2

So $r'(t) = C \times r(t)$, where $r(t)$ is vector function and C is constant vector and $\times$ is cross product. What is then special about $r(t)$?

VECT
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1 Answers1

3

Notice

  1. $$\frac12 \frac{d}{dt}|\vec{r}|^2 = \vec{r} \cdot \vec{r}' = \vec{r} \cdot ( \vec{C} \times \vec{r} ) = 0 \quad\implies\quad |\vec{r}| = \text{ constant. }$$ i.e. $\vec{r}$ is constrained to move on some sphere centered at origin.

  2. $$ \frac{d}{dt} \vec{C}\cdot\vec{r} = \vec{C} \cdot \vec{r}' = \vec{C} \cdot ( \vec{C} \times \vec{r} ) = 0\quad\implies\quad \vec{C} \cdot \vec{r} = \text{ constant. }$$ i.e. $\vec{r}$ is constrained to on some plane whose normal vector is in the direction of $\vec{C}$.

Without solving any ODE, $1.$ an $2.$ tell us $\vec{r}(t)$ is moving along a circle which lies on a plane whose normal is in the direction of $\vec{C}$. Furthermore, if you join a line passing through the origin and the center of the circle, that line will be perpendicular to the plane holding the circle.

achille hui
  • 122,701
  • But $C \cdot r$ is not zero... So how is normal vector in direction of $C$? – VECT Sep 18 '13 at 06:35
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    $\vec{C}\cdot\vec{r}$ is not zero but a constant, so the locus of $\vec{r}(t)$ lies in a plane and the normal vector of that "plane" is in the direction of $\vec{C}$. In the special case when $\vec{C}\cdot\vec{r} = 0$, the plane holding $\vec{r}(t)$ will crosses the origin. Under this special situation, $\vec{r}$ itself will be perpendicular to $\vec{C}$. See the difference? – achille hui Sep 18 '13 at 06:43
  • @vect decompose $r$ into two vectors, one orthogonal to the normal and one parallel to it. Do you understand now? – obataku Sep 18 '13 at 16:44
  • @achille hui nice solution – obataku Sep 18 '13 at 16:45
  • I get that. But why is "if you join a line passing through the origin and the center of the circle, that line will be perpendicular to the plane holding the circle." true? – VECT Sep 18 '13 at 19:05
  • That's pure geometry. If you draw a line $\ell$ passing through the center $c$ of the circle perpendicular to the plane, then for any point $a$ on $\ell$ and point $b$ on the circle, the three points $a, b, c$ forms a right angled triangle. We have $|ab|^2 = |ac|^2+|bc|^2$ and this value is independent of choice of $b$. i.e. $a$ is equi-distant to any point $b$ on the circle. It is easy to see any points that are equi-distant to all points on the circle lies on $\ell$. By 1. the origin is equi-distant to all points on the circle and hence lies on $\ell$. i.e. the $\ell$ is the line we seek. – achille hui Sep 19 '13 at 10:04