vec{c} + \vec{d}$ $= - 2 \vec{a} + 2 \vec{b} - 2 \vec{c} + 2 \vec{d}$

$= - 2 (\vec{a} - \vec{b} + \vec{c} - \vec{d}) = - 4 \vec{F E}$

$Hence, k = - 4$

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