Question
If $ ABCDEF$ is a regular hexagon and $\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} = \lambda \,\overrightarrow {AD} ,$ then $\lambda = $
✓
Answer
b
(b) By triangle law, $\overrightarrow {AB} = \overrightarrow {AD} - \overrightarrow {BD} ,$
(b) By triangle law, $\overrightarrow {AB} = \overrightarrow {AD} - \overrightarrow {BD} ,$
$\overrightarrow {AC} = \overrightarrow {AD} - \overrightarrow {CD} $
Therefore, $\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} $
= $3\overrightarrow {AD} + (\overrightarrow {AE} - \overrightarrow {BD} ) + (\overrightarrow {AF} - \overrightarrow {CD} ) = 3\overrightarrow {AD} $
Hence $\lambda = 3$, [Since $\overrightarrow {AE} = \overrightarrow {BD,} \,\overrightarrow {AF} \, = \overrightarrow {CD} ]$.
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