Why's the third side of this vector triangle $|| v - u ||$, on page 145 of Kuldeep Singh's Linear Algebra: Step by Step (2013)?

Question

While talking about vector dot products, Kuldeep Singh shows this diagram on page 145 of Linear Algebra: Step by Step (2013).

What I do not understand is why in figure 2.14 below, the third side (the "$a$" side) equals $|| v - u ||$. Singh uses this fact to derive a formula for angles between vectors, but doesn't explain his initial premise that the third side equals $|| v - u ||$. Can someone explain it to me?

Answer 1

Consider this.

You start from point A. Then go to point B. Then point C from there and then back to point A.

A $\rightarrow$ B then B $\rightarrow$ C then C $\rightarrow$ A

A $\rightarrow$ B = ||u||

B $\rightarrow$ C = ||x|| (we have to find x)

C $\rightarrow$ A = -||v|| (minus because direction is opposite. We are going from C $\rightarrow$ A and not A $\rightarrow$ C)

Since it start and comes back to same point, their sum should be 0.

||u|| + ||x|| - ||v|| = 0

||x|| = ||v - u||

Hope it helps

Answer 2

$\mathbf{u}$ and $\mathbf{v}$ are the vectors representing sides $\overline{AB}$ and $\overline{AC}$ respectively. The difference, $\mathbf{v}-\mathbf{u}$, is just the vector you'd have to add to $\mathbf{u}$ to get $\mathbf{v}$. That is, $\mathbf{u}+(\mathbf{v}-\mathbf{u})=\mathbf{v}$.

If you're aware that vector addition is like attaching each vector together head-to-tail, then it should be clear that $\mathbf{v}-\mathbf{u}$ is the vector for $\overline{BC}$.

$\|\mathbf{v}-\mathbf{u}\|$ is just the magnitude of the difference vector, which makes it the length of $\overline{BC}$.