|a+b| = |a| + |b| implies a and b parallel?

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Is the following theorem true:

Theorem: Suppose [itex]a, \, b \in \mathbb{R}^k[/itex]. If [itex]|a| + |b| = |a + b|[/itex], then [itex]|a|[/itex] and [itex]|b|[/itex] are parallel to each other in the same direction.

I proved the converse, but I couldn't prove the theorem above. Please post the proof or the disproof of it, or a link of them. Thanks.
 
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Thank you so much, micromass.

So, by using the trick, I derived that

[itex]2|a||b| = 2<a, \, b>[/itex], and by the definition of inproduct

[itex]|a||b| = |a||b|cos\theta[/itex]

and therefore, [itex]\theta = 0[/itex]

where [itex]\theta[/itex] is the angle between.
 
julypraise said:
Is the following theorem true:

Theorem: Suppose [itex]a, \, b \in \mathbb{R}^k[/itex]. If [itex]|a| + |b| = |a + b|[/itex], then [itex]|a|[/itex] and [itex]|b|[/itex] are parallel to each other in the same direction.

I proved the converse, but I couldn't prove the theorem above. Please post the proof or the disproof of it, or a link of them. Thanks.

If `a` and `b` are parallel then ∃c∈ℝ s.t., a=cb. So then, ||a+b||=||cb+b||=||b||(|c+1|)≠||a||+||b||. Since ||a||=||cb||=(|c|)||b||.
 
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