If the position vectors of A and B are \(\mathop a\limits^ \to\) and \(\mathop b\limits^ \to\) respectively, then the position vector of mid-point of AB is:

Concept: Section formula is GIVEN by \(mA + nB\over m+n\)Calculation: Given: Position vector of A and B are \(\mathop a\limits^ \to\) and \(\mathop b\limits^ \to\) respectivelyLet M be the mid-point of AB, such that AM = MB, it means that M DIVIDES AB in the ratio 1:1, so by section formula, we get,\((1)\mathop a\limits^ \to+ (1)\mathop b\limits^ \to\over1+1\) = \(\frac{1}{2}\left( {\mathop a\limits^ \to + \mathop b\limits^ \to } \right)\)Hence, we conclude that If the position vectors of A and B are \(\mathop a\limits^ \to\) and \(\mathop b\limits^ \to\) RESPECTIVELY, then the position vector of mid-point of AB is \(\frac{1}{2}\left( {\mathop a\limits^ \to + \mathop b\limits^ \to } \right)\)