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

[HF08]

Let a,b\inR^{m}, b\neq0 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},\inR 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