Question

\(M\) and \(N\) are the mid points of the sides \(BC\) and \(CD\) of a parallelogram \(ABCD\) respectively, then \[ \overrightarrow{AM}+\overrightarrow{AN}= \]

• A. \(\frac{1}{3}\overrightarrow{AC}\)
• B. \(\frac{2}{3}\overrightarrow{AC}\)
• C. \(\frac{3}{4}\overrightarrow{AC}\)
• D. \(\frac{3}{2}\overrightarrow{AC}\)

Hint:

In a parallelogram, if \[ \overrightarrow{AB}=\vec{u},\quad \overrightarrow{AD}=\vec{v}, \] then \[ \overrightarrow{AC}=\vec{u}+\vec{v}. \] Midpoint vectors can be found by taking the average of endpoint position vectors.

Answer 1

The Correct Option is D

Step 1: Set up position vectors.
In parallelogram $ABCD$, take $A$ as origin. Let $\overrightarrow{AB}=\vec{u}$, $\overrightarrow{AD}=\vec{v}$. Then $B=\vec{u}$, $D=\vec{v}$, $C=\vec{u}+\vec{v}$, and $\overrightarrow{AC}=\vec{u}+\vec{v}$.

Step 2: Find $M$ (midpoint of $BC$).
$B = \vec{u}$, $C = \vec{u}+\vec{v}$. Midpoint: $M = \frac{\vec{u}+(\vec{u}+\vec{v})}{2} = \vec{u}+\frac{\vec{v}}{2}$. So $\overrightarrow{AM}=\vec{u}+\frac{\vec{v}}{2}$.

Step 3: Find $N$ (midpoint of $CD$).
$C=\vec{u}+\vec{v}$, $D=\vec{v}$. Midpoint: $N = \frac{(\vec{u}+\vec{v})+\vec{v}}{2} = \frac{\vec{u}+2\vec{v}}{2} = \frac{\vec{u}}{2}+\vec{v}$. So $\overrightarrow{AN}=\frac{\vec{u}}{2}+\vec{v}$.

Step 4: Add $\overrightarrow{AM}+\overrightarrow{AN}$.
\[ \overrightarrow{AM}+\overrightarrow{AN} = \vec{u}+\frac{\vec{v}}{2}+\frac{\vec{u}}{2}+\vec{v} = \frac{3\vec{u}}{2}+\frac{3\vec{v}}{2} = \frac{3}{2}(\vec{u}+\vec{v}) = \frac{3}{2}\overrightarrow{AC} \]

Step 5: Confirm the answer.
$\overrightarrow{AM}+\overrightarrow{AN} = \frac{3}{2}\overrightarrow{AC}$. This makes geometric sense since both $M$ and $N$ are on sides adjacent to $C$.

Step 6: State the answer.
\[ \boxed{\dfrac{3}{2}\overrightarrow{AC}} \]