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

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In summary: 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
  • #1
HF08
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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,
HF08
 
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  • #2
[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].
 
  • #3
Hf08

HallsofIvy said:
[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].

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
 

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

1. What does "phi(t1)-phi(t0)" represent?

The expression "phi(t1)-phi(t0)" represents the difference between the angles phi at times t1 and t0.

2. Why is it important to show the angle between phi(t1)-phi(t0) is 0 or pi?

This is important because it allows us to determine if the angle between phi at two different times is either 0 (meaning the angles are the same) or pi (meaning the angles are opposite). This information can be useful in various scientific and mathematical calculations and analyses.

3. How is the angle between phi(t1)-phi(t0) calculated?

The angle between phi(t1)-phi(t0) is calculated using the formula: phi(t1)-phi(t0) = phi(t1) - phi(t0), where phi represents the angle at a specific time.

4. Can the angle between phi(t1)-phi(t0) be a value other than 0 or pi?

Yes, it is possible for the angle between phi(t1)-phi(t0) to be a value other than 0 or pi. This would occur if the angles phi at times t1 and t0 are not equal or opposite, but rather any other value in between.

5. How does the angle between phi(t1)-phi(t0) relate to the concept of phase difference?

The angle between phi(t1)-phi(t0) is closely related to the concept of phase difference. In fact, the phase difference between two waves can be calculated by finding the angle between phi at two different times. If the angle is 0, the waves are in phase, and if the angle is pi, the waves are out of phase.

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