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(C) In a regular hexagon $PQRSTU$,the sides are vectors. Let $\vec{a} = \overline{PQ}$. Then $\overline{RS}$ is parallel to $\overline{UP}$ and $\over

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is False.

Vedclass · Answer

(C) In a regular hexagon $PQRSTU$,the sides are vectors. Let $\vec{a} = \overline{PQ}$. Then $\overline{RS}$ is parallel to $\overline{UP}$ and $\overline{ST}$ is parallel to $\overline{PQ}$.Specifically,$\overline{RS} = -\overline{PQ} = -\vec{a}$ is incorrect; rather,$\overline{RS}$ is parallel to $\overline{PQ}$ but in the opposite direction,so $\overline{RS} = -\overline{PQ}$ is false. In a regular hexagon,$\overline{PQ} = \overline{UT}$ and $\overline{RS} = \overline{PQ}$ is not true. Actually,$\overline{PQ} \parallel \overline{ST}$ and $\overline{RS} \parallel \overline{UP}$.Since $\overline{PQ}$ is not parallel to $\overline{RS}$,$\overline{PQ} 	imes \overline{RS} 
eq \overrightarrow{0}$.Thus,$Statement-2$ is False because it claims $\overline{PQ} 	imes \overline{RS} = \overrightarrow{0}$.For $Statement-1$,$\overline{PQ} 	imes (\overline{RS} + \overline{ST}) = \overline{PQ} 	imes \overline{RS} + \overline{PQ} 	imes \overline{ST}$. Since $\overline{PQ}$ is not parallel to the resultant $\overline{RS} + \overline{ST}$,the cross product is non-zero. Thus,$Statement-1$ is True.

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