The position and the velocity of two particles are \(\vec{r}_1\), \(\vec{r}_2\) and \(\vec{v}_1\), \(\vec{v}_2\). Prove that they collide only if : \((\vec{r}_1 - \vec{r}_2) \times (\vec{v}_1 - \vec{v}_2) = 0\).

For two particles to collide, their relative position and velocity vectors \(\vec{r}_1 - \vec{r}_2\) and \(\vec{v}_1 - \vec{v}_2\) must result in a zero cross product.

ANSWER : To prove that two particles collide if and only if the cross product of the position and velocity vectors equals zero, we can use the concept of relative motion.

Let's consider two particles with positions \(\mathbf{r}_1\) and \(\mathbf{r}_2\), and velocities \(\mathbf{v}_1\) and \(\mathbf{v}_2\). We want to show that they collide if and only if \((\mathbf{r}_1 - \mathbf{r}_2) \times (\mathbf{v}_1 - \mathbf{v}_2) = 0\).

The condition for the particles to collide is that their separation distance decreases to zero at some point in time. This implies that \(\mathbf{r}_1 - \mathbf{r}_2\) and \(\mathbf{v}_1 - \mathbf{v}_2\) are parallel.

Now, let's expand the cross product using the properties of the cross product and dot product :

\((\mathbf{r}_1 - \mathbf{r}_2) \times (\mathbf{v}_1 - \mathbf{v}_2) = (\mathbf{r}_1 - \mathbf{r}_2) \times \mathbf{v}_1 - (\mathbf{r}_1 - \mathbf{r}_2) \times \mathbf{v}_2\)

Using the properties of the cross product, we have :

\((\mathbf{r}_1 - \mathbf{r}_2) \times \mathbf{v}_1 = \mathbf{r}_1 \times \mathbf{v}_1 - \mathbf{r}_2 \times \mathbf{v}_1\)

\((\mathbf{r}_1 - \mathbf{r}_2) \times \mathbf{v}_2 = \mathbf{r}_1 \times \mathbf{v}_2 - \mathbf{r}_2 \times \mathbf{v}_2\)

Now, if \(\mathbf{r}_1 - \mathbf{r}_2\) and \(\mathbf{v}_1 - \mathbf{v}_2\) are parallel, it means that \(\mathbf{r}_1 \times \mathbf{v}_1\) and \(\mathbf{r}_2 \times \mathbf{v}_1\) are parallel, and \(\mathbf{r}_1 \times \mathbf{v}_2\) and \(\mathbf{r}_2 \times \mathbf{v}_2\) are also parallel.

Hence, we can write :

\((\mathbf{r}_1 - \mathbf{r}_2) \times (\mathbf{v}_1 - \mathbf{v}_2) = (\mathbf{r}_1 \times \mathbf{v}_1 - \mathbf{r}_2 \times \mathbf{v}_1) - (\mathbf{r}_1 \times \mathbf{v}_2 - \mathbf{r}_2 \times \mathbf{v}_2)\)

Since \(\mathbf{r}_1 \times \mathbf{v}_1\) and \(\mathbf{r}_2 \times \mathbf{v}_1\) are parallel, and \(\mathbf{r}_1 \times \mathbf{v}_2\) and \(\mathbf{r}_2 \times \mathbf{v}_2\) are parallel, the subtraction of these parallel vectors will result in a zero vector :

\((\mathbf{r}_1 - \mathbf{r}_2) \times (\mathbf{v}_1 - \mathbf{v}_2) = 0\)

Therefore, we have proven that the particles collide if and only if :

\((\mathbf{r}_1 - \mathbf{r}_2) \times (\mathbf{v}_1 - \mathbf{v}_2) = 0\).

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