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I came across this question in preparation for my mid semester exam:

Suppose that $m(t)$, where t is the parameter, $ t ∈ \mathbb{R}$, is a position vector with the properties as follows: $$m(t) \times \frac{dm(t)}{dt} = 0$$ Proof that this vector has a fixed direction.

I have no idea how to approach this question, any help would be highly appreciated.

Joshua
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1 Answers1

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Method 1: Geometric Intuition

For any vectors $a,b\in\mathbb{R}^3$, we have that $a\times b = 0$ only if either

  • $a=0$ or $b=0$
  • $a||b$ or $a||-b$

Here $a=m(t)$ and $b=m'(t)$. Then:

  • If $m(t)=0$, then the direction never changes (the position is always zero)
  • If $m'(t)=0$, then the tangent vector never changes (i.e. it always points in the same direction), hence the curve must be a straight line, aligned with $m'(t)$.
  • If the $m$ and $m'$ are parallel (or anti-parallel) then the curve must again be a line since the position vector must always be aligned with the tangent vector (if the curve ever turned, $m$ and $m'$ would no longer be aligned).

Hence the line must be straight, and so the direction never changes.

Method 2: Definition of Curvature

By definition, the curvature of a space curve can be written: $$ \kappa = \frac{|| T(t) \times T'(t) ||}{|| T'(t) ||^3} $$ where $T(t)=m'(t)$. But, it is given that $T(t) \times T'(t) = 0$.

Hence, $\kappa=0$. Thus the curve must be straight, because any bending (i.e. change in direction) would give it non-zero curvature.

(Note: because the unit normal is given by $N(t)=T'(t)/||T'(t)||=m''(t)/||m''(t)$ when we parametrize by arclength, this is equivalent to the comment above suggesting to show that $m''(t)=0$. See also this. And yes, I know its the same as part 3 of method 1 too.)

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