question_answer

If ABCDEF is regular hexagon, the length of whose side is a, then \[\overrightarrow{AB}\,\,.\,\overrightarrow{AF}+\frac{1}{2}\,{{\overrightarrow{BC}}^{2}}=\]

A)             a

B)             \[{{a}^{2}}\]

C)             \[2\,{{a}^{2}}\]

D)             0

Correct Answer: D

Solution :

           \[\overrightarrow{AB}\,.\,\overrightarrow{AF}=|\mathbf{a}|\,\,|\mathbf{a}|\,\cos 120{}^\circ =\frac{-1}{2}{{a}^{2}}\] and \[\frac{1}{2}{{\overrightarrow{BC}}^{2}}=\frac{1}{2}{{a}^{2}}\] Therefore, \[\overrightarrow{AB}\,.\,\overrightarrow{AF}+\frac{1}{2}{{\overrightarrow{BC}}^{2}}=\frac{1}{2}{{a}^{2}}-\frac{1}{2}{{a}^{2}}=0.\]