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: Assume position vectors.
Let \[ \overrightarrow{AB}=\vec{u} \] and \[ \overrightarrow{AD}=\vec{v} \] Taking \(A\) as origin, we get \[ B=\vec{u},\quad D=\vec{v} \] Since \(ABCD\) is a parallelogram, \[ C=\vec{u}+\vec{v} \] Thus, \[ \overrightarrow{AC}=\vec{u}+\vec{v} \]

Step 2: Find position vector of \(M\).
\(M\) is the midpoint of \(BC\).
So, \[ M=\frac{B+C}{2} \] \[ M=\frac{\vec{u}+(\vec{u}+\vec{v})}{2} \] \[ M=\frac{2\vec{u}+\vec{v}}{2} \] \[ \overrightarrow{AM}=\vec{u}+\frac{\vec{v}}{2} \]

Step 3: Find position vector of \(N\).
\(N\) is the midpoint of \(CD\).
So, \[ N=\frac{C+D}{2} \] \[ N=\frac{(\vec{u}+\vec{v})+\vec{v}}{2} \] \[ N=\frac{\vec{u}+2\vec{v}}{2} \] \[ \overrightarrow{AN}=\frac{\vec{u}}{2}+\vec{v} \]

Step 4: Add \(\overrightarrow{AM}\) and \(\overrightarrow{AN}\).
\[ \overrightarrow{AM}+\overrightarrow{AN} = \left(\vec{u}+\frac{\vec{v}}{2}\right) + \left(\frac{\vec{u}}{2}+\vec{v}\right) \] \[ = \frac{3\vec{u}}{2}+\frac{3\vec{v}}{2} \] \[ = \frac{3}{2}(\vec{u}+\vec{v}) \] Since \[ \vec{u}+\vec{v}=\overrightarrow{AC}, \] we get \[ \overrightarrow{AM}+\overrightarrow{AN} = \frac{3}{2}\overrightarrow{AC} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{3}{2}\overrightarrow{AC}} \]