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

[julypraise]

Is the following theorem true:

Theorem: Suppose a, \, b \in \mathbb{R}^k. If |a| + |b| = |a + b|, then |a| and |b| 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.

 

[micromass]

Hi julypraise!

The trick to proving this is to calculate the inproduct

<a+b,a+b>

in two different ways. The first way involves

<a+b,a+b>=|a+b|^2=(|a|+|b|)^2

The other way starts as

<a+b,a+b>=|a|^2+|b|^2+2<a,b>

Can you finish it?

 

[julypraise]

Thank you so much, micromass.

So, by using the trick, I derived that

2|a||b| = 2<a, \, b>, and by the definition of inproduct

|a||b| = |a||b|cos\theta

and therefore, \theta = 0

where \theta is the angle between.

 

[matphysik]

Is the following theorem true:

Theorem: Suppose a, \, b \in \mathbb{R}^k. If |a| + |b| = |a + b|, then |a| and |b| 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||.