[SOLVED]vector manipulation

dwsmith

Well-known member
Is it possible to write $$\lvert\dot{\mathbf{r}}\rvert\lvert\ddot{\mathbf{r}} - \dot{\mathbf{r}}\cdot\ddot{\mathbf{r}}\rvert$$ as $$\lvert\dot{\mathbf{r}}\rvert\lvert\dot{\mathbf{r}}\times\ddot{\mathbf{r}}\rvert$$?

I want to show
$$\lvert\left(\dot{\mathbf{r}}\times\ddot{\mathbf{r}}\right) \times \dot{\mathbf{r}}\rvert = \lvert\dot{\mathbf{r}}\rvert \lvert\dot{\mathbf{r}} \times\ddot{\mathbf{r}}\rvert$$

Ackbach

Indicium Physicus
Staff member
The first expression, as it stands, doesn't make sense. You cannot add $\ddot{\mathbf{r}}$ to $-\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}$, as the first is a vector and the second is a scalar. To show your identity, I'd try to look at the angle between certain vectors. That is, you know by the cross product rules that
$$\left|( \dot{ \mathbf{r}} \times \ddot{ \mathbf{r}}) \times \dot{ \mathbf{r}} \right| = \left| \dot{ \mathbf{r}} \right| \, \left| \dot{ \mathbf{r}} \times \ddot{ \mathbf{r}} \right| \sin(\theta),$$
where $\theta$ is the angle between $\dot{\mathbf{r}}$ and $\dot{\mathbf{r}} \times \ddot{ \mathbf{r}}$. If you can show that $\dot{\mathbf{r}}$ and $\dot{\mathbf{r}} \times \ddot{\mathbf{r}}$ are orthogonal, you'd be done, right?