Distance of points in $\mathbb R^3$ using vectors

Question

Why is this $\mathbf{w}-\mathbf{v}$ and not $\mathbf{v}+\mathbf{w}$? I understand $\mathbf{v}+\mathbf{w}$ means $\mathbf{w}$ vector is placed at end of $\mathbf{v}$. However book didn't explain well how $\mathbf{w}-\mathbf{v}$ works here. I can think of it as $c^2 - a^2 = b^2$ which is essentially same thing as $\mathbf{w}-\mathbf{v}$. I can't however see how the vector property works here for $\mathbf{w}-\mathbf{v}$ or $\mathbf{w}+(-\mathbf{v})$.

Answer 1

$w-v = w+(-v)$

Note that $-v$ is $v$ with the head and tail of the arrow swapped. Then, you can add them using the tail to head method to get the vector $w-v$.

Answer 2

There are two ways to get from $0$ to $Q$:

1. first travel to $P$ along the vector $v$, then travel from $P$ to $Q$ along the vector $\vec{PQ}$, or
2. travel along the vector $w$.

We can write the first way as $v + \vec{PQ}$, so we have

$$v + \vec{PQ} = w,$$

and therefore $\vec{PQ} = w - v$.

Answer 3

First, notice that $\vec{PQ} = \vec{PO}+\vec{OQ}$.

To go from $P$ to $O$ we have to travel backwards along ${\bf v}$, i.e. $\vec{PO}=-{\bf v}$

To go from $O$ to $Q$ we travel along ${\bf w}$, i.e. $\vec{OQ} = {\bf w}$.

Clearly, $\vec{PQ} = -{\bf v}+{\bf w} = {\bf w}-{\bf v}$.