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I have $\frac{d}{dt} (\vec{r} \cdot (\vec{r}' \times \vec{r}'') \\ = \vec{r}' \cdot (\vec{r}' \times \vec{r}'') + \vec{r} \cdot (\vec{r}' \times \vec{r}'')'\\ =\vec{r}' \cdot (\vec{r}' \times \vec{r}'')+ \vec{r} \cdot (\vec{r}' \times \vec{r}''')$

Now I don't know what to do because i think when a vector is in form $a \cdot (b \times c)$, then I need to use triple determinant. But I can't because I don't have components.

3 Answers3

1

Think geometrically. $$\vec{r}' \cdot (\vec{r}' \times \vec{r}'')$$

The cross product results in a new vector, say $\vec{v}$, which is by construction perpendicular to $\vec{r}' $and $\vec{r}''$.

What can you say about $\vec{r}'\cdot \vec{v}$ then?

zahbaz
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  • They equal 0 because $\vec{v}$ is perpendicular to the plane of $\vec{r}'$? –  Feb 10 '16 at 04:24
  • Yeah, and that stems from the definition of a dot product. $\vec{v}\cdot\vec{u}=||v|| ||u||\cos\theta$ – zahbaz Feb 10 '16 at 04:28
  • Awesome, thanks! –  Feb 10 '16 at 04:29
  • @Pallas. Sure thing. Also, check out Vikasy's answer. He or she makes a good point about using the product rule. Your work checks out, though, as one of the product rule terms will vanish. – zahbaz Feb 10 '16 at 05:14
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Note that $$\vec{r}' \cdot (\vec{r}' \times \vec{r}'')=\det(\vec{r}',\vec{r}',\vec{r}'')=0$$ because there are two rows (or columns) equal.

mfl
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Yes, as others said $u.(u \times w) = 0$. Also, $(a \times b)' = (a' \times b) + (a \times b')$, where $u, v, w, a, b$ are vectors. Using these you can fix the third line of your answer.

JKnecht
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Vikasy
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