If ABCDEF is a regular hexagon and vec{AB} + | vedclass

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(C) Let the origin be at the center $O$ of the regular hexagon. The position vectors of the vertices are $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e},

Vedclass · Accepted Answer

$3, 6$

Vedclass · Answer

(C) Let the origin be at the center $O$ of the regular hexagon. The position vectors of the vertices are $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e}, \vec{f}$.Since $O$ is the center, $\vec{a} + \vec{d} = \vec{0}$, $\vec{b} + \vec{e} = \vec{0}$, and $\vec{c} + \vec{f} = \vec{0}$.Also, $\vec{d} = -\vec{a}$.The given sum is $\vec{S} = (\vec{b}-\vec{a}) + (\vec{c}-\vec{a}) + (\vec{d}-\vec{a}) + (\vec{e}-\vec{a}) + (\vec{f}-\vec{a})$.$\vec{S} = (\vec{a} + \vec{b} + \vec{c} + \vec{d} + \vec{e} + \vec{f}) - 6\vec{a} - \vec{a} = \vec{0} - 6\vec{a} = -6\vec{a} = 6\vec{d}$.Given $\vec{S} = p\vec{AD} = p(\vec{d}-\vec{a}) = p(2\vec{d}) = 2p\vec{d}$.Comparing $6\vec{d} = 2p\vec{d}$, we get $p = 3$.Given $\vec{S} = q\vec{AO} = q\vec{a} = q(-\vec{d})$.Since $\vec{S} = 6\vec{d}$, then $q(-\vec{d}) = 6\vec{d}$, so $q = -6$. However, checking the vector direction $\vec{S} = 6\vec{d} = 6\vec{AO}$ is incorrect as $\vec{AO} = -\vec{a} = \vec{d}$.Thus $\vec{S} = 6\vec{d} = 6\vec{AO}$. So $p=3, q=6$.

If $ABCDEF$ is a regular hexagon and $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = p\vec{AD} = q\vec{AO}$, where $O$ is the center of the hexagon, then the values of $p$ and $q$ respectively are

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