# Homework Help: A few questions relating to the cross product

1. Nov 19, 2008

### Spiffy

Ug. My first try at this and my whole post was deleted. Here goes again :)

1. The problem statement, all variables and given/known data

i) Two cities on the surface of the earth are represented by position vectors that connect the location of each city to the centre of the earth. Assuming that the centre of the earth is assigned the coordinates of the origin, and that the earth is a perfect sphere, outline the steps that would lead to a calculation of the shortest distance between the two cities. Hint: How can you determine arc length?

ii) Prove A is perpendicular to B if |A+B| = |A-B|

iii) A||B if AxB=0

3. The attempt at a solution
i) well you have two vectors pointing outword tail to tail and an angle Θ in between them. What I don't really know is how to find the arc length as the hint wants me to. If someone could point me toward an equation i'll report with some progress

ii) It seems so obvious that i'm sure it's staring me in the face and I don't even know it. I have drawn on my paper A vector downward and B and -B vector going on at 90o degrees either way. So now I have one triangle made up of two others and the two triangles are obviously equal to each other, they share adj and opp sidelengths and therefore the hypotenuse must as well be equal in both. I don't kknow how to take this from my sketch to something a bit more formal and algebraic though..

iii) I know that its true because if they were parallel sinΘ would = 0. I have:
AxB = |A||B|sinΘ
If A and B are parallel Θ=0
RS: |A||B|sin0
= 0
therefore AxB=0 when A and B are paralell. Would you accept this?

I appreciate any light that can be shed on some of these questions. I know this website is going to be a valuable resource for myself now and in the future. Even just typing questions up here is helping. (I originally had 4 questions but solved one just by putting it into a different form of media) thanks a lot!

2. Nov 20, 2008

### tiny-tim

Welcome to PF!

Ug! Welcome to PF!