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[SOLVED] vector manipulation

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Jan 26, 2012
4,192
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?