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Can you guys give me a hint for this problem.

ABCDEF is a regular hexagon.

Express in terms of a single vector the sum of the vectors, $4\overrightarrow{AB}$, $2\overrightarrow{AC}$, $\overrightarrow{AD}$, $\overrightarrow{AE}$, and $5\overrightarrow{AF}$.

I was able to combine $4\overrightarrow{AB} + 4\overrightarrow{AF} = 4\overrightarrow{AO} = 2\overrightarrow{AD}$. I am a little stuck with the other vectors.

Thanks for all your help.

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  • $\begingroup$ Would it help to draw a picture? (Hint: Put A at the origin. You can also apply a linear transformation to deform the hexagon so that e.g. B = (1,0), C = (2,1), D = (2,2), E = (1,2) and F = (0,1). This makes it easier to draw on graph paper.) $\endgroup$ – Ilmari Karonen Jul 19 '11 at 15:47
  • $\begingroup$ @Ilmari: I'd put the centre at the origin. The problem gets a lot easier with the identities in my answer, and these suggest themselves most readily by symmetry if the centre is at the origin. $\endgroup$ – joriki Jul 19 '11 at 17:03
  • $\begingroup$ @joriki: De gustibus non est disputandum, I suppose, and certainly the problem can be solved with the origin placed anywhere. But when all the vectors in the problem statement are rooted at the same point, it makes sense to me to define that point to be the origin. Still, it's always nice to see multiple ways to approach the same problem. $\endgroup$ – Ilmari Karonen Jul 19 '11 at 18:39
  • $\begingroup$ Thanks, drawing an accurate picture helped. $\endgroup$ – mathguy80 Jul 20 '11 at 8:07
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It will make things a whole lot easier when you notice that

$$\overrightarrow{AA}+\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{AF}=2\overrightarrow{AO}+\overrightarrow{OC}+\overrightarrow{OF}=2\overrightarrow{AO}\;,$$

and in particular

$$\overrightarrow{AA}+\overrightarrow{AD}+\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{AC}+\overrightarrow{AF}=6\overrightarrow{AO}\;.$$

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  • $\begingroup$ Thanks, Using the center of the hexagon as the orgin helped. I ended up with something like this, $$ \begin{align} 4\overrightarrow{AB} + 2\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + 5\overrightarrow{AF} &= 4\overrightarrow{AB} + (2\overrightarrow{AB} + 2\overrightarrow{AO}) + 2\overrightarrow{AO} + \overrightarrow{AF} + \overrightarrow{AO} + 5\overrightarrow{AF} \\ &= 6\overrightarrow{AB} + 5\overrightarrow{AO} + 6\overrightarrow{AF} \\ &= 6\overrightarrow{AO} + 5\overrightarrow{AO} \\ &= 11\overrightarrow{AO} \\ &= \dfrac{11}{2}\overrightarrow{AD} \end{align} $$ $\endgroup$ – mathguy80 Jul 20 '11 at 8:08

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