Find Closest Approach Time in Equation of Path

[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.

 

[HallsofIvy]

Closest approach to what?

 

[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.

 

[HallsofIvy]

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?

 

[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}