ANGLES

[vea]

While we are on the subject, I don't get 4d.

Why would the closest point to the base of the tower be when r.v = 0 ??

Would also be interested to know this. Personally, I took the modulus of displacement to find distance and then used calculus to find the time of the minimum distance.

[b^3]

While we are on the subject, I don't get 4d.

Why would the closest point to the base of the tower be when r.v = 0 ??

The closest point will be when the velocity is perpendicular to the position vector because. Try to draw it out, the velocity vector gives you the direction that you are travelling in and since it is not time dependent, you are travelling in a straight line. It's kinda like the minimisation problems from other questions, arrrrrrrrrrrgggggggggggg(z) I don't think I can explain it well. Try to draw it out. Sorry I can't be of more help. dc or someone might come along though.

[dc302]

Kinda hard to explain without a picture. But picture this:

You are walking on one side of the road. On the other side there is a bank that you wish to go to. Where would you cross so that you would cross the road in the least distance possible? (ie. where would you cross so that you were the closest to the bank?). Well, when you get directly opposite it so that your 'position' relative to the bank is perpendicular to your 'direction' (which is down the road you're walking).

This is the same concept. When your position relative to the point you are concerned about, is perpendicular with your direction, then you are closest to it (or furthest, but let's not worry about that).

So position =r, direction =v. When r.v=0, they are perpendicular and so you are closest to it at that point.

[jane1234]

Kinda hard to explain without a picture. But picture this:

You are walking on one side of the road. On the other side there is a bank that you wish to go to. Where would you cross so that you would cross the road in the least distance possible? (ie. where would you cross so that you were the closest to the bank?). Well, when you get directly opposite it so that your 'position' relative to the bank is perpendicular to your 'direction' (which is down the road you're walking).

This is the same concept. When your position relative to the point you are concerned about, is perpendicular with your direction, then you are closest to it (or furthest, but let's not worry about that).

So position =r, direction =v. When r.v=0, they are perpendicular and so you are closest to it at that point.

Ah yes, get it now. Thank you (and b^3)!