# Equation of path

1. Oct 12, 2006

### Reshma

The equation of a path is of the form: $\vec r = \vec r_0 + \vec A t$
If 't' represents time, show that the time of closest approach is:
$$t = -\frac{\vec r_0 \cdot \vec A}{|\vec A|^2}$$

I am not really sure on how to proceed about this, but I made a crude approach by assuming $\vec r$ and $\vec r_0$ to be perpendicular. I took the dot product with r0 on both sides of given equation.
$$-\vec r_0^2 = \vec A \cdot \vec r_0 t$$

I don't think this a right way to solve, please give some suggestions.

Last edited: Oct 12, 2006
2. Oct 12, 2006

### HallsofIvy

Staff Emeritus
Closest approach to what?

3. Oct 15, 2006

### Reshma

Oh sorry, I forget to add that. Find the distance of closest approach to the origin i. e. the distance from the origin to the line.

4. Oct 15, 2006

### HallsofIvy

Staff Emeritus
It's not $\vec r_0$ that's perpendicular to $\vec r$, it's $\vec A$.
At the point of closest approach, the vector in the direction of the line, that is, $\vec A$, is perpendicular to the position vector, $\vec r_0+ \vec At$ itself. That is $\vec A \cdot (\vec r_0+ \vec At= 0$. That is, $\vec A \cdot \vec r_0+ \vec A \cdot \vec A t= 0$. Can you solve that for t?

5. Oct 16, 2006

### Reshma

Wow, thanks! That makes sense.
$$-\vec A \cdot \vec r_0 = |\vec A|^2t$$

$$t = -\frac{\vec r_0 \cdot \vec A}{|\vec A|^2}$$