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

  1. Feb 14, 2008 #1
    Let a,b[tex]\in[/tex]R[tex]^{m}[/tex], b[tex]\neq[/tex]0 and set [tex]\phi[/tex](t)=a+tb.

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

    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 [tex]\phi[/tex](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 [tex]\pi[/tex]? 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,
  2. jcsd
  3. Feb 14, 2008 #2


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    [itex]\vec{a}+ t\vec{b}[/itex] is a straight line. The lines from the point [itex]\phi(t_0)[/itex] and any other points on that same line are exactly the same line. If the two points are the same side of [itex]\phi(t_0)[/itex] then the angle is 0, if on opposite sides, [itex]\pi[/itex].
  4. Feb 14, 2008 #3

    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?

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