# Algebra of Vectors

Is it possible that there are $3$ vectors, $a, b, c$, such that $a + b + c = 0$ but $|a| = 1$, $|b| = 2$ and $|c| = 4$?

If yes why? and if no why?. I'm trying to get the solution since last $2$ days, so kindly help me out...

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or $4=|c|=|-(a+b)|=|a+b|\leq |a|+|b|=1+2=3$ –  enzotib Oct 8 '12 at 7:16
$4=|c|=|-a-b|\le |a|+|b|=3$. Contradiction –  PAD Oct 8 '12 at 7:16
The triangle inequality for vectors says that $|x+y| \le |x|+|y|$ for all $x,y$. We also have $|\lambda x|=|\lambda|\cdot |x|$ for any number $\lambda$.
So, if $a+b+c=0$, then $c=-a-b = (-1)(a+b)$. Then, $$|c|=|-1|\cdot |a+b|=|a+b|\le |a|+|b|$$ So, if $|a|=1$ and $|b|=2$, then $|c|\le 3$ and cannot be $4$.