Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\vec a\cdot\vec b+ \vec b\cdot\vec c+ \vec a\cdot\vec c$.

share|improve this question
    
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
show 2 more comments

3 Answers

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

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

Then

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

share|improve this answer
1  
This answer is absolutely correct with the explanation that $\vec x . \vec x $ = $ |\vec x |^2 $ –  Souvik Dey Nov 9 '12 at 8:56
add comment

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 $a+b+c=0$.

share|improve this answer
    
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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.