Questions regarding dot product (possible textbook mistake)

Question

I am given the following exercise:

Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and point to the same direction. Also, show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} - \overrightarrow{b} \Vert$ if and only if $\overrightarrow{a} \cdot \overrightarrow{b} = 0$

Regarding the first question:

Firstly, i feel like the first question is wrong since $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ for $\theta = 0$ or $\pi$ (so they must not necessarily point on the same direction). Is that correct?

if so, I proceeded the following way:

$$ \Vert \overrightarrow{a} + \overrightarrow{b} \Vert^2 = \Vert \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\ \Vert \overrightarrow{a} \Vert^2 + 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert +\Vert \overrightarrow{b} \Vert^2 = \Vert \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\ \cos(\theta) = \pm 1\\ \theta = 0 \text{ or } \theta = \pi $$

Regarding the second question:

$$ \Vert \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 +2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta) = \Vert \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 -2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\ - \cos(\theta) = \cos(\theta)\\ \cos(\theta) = 0\\ \theta = \pi/2 $$

Can someone help me with these?

Answer 1

Both of your questions can be answered using the fact that $$ \lVert a\rVert^2 = a\cdot a $$ So, $$ \begin{align} \lVert a + b\rVert^2 &= (a + b)\cdot (a + b) = \lVert a\rVert^2 + \lvert b\rVert^2 + 2a\cdot b. \end{align} $$

Answer 2

The first question is valid. For instance, if we let $0\neq b = -a$, then $||a+b|| = 0 \neq ||a||+||b||.$ So, $a$ and $b$ must point in the same direction.

As for a proof of the first statement: going backwards, assume $a$ and $b$ point in the same direction, i.e. (without loss of generality) $b = ca$ for some scalar $c\geq 0$, and so $$||a+b|| = ||a+ca|| = (1+c)\cdot||a|| = ||a||+||ca|| = ||a||+||b||$$

Now assume $||a+b||=||a||+||b||.$ The left hand side is equal to $\sqrt{||a||^2+||b||^2+2(a\cdot b)}$. So squaring both sides of the original equation and cancelling like terms, we get $a\cdot b = ||a||\cdot ||b||$, and so $b = ca$ where $c \geq 0$.

As for the second question, we can similarly expand $||a+b||=||a-b||$ to obtain $\sqrt{||a||^2+||b||^2+2(a\cdot b)} = \sqrt{||a||^2+||b||^2-2(a\cdot b)}$, and so $a\cdot b = 0$. And for proving this in the other direction, simply follow those steps backwards.

Your approach is more or less valid (depending on what sort of math class you're in), but the angles are unnecessary and should be avoided if possible.