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Question: If ABCDEF is regular hexagon, the length of whose side is a, then \(\overset{\rightarrow}{AB}.\overs...

If ABCDEF is regular hexagon, the length of whose side is a, then AB→.AF→+12BC→2=\overset{\rightarrow}{AB}.\overset{\rightarrow}{AF} + \frac{1}{2}{\overset{\rightarrow}{BC}}^{2} =

A

a

B

a2a^{2}

C

2a22a^{2}

D

0

Answer

0

Explanation

Solution

AB→.AF→=∣a∣∣a∣cos⁡120∘=−12a2\overset{\rightarrow}{AB}.\overset{\rightarrow}{AF} = |\mathbf{a}||\mathbf{a}|\cos 120{^\circ} = \frac{- 1}{2}a^{2} and 12BC→2=12a2\frac{1}{2}{\overset{\rightarrow}{BC}}^{2} = \frac{1}{2}a^{2}

Therefore, AB→.AF→+12BC→2=12a2−12a2=0.\overset{\rightarrow}{AB}.\overset{\rightarrow}{AF} + \frac{1}{2}{\overset{\rightarrow}{BC}}^{2} = \frac{1}{2}a^{2} - \frac{1}{2}a^{2} = 0.