UPSC NDA II 2015 Mathematics Paper

Concept:In a parallelogram, diagonals bisect each other, so the intersection point P is the midpoint of both AC and BD.Using vector addition from the origin O to the vertices, we sum the four position vectors.Explanation:Let O be the origin and A, B, C, D be the vertices of the parallelogram ABCD with diagonals intersecting at P.Apply the triangle law of vector addition to each vertex:$\overrightarrow{OA} + \overrightarrow{AP} = \overrightarrow{OP}$$\overrightarrow{OB} + \overrightarrow{BP} = \overrightarrow{OP}$$\overrightarrow{OC} + \overrightarrow{CP} = \overrightarrow{OP}$$\overrightarrow{OD} + \overrightarrow{DP} = \overrightarrow{OP}$Add all four equations:$(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}) + (\overrightarrow{AP}+\overrightarrow{BP}+\overrightarrow{CP}+\overrightarrow{DP}) = 4\overrightarrow{OP}$Since P is the midpoint of AC, we have $\overrightarrow{AP} = -\overrightarrow{CP}$.Similarly, P is the midpoint of BD, so $\overrightarrow{BP} = -\overrightarrow{DP}$.Thus $\overrightarrow{AP} + \overrightarrow{BP} + \overrightarrow{CP} + \overrightarrow{DP} = \mathbf{0}$.Substitute into the sum: $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} = 4\overrightarrow{OP}$.Answer:$4\;\overrightarrow{OP}$ (option A).