Question

If ABCDEF is a regular hexagon and

$\overrightarrow { A B } + \overrightarrow { A C } + \overrightarrow { A D } + \overrightarrow { A E } + \overrightarrow { A F } = \lambda \overrightarrow { A D }$ then $\lambda =$

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (B)

3

Answer 2

By triangle law, $\overrightarrow { A B } = \overrightarrow { A D } - \overrightarrow { B D }$ $\overrightarrow { A C } = \overrightarrow { A D } - \overrightarrow { C D }$

Therefore, $\overrightarrow { A B } + \overrightarrow { A C } + \overrightarrow { A D } + \overrightarrow { A E } + \overrightarrow { A F }$

= $3 \overrightarrow { A D } + ( \overrightarrow { A E } - \overrightarrow { B D } ) + ( \overrightarrow { A F } - \overrightarrow { C D } ) = 3 \overrightarrow { A D }$

Hence $\lambda = 3$, [Since $\overrightarrow { A E } = \overrightarrow { B D } , \overrightarrow { A F } = \overrightarrow { C D } ]$.