# I am trying to understand dot and cross products from a physics perspective. If a space curve r(t) satisfies the equation r′(t) xx r′′(t)=0 for all t, I understand that the derivative of r′(t) is parallel to r′(t), so the velocity vector r′(t) does not change direction. Thus, this curve moves along a line.

I am trying to understand dot and cross products from a physics perspective. If a space curve r(t) satisfies the equation ${r}^{\prime }\left(t\right)×{r}^{″}\left(t\right)=0$ for all t, I understand that the derivative of r′(t) is parallel to r′(t), so the velocity vector r′(t) does not change direction. Thus, this curve moves along a line.
However, I'm not sure what ${r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)=0$ means. My intuition is that it represents motion along a circle, or part of a circle, since the velocity and acceleration vectors should be perpendicular in that case.
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Haylee Branch
Observe the identity
$\frac{d}{dt}|{r}^{\prime }\left(t\right){|}^{2}=2{r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)$
Thus, ${r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)\equiv 0$ if and only if the motion r has constant speed.
P.S. The circle you mentioned is somewhat relevant to this, since the velocity vector r′(t) would lie on a fixed circle centered at the origin.
P.S.(again) Actually, ${r}^{\prime }\left(t\right)×{r}^{″}\left(t\right)\equiv 0$ does not imply that r(t) lies on a fixed line in general. You may consider r that stalls at a point, stays at that point for a while(say 1 second), then changes direction, and re-accelerates. Still, your intuition is correct if we add condition that r′(t) is never zero. (the proof is clear; the notion of direction you mentioned now makes sense)
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Drew Williamson
You can integrate both sides:
${\int }_{t=0}^{t={t}_{f}}\stackrel{\to }{r}\cdot \stackrel{˙}{\stackrel{\to }{r}}dt=0$
But, $\frac{d\stackrel{\to }{r}\cdot \stackrel{\to }{r}}{dt}=2\left(\stackrel{\to }{r}\cdot \stackrel{˙}{\stackrel{\to }{r}}\right)$
$\left[\stackrel{\to }{r}\cdot \stackrel{\to }{r}{\right]}_{t=0}^{t={t}_{f}}=0$
Or,
$\stackrel{\to }{r}\cdot \stackrel{\to }{r}{|}_{{t}_{f}}=\stackrel{\to }{r}\cdot \stackrel{\to }{r}{|}_{t=0}=C$
Meaning that the length of vector is constant. If the length of vector is constant, we can imagine it to be spinning about a origin. A physics example would be to keep origin at the center of a circle and track how a particle spins around a circular loop centered at the same origin.