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  4. Consider the regular hexagon ABCDEF with centre at O (origin). AD + EB + FC is equal to

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Q. Consider the regular hexagon $ABCDEF$ with centre at $O$ (origin).
$\overrightarrow{ AD }+\overrightarrow{ EB }+\overrightarrow{ FC }$ is equal to

Vector Algebra

Solution:

Consider the regular hexagon $ABCDEF$ with centre at $O$ (origin).
image
$=\overrightarrow{ AD }+\overrightarrow{ EB }+\overrightarrow{ FC }=2 \overrightarrow{ AO }+2 \overrightarrow{ OB }+2 \overrightarrow{ OC }$
$=2(\overrightarrow{ AO }+\overrightarrow{ OB })+2 \overrightarrow{ OC }=2 \overrightarrow{ AB }+2 \overrightarrow{ AB }$
$=(\because \overrightarrow{ OC }=\overrightarrow{ AB })=4 \overrightarrow{ AB }$
$=\overrightarrow{ R }=\overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }$
$=\overrightarrow{ ED }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ CD }$
$=(\because \overrightarrow{ AB }=\overrightarrow{ ED }$ and $\overrightarrow{ AF }=\overrightarrow{ CD })$
$=(\overrightarrow{ AC }+\overrightarrow{ CD })+(\overrightarrow{ AE }+\overrightarrow{ ED })+\overrightarrow{ AD }$
$=\overrightarrow{ AD }+\overrightarrow{ AD }+\overrightarrow{ AD }=3 \overrightarrow{ AD }=6 \overrightarrow{ AO }$