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

1. Jul 30, 2011

### 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.

2. Jul 30, 2011

### micromass

Staff Emeritus
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?

3. Jul 30, 2011

### 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.

4. Aug 4, 2011

### matphysik

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||.

Last edited: Aug 4, 2011