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 September 23rd, 2013, 10:41 AM #1 Newbie   Joined: Sep 2013 Posts: 1 Thanks: 0 Proof - Two dimensional vectors Hi all, Could you please give me guidance on the following proof please? Suppose we have three two-dimensional vectors: a, b, and c. Given that the relation a+b+c=0 where 0 is the 0 vector. Could you prove that mod(c) <= mod(a) + mod(b) using a necessary and/or a sufficient condition? Thanks!
 September 23rd, 2013, 01:15 PM #2 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 716 Re: Proof - Two dimensional vectors I am not familiar with mod. Is this the same as length? In that case, the expressions is the triangle inequality. The condition that the sum is 0 means the vectors form a triangle.
 September 23rd, 2013, 01:23 PM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Proof - Two dimensional vectors Applying the law of cosines, $|c|^2= |a|^2 + |b|^2 - 2|a|.|b|cos\theta$ where $\theta$ is the angle between vectors $a$ and $b.$ On the other hand $(|a| + |b|)^2= |a|^2 + |b|^2 + 2|a|.|b|$. Since $|cos\theta | \leq 1$, we may conclude that $(|a| + |b|)^2 \geq |c|^2 \ \Rightarrow \ |a| + |b| \geq |c|$.

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