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, \[\overrightarrow{AB}\,.\,\overrightarrow{AF}+\frac{1}{2}{{\overrightarrow{BC}}^{2}}=\frac{1}{2}{{a}^{2}}-\frac{1}{2}{{a}^{2}}=0.\]You need to login to perform this action.
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