Let ABCD be a quadrilateral. If E and F are the midpoints of the… — Solution | JEE Mathematics

MCQMediumJEE 2023Basics of Vectors

Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$ respectively and

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

then $k$ is equal to

Answer

Correct answer:A

Standard Method

Given: $ABCD$ is a quadrilateral. $E$ and $F$ are the midpoints of diagonals $AC$ and $BD$ respectively.

Find: The value of $k$ in

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

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

Then,

\overrightarrow{AB}=\vec b-\vec a,\quad \overrightarrow{BC}=\vec c-\vec b

and

\overrightarrow{AD}=\vec d-\vec a,\quad \overrightarrow{DC}=\vec c-\vec d

So,

(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})

=(\vec b-\vec a-\vec c+\vec b)+(\vec d-\vec a-\vec c+\vec d)

=2(\vec b+\vec d-\vec a-\vec c)

Since $E$ and $F$ are the midpoints of $AC$ and $BD$ ,

\vec E=\frac{\vec a+\vec c}{2},\quad \vec F=\frac{\vec b+\vec d}{2}

Hence,

\overrightarrow{FE}=\vec E-\vec F=\frac{\vec a+\vec c-\vec b-\vec d}{2}

Therefore,

\vec a+\vec c-\vec b-\vec d=2\,\overrightarrow{FE}

Now,

2(\vec b+\vec d-\vec a-\vec c)=-2(\vec a+\vec c-\vec b-\vec d)

=-2\times 2\,\overrightarrow{FE}

=-4\,\overrightarrow{FE}

Comparing with

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

we get $k=-4$ .

Therefore, the correct option is A.

Midpoint Vector Shortcut

Given: Midpoints $E$ of $AC$ and $F$ of $BD$ .

Find: The constant $k$ .

Use midpoint vectors directly:

\overrightarrow{FE}=\frac{1}{2}\left((\vec a+\vec c)-(\vec b+\vec d)\right)

So,

\vec b+\vec d-\vec a-\vec c=-2\,\overrightarrow{FE}

From the given left-hand side,

(\overrightarrow{AB}-\overrightarrow{BC})+(\overrightarrow{AD}-\overrightarrow{DC})=2(\vec b+\vec d-\vec a-\vec c)

Thus,

2(\vec b+\vec d-\vec a-\vec c)=2(-2\,\overrightarrow{FE})=-4\,\overrightarrow{FE}

Hence, $k=-4$ , so the correct option is A.

Taking $\overrightarrow{EF}$ instead of $\overrightarrow{FE}$ . These vectors differ by a negative sign, so this changes the value of $k$ . Always check the direction carefully before comparing coefficients.
Using midpoint coordinates incorrectly. For a midpoint, the position vector is the average of the endpoint position vectors, so $\vec E=\frac{\vec a+\vec c}{2}$ and $\vec F=\frac{\vec b+\vec d}{2}$ . Do not add them without dividing by $2$ .
Expanding $\overrightarrow{AB}-\overrightarrow{BC}$ or $\overrightarrow{AD}-\overrightarrow{DC}$ with wrong signs. Write each vector first in position-vector form, then substitute carefully to avoid sign errors.

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