If ABCDEF is a regular hexagon, then AB+AC+AD+AE+AF is equal to

The correct answer is Option (4) → \(3\,\overrightarrow{AD}\)

Let the centre O of the hexagon be the origin and let \(\vec{a}, \vec{b}, \dots, \vec{f}\) be the position vectors of A, B, …, F. Then each vector from A is a difference of position vectors, for example \(\overrightarrow{AB} = \vec{b}-\vec{a}\). So the sum is

    \((\vec{b}+\vec{c}+\vec{d}+\vec{e}+\vec{f})-5\vec{a}\)

Opposite vertices of a regular hexagon are symmetric about the centre, so

    \(\vec{d} = -\vec{a}\),  \(\vec{e} = -\vec{b}\),  \(\vec{f} = -\vec{c}\)

Hence

    \(\vec{b}+\vec{c}+\vec{d}+\vec{e}+\vec{f} = \vec{b}+\vec{c}-\vec{a}-\vec{b}-\vec{c} = -\vec{a}\)

and the sum becomes

    \(-\vec{a}-5\vec{a} = -6\vec{a}\)

Also

    \(\overrightarrow{AD} = \vec{d}-\vec{a} = -2\vec{a}\)

So the sum is

    \(-6\vec{a} = 3(-2\vec{a}) = 3\,\overrightarrow{AD}\)