# Geometric consequence of absolute vectors

If $$| \mathbf{a} + \mathbf{b} | = | \mathbf{a} | + | \mathbf{b} |$$, where $$\mathbf{a}$$ and $$\mathbf{b}$$ are vectors, what is the geometrical significance of this?

My first thought was that the vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ must be in the first quadrant, only having positive components.

• Welcome to Math Stack Exchange. It means they are collinear. Are you familiar with the triangle inequality? – J. W. Tanner Jun 18 at 10:23
• @J.W.Tanner But this is not enough. Consider the vectors $x$ and $-x$ – Peter Jun 18 at 10:24
• It means they're pointing in the same direction, essentially. One is a resize of the other (by a positive factor). – vrugtehagel Jun 18 at 10:25
• @J.W.Tanner Why is this the case? If two vectors are in the first quadrant then does it matter that they are collinear? I am familiar with the inequality. – Physics Jun 18 at 10:26
• @vrugtehagel : Nonnegative, to be pedantic ;) – MPW Jun 18 at 11:20

Let's see what $$a+b$$ is, geometrically. The dark blue arrows are $$a$$ and $$b$$. The orange vector is $$a+b$$. Essentially, you just stick one on top of the other. Now, we know that the absolute value of a vector just represents the length; can you see what happens when $$|a+b|=|a|+|b|$$?

Essentially, they would have to be pointing in the same direction! It has nothing to do with what quadrant they're in, it only says they're the same direction. The converse is also true; if $$a$$ and $$b$$ point in the same direction, then $$|a+b|=|a|+|b|$$ (do you see why?).

• That's not true for a generic norm. Take $(1,0)$ and $(0,1)$ regarding the Manhattan-norm. We have $|(1,0)|=|(0,1)|=1$, and $|(1,1)|=2$, but the vectors don't have the same direction. – Michael Hoppe Jun 18 at 11:59
• Yes, but the question was phrased in a way that led me to believe that the OP was talking about the regular, Euclidean norm. Feel free to add your own answer based on another norm, or norms in general. – vrugtehagel Jun 18 at 14:46

Let $$\theta$$ be the angle between $$\mathbf a$$ and $$\mathbf b$$. Using properties of the dot product,

$$|\mathbf a+\mathbf b|=|\mathbf a|+|\mathbf b|$$ $$\implies|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2+2|\mathbf a||\mathbf b|$$ $$\implies(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)=\mathbf a\cdot \mathbf a+\mathbf b\cdot \mathbf b+2|\mathbf a||\mathbf b|.$$

But the left side is $$\mathbf a\cdot \mathbf a + \mathbf b\cdot \mathbf b + 2\mathbf a\cdot \mathbf b = \mathbf a\cdot \mathbf a +\mathbf b \cdot \mathbf b+2|\mathbf a||\mathbf b|\cos\theta$$, so this means $$\cos\theta=1$$.

Squaring of the both sides gives: $$\vec{a}\vec{b}=|\vec{a}||\vec{b}|.$$ Now, can you understand what happens here?

• Yes, thank you. – Physics Jun 18 at 10:31
• @Physics You are welcome! – Michael Rozenberg Jun 18 at 10:31

A norm for which is valid $$|a+b|=|a|+|b|\Rightarrow \text{a and b are parallel}$$ is called strict. (All norms deriving from an inner product are strict, e.g.)

An example for a non-strict norm is the Manhattan-norm, see https://en.wikipedia.org/wiki/Taxicab_geometry. Here $$|(1,0)|=|(0,1)|=1\text{ and } |(1,1)|=2,$$ hence $$|(1,0)+(0,1)|=|(1,0)|+|(0,1)|,$$ but the vectors aren't parallel.