I am trying to understand dot and cross products from a physics perspective. If a space curve r(t) satisfies the equation r ′ ( t ) × 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.However, I'm not sure what r ′ ( t ) ⋅ r ″ ( t ) = 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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I am trying to understand dot and cross products from a physics perspective. If a space curve r(t) satisfies the equation       r    ′    (  t  )  ×      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.However, I&#039;m not sure what       r    ′    (  t  )  ⋅      r    ″    (  t  )  =  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.

Haylee Branch · Accepted Answer

Observe the identity      d          d      t            |        r    ′    (  t  )            |        2    =  2      r    ′    (  t  )  ⋅      r    ″    (  t  )Thus,       r    ′    (  t  )  ⋅      r    ″    (  t  )  ≡  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    ′    (  t  )  ×      r    ″    (  t  )  ≡  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)

Drew Williamson · Accepted Answer

You can integrate both sides:      ∫          t      =      0              t      =              t        f                        r      →        ⋅                              r          →                    ˙        d  t  =  0But,             d                        r          →                    ⋅                        r          →                            d      t        =  2  (            r      →        ⋅                              r          →                    ˙        )  [            r      →        ⋅            r      →            ]          t      =      0              t      =              t        f              =  0Or,            r      →        ⋅            r      →                  |                      t        f              =            r      →        ⋅            r      →                  |              t      =      0        =  CMeaning 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.

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