Vector Algebra Question 201
Question: If ABCDEF is regular hexagon, the length of whose side is a, then $ \overrightarrow{AB}.,\overrightarrow{AF}+\frac{1}{2},{{\overrightarrow{BC}}^{2}}= $
Options:
A) a
B) $ a^{2} $
C) $ 2,a^{2} $
D) 0
Show Answer
Answer:
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. $
BETA
