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por **stuart clark** » Seg Abr 09, 2012 12:16

If $\vec{a}\;\;,\vec{b}$ and $\vec{c}$ are three vectors such that

$\mid \vec{a} \mid = \mid \vec{b} \mid = \mid \vec{c} \mid = 1$ and $\mid \vec{a}-\vec{b}\mid^2+\mid\vec{b}-\vec{c}\mid^2+\mid\vec{c}-\vec{a}\mid^2 = 9$

then $\mid 2\vec{a}+5\vec{b}+5\vec{c}\mid =$

por **stuart clark** » Seg Abr 09, 2012 12:19

I have Tried like in this way

$\mid \vec{a}-\vec{b}\mid^2+\mid\vec{b}-\vec{c}\mid^2+\mid\vec{c}-\vec{a}\mid^2 = 9$

$6-2.\left(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\right) = 9$

$\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a} = -\frac{3}{2}$

por **stuart clark** » Qua Abr 11, 2012 23:44

Thanks Moderator Got it

$\mid \vec{a}+\vec{b}+\vec{c} \mid^2 = \left(\vec{a}+\vec{b}+\vec{c}\right).\left(\vec{a}+\vec{b}+\vec{c}\right) = 0$

So $\vec{a}+\vec{b}+\vec{c} = 0\Leftrightarrow \vec{b}+\vec{c} = -\vec{a}$

So $\mid 2\vec{a}+5(\vec{b}+\vec{c})\mid = 3\mid \vec{a} \mid = 3$

por **LuizAquino** » Qui Abr 12, 2012 11:13

stuart clark escreveu:Thanks Moderator Got it

$\mid \vec{a}+\vec{b}+\vec{c} \mid^2 = \left(\vec{a}+\vec{b}+\vec{c}\right).\left(\vec{a}+\vec{b}+\vec{c}\right) = 0$

So $\vec{a}+\vec{b}+\vec{c} = 0\Leftrightarrow \vec{b}+\vec{c} = -\vec{a}$

So $\mid 2\vec{a}+5(\vec{b}+\vec{c})\mid = 3\mid \vec{a} \mid = 3$

Ok.

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