Show the angle between phi(t1)-phi(t0) is 0 or pi

[HF08]

Let a,b$$\in$$R$$^{m}$$, b$$\neq$$0 and set $$\phi$$(t)=a+tb.

We must show that the angle between $$\phi$$(t$$_{1}$$)-$$\phi$$(t$$_{0}$$) and $$\phi$$(t$$_{2}$$)-$$\phi$$(t$$_{0}$$)
is 0 or $$\pi$$ for any t$$_{0}$$,t$$_{1}$$,t$$_{2}$$,$$\in$$R with t$$_{1}$$,t$$_{2}$$$$\neq$$.

The 0,1, and 2 are supposed to be subscripts next to t, but my Latex is showing is as superscripts? I don't know why.

Here is my work:
I don't know where to start. My guess is we start with $$\phi$$(t)=a+tb. Do we use the law of cosines? How can I even begin to show what looks like arbitrary definitions, arbitrary values and show either 0 or $$\pi$$? Please
help me. It isn't lack of trying, I need to attack this problem, but I feel more like I am drowning.

Thank You,
HF08

 

[HallsofIvy]

$\vec{a}+ t\vec{b}$ is a straight line. The lines from the point $\phi(t_0)$ and any other points on that same line are exactly the same line. If the two points are the same side of $\phi(t_0)$ then the angle is 0, if on opposite sides, $\pi$.

 

[HF08]

Hf08

$\vec{a}+ t\vec{b}$ is a straight line. The lines from the point $\phi(t_0)$ and any other points on that same line are exactly the same line. If the two points are the same side of $\phi(t_0)$ then the angle is 0, if on opposite sides, $\pi$.

Holy smokes, freaking dumb. I am today's biggest stupid person. I really don't think I am stupid, I just feel that way. I do want to make things more formal.

So could I say given t0, t1, if t1<t0, then it follows that we will get pi? Otherwise,
if t0<t1, we will get 0. I think this is "obvious". I don't know how formal I should make it?

Perhaps I should say, if phi(t1)<phi(t0). This is more concrete. Since they are on the same line, then it follows that our angle is pi. Agreed?

Thanks,
HF08