QJEEM23Mac649: JEE Main 2023 Physics - Class 11 JEEM23Mac649 Solution: Let be a quadrilateral. If and are the mid points of the…

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

Then the position vector of $E$ is:

$\vec{E} = \frac{\vec{a} + \vec{c}}{2}$

And the position vector of $F$ is:

$\vec{F} = \frac{\vec{b} + \vec{d}}{2}$

Now, we are given the equation:

$(\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE}$

We can rewrite this equation using the position vectors:

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

Simplifying the equation, we get:

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

Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:

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

Simplifying the equation:

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

Since both sides of the equation are equal:

$k = -4$