# 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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This is not possible due to the triangle inequality. To be specific, if you add three vectors and obtain 0, that means, when they're attached at the end in sequence, they constitute a triangle. The triangle inequality states that one of the sides of a triangle can not be longer than the sum of the other two.

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or $4=|c|=|-(a+b)|=|a+b|\leq |a|+|b|=1+2=3$ –  enzotib Oct 8 '12 at 7:16
Yes, I see now. Thank you for the clarification. –  EuYu 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 $4$.