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We have the vectors $\vec a$ ,$\vec b$ and $\vec c$ of lengths $|\vec a|=2 , |\vec b|=5$ and $|\vec c|=7$. If $\vec a+ \vec b + \vec c=0$ , find $\vec a\cdot\vec b+ \vec b\cdot\vec c+ \vec a\cdot\vec c$.

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By 'modules' do you mean length, and what do you mean by $a*b+c*c+a*c$? –  copper.hat Nov 9 '12 at 8:35
yes,i mean length by modules.Find the value of a*b +b*c+c*a,thats all. "*" means multiplication –  nicegirl Nov 9 '12 at 8:36
And *' means '.' (dot/scalar product) multiplication, or $\times$' (cross/vector product) product? –  Tapu Nov 9 '12 at 8:40
scalar my dear,scalar. –  nicegirl Nov 9 '12 at 8:40
Are $a,b,c$ scalars or vectors? –  copper.hat Nov 9 '12 at 8:43
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3 Answers 3

let $P=\vec a.\vec b+\vec b.\vec c+\vec a.\vec c$

Then $P=\vec a.(\vec b+\vec c)+\vec b.\vec c$

$\vec a+\vec b+\vec c=0$ then $\vec b+\vec c=-\vec a$

So $P=\vec a.(-\vec a)+\vec b.\vec c=-|\vec a|^2+\vec b.\vec c$ $\space\space\space\space\space\space\space(1)$

Similarily :

$P=-|\vec b|^2 + \vec a.\vec c$ $\space\space\space\space\space\space\space(2)$

$P=-|\vec c|^2+ \vec b.\vec a$ $\space\space\space\space\space\space\space(3)$

Then summing $(1)+(2)+(3)$ you get

$3P = -(|\vec a|^2+|\vec b|^2+|\vec c|^2) + P$


$P= -\frac12(|\vec a|^2+|\vec b|^2+|\vec c|^2) $

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This answer is absolutely correct with the explanation that $\vec x . \vec x $ = $ |\vec x |^2 $ –  Souvik Dey Nov 9 '12 at 8:56
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Well, we have the formula: $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$ Now use the given values of $a,b,c$ and $a+b+c=0$.

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they are vectors... –  nicegirl Nov 9 '12 at 8:43
No, $|a|$'s are scalar and so is $(a+b+c)^2=(a+b+c).(a+b+c)$. –  Tapu Nov 9 '12 at 8:44
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Since $|a|+|b|=|c|$, they must be collinear. Let $u$ be a unit vector in the direction of $a$, then we have $a = 2u$, $b = \pm 5 u$ and $c = \pm 7u$. Since $a+b+c =0$, we must have $b=5u, c=-7u$.

Then $a \cdot b+b \cdot c + c \cdot a = 2u \cdot 5u+5u \cdot (-7 u) + (-7u) \cdot 2u =10 -35-14 = -39$.

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