If A, B, C, D be any four points and E and F be the middle points of AC and BD respectively, then $\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD}$ is equal to
Answer & explanation
Correct answer: option 2
Since F is the middle point of BD. Therefore,
$\vec{AB}+\vec{AD}=2\vec{AF}$ ...(i)
Similarly, we have
$\vec{CB}+\vec{CD}=2\vec{CF}$ ...(ii)
Adding (i) and (ii), we get
$\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=2(\vec{AF}+\vec{CF})=-2(\vec{FA}+\vec{FC})$
$⇒\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=-2(2\vec{FE})$
$⇒\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}=4\vec{EF}$ [∵ E is the mid-point of AC]