# Position Vector

1. Sep 11, 2013

### tchouhan

1. The problem statement, all variables and given/known data
The distance travelled by an airplane flying from San Francisco International airport (SFO) to San Jose International (SJC) is 30 nautical miles 36° south of east. Flying from SJC to Tracy, the plane's displacement is 36 nautical miles 52° east of north. What is Tracy's position vector in nautical miles relative to SFO?

2. Relevant equations

$a^2+b^2=c^2$

3. The attempt at a solution

I've tried to draw a triangle, but the positions are weird. I know I need to use trig, but I can figure out what formula to use, or where to start even. Someone help!

2. Sep 11, 2013

### SteamKing

Staff Emeritus
What's weird about the positions? Have you checked your sketch with a map of California? After all, Google Maps has covered the globe.

3. Sep 11, 2013

### tchouhan

Yeah I have, but I don't think I'm getting the angles correctly.

Here's a photo, SF at the top left, SJ south, Tracy east: http://imgur.com/LSEgciu

I'm also not entirely sure what sort of trig to use, only just learned what vectors were yesterday.

4. Sep 11, 2013

### SteamKing

Staff Emeritus
Well, we can't really provide any suggestions unless we see your work.

5. Sep 14, 2013

### tchouhan

I went to office hours, and I learned that it was just the Law of Cosines.

such that:

c^2 = (30^2+36^2-2(30*36)cos(ab).