Proof of cross product and dot product relationship?

Question

Let $r(t)$ be a smooth vector function with values in $\mathbb R^3$ such that $|r(t)| = 2$ for all $t$. Why does the dot product of $|r(t)|$ and $|r'(t)|$ = the cross product of $|r(t)|$ and $|r'(t)|$?

Answer 1

So, I've done my best to interpret your question, and I believe that you wanted your question to state:

Let $r(t)$ be a smooth vector function with values in $\mathbb{R}^3$ such that $|r(t)|=2$ for all $t$. Why does $|r(t) \cdot r'(t)| = |r(t) \times r'(t)|$?

It's important to know that taking the modulus of a vector always yields a scalar. The dot product and cross product ONLY can be applied to vectors.

Now to the restated version of your question, we know $u \cdot v = |u||v|\cos(\theta)$ and $u \times v = |u||v|\sin(\theta)$ where $\theta$ is the angle between u and v. Thus, $|r(t) \cdot r'(t)| = |r(t) \times r'(t)|$ can be restated as $|\cos(\theta)| = |\sin(\theta)|$ where $\theta$ is the angle between $r(t)$ and $r'(t)$. So for this statement to be true, we're looking for the angle between your function and your derivative to be $\pi/4$ (or some sister angle).

Now since the modulus of $r(t)$ is 2 for all t, $r(t)$ has a constant distance of 2 from the origin. Thus, r lies on a sphere of radius 2. However, I do not see how this condition would enforce the necessary conditions. For example, let $r(t) = < 2\cos(t),2\sin(t),0 >$ has modulus 2 for all t, lies on the sphere of radius 2, but $r(t)$ is orthonormal to $r'(t)$ ($\theta = \pi/2$).

So, all of that to say, that either I've misinterpreted your question or that your hypothesis is false.