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  4. In a regular hexagon A B C D E F, A B+A C+ A D+A E+A F=k A D, where k is equal to

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Q. In a regular hexagon $A B C D E F, \quad \overrightarrow{A B}+\overrightarrow{A C}+$ $\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}=k \overrightarrow{A D}$, where $k$ is equal to

Vector Algebra

Solution:

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$\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}$
$=\overrightarrow{E D}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{C D} $
$[\because \overrightarrow{A B}=\overrightarrow{E D}, \overrightarrow{A F}=\overrightarrow{C D}]$
$=(\overrightarrow{A C}+\overrightarrow{C D})+(\overrightarrow{A E}+\overrightarrow{E D})+\overrightarrow{A D}$
$=\overrightarrow{A D}+\overrightarrow{A D}+\overrightarrow{A D}=3 \overrightarrow{A D}$