$\|u+v\| = \|u\| - \|v\|$ under what conditions in $\mathbb{R}^2$ and $\mathbb{R}^3$?

Question

I know that $\renewcommand{vec}{\mathbf}\|\vec{u}+\vec{v}\| = \|\vec{u}\|+\|\vec{v}\|$ when $\vec{u} = \vec{v}$, but I don't understand for $\|\vec{u}+\vec{v}\| = \|\vec{u}\|-\|\vec{v}\|$.

Answer 1

$\left\lVert \mathbf u + \mathbf v \right\rVert = \left\lVert \mathbf u \right\rVert - \left\lVert \mathbf v \right\rVert$ is true when the angle between vectors $\mathbf u $ and $\mathbf v$ is $ \pi $ radians, meaning the vectors are parallel and go in opposite directions, or when vector $\mathbf v$ is the zero vector.

We can deduce this by squaring both sides of the equation. $$ \left\lVert \mathbf u + \mathbf v \right\rVert ^2 = (\left\lVert \mathbf u \right\rVert - \left\lVert \mathbf v \right\rVert)^2$$

equivalently, because $ \left\lVert \mathbf a \right\rVert ^2 = \mathbf a \cdot \mathbf a $, $$ (\mathbf u + \mathbf v) \cdot (\mathbf u + \mathbf v) = \left\lVert \mathbf u \right\rVert ^2 -2\left\lVert \mathbf u \right\rVert\left\lVert \mathbf v \right\rVert + \left\lVert\mathbf v \right\rVert ^2$$

equivalently, $$ \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf v + \mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u -2 \left\lVert \mathbf u \right\rVert\left\lVert \mathbf v \right\rVert + \mathbf v \cdot \mathbf v $$

equivalently, $$ \mathbf u \cdot \mathbf v = - \left\lVert \mathbf u \right\rVert \left\lVert \mathbf v \right\rVert $$

equivalently, $$\frac{\mathbf u \cdot \mathbf v}{\left\lVert \mathbf u \right\rVert \left\lVert \mathbf v \right\rVert} = -1$$

Since we know that the angle between the two vectors is described by $\theta$ in $\cos(\theta) = \frac{\mathbf u \cdot \mathbf v}{\left\lVert \mathbf u \right\rVert \left\lVert \mathbf v \right\rVert}$, by substitution,

$$\cos(\theta) = -1$$

Thus,

$$\theta = \pi$$

Answer 2

Guide:

We want to understand when does

$$\|u + v\|=\|u\| - \|v\|$$

equivalently,

$$\|u\|= \|u + v\|+ \|v\|$$

or equivalently,

$$\|(u+v) + (- v)\|= \|u + v\|+ \|-v\|$$

Hence, if you answer the first question correctly, you can answer this question too.