Recent content by andylatham82

Thanks for the link anyway, I was also wondering what the cross product would represent, so you've answered that question for me!

Thanks very much for the response. So the resultant area doesn't really represent anything tangible with regards to the two displacements? It's just an abstract 'area' that you couldn't, say, draw on a diagram of the two vectors?

Hi, This feels like such a stupid question, but it's bugging me. Two displacements can be represented with two vectors. Let's say their magnitudes are expressed in metres. The scalar (dot) product of the two vectors results in a value with the units of square metres, which must be an area. Can...

Well I finally got to the bottom of it then thanks to learning that tan = sin/cos. sin θ/cos θ = |AxB|/A.B ⇒ tan θ = |AxB|/A.B = 9/-7 ⇒ θ = 52° or 180° - 52° = 128° Since the A.B < 0, the angle between the vectors must be greater than 90°, thus θ = 128°. Phew! Thanks so much for your help...

I can combine the two equations to give (sin θ)/(cos θ) = (|AxB|)/AB) But I'm not sure where to go next. If we were talking about a right-angled triangle, we could say sin/cos = tan, but that's not the case here is it? I could do cos θ = sin (90-θ) perhaps but my algebra is failing me in...

I know that: A.B = ABcosθ and |AxB| = ABsinθ I'm just not sure where to go from there with no magnitudes or components to work with. I have a value of -7 for the scalar product (so I know the angle is greater than 90°) and a magnitude of 9 for the vector product. I'm sure there's something...

Hi, hopefully a quick question here...how do you calculate the angle between two vectors if the only information you have is the value of their scalar product and the magnitude of their cross product? Thanks! Andy

Ah! I'm correct? Thank heavens for that! I thought my brain had stopped working! Thanks for putting my mind at ease! As for a diagram, I always draw them, I just didn't include my diagram here. I'd be lost without a diagram haha.

I've been racking my brain for a while now over what I feel should be a simple problem to solve, but my answer is wrong. I'm not wildly wrong, but wrong enough to think I've made a proper error and it's not just a rounding error or something. The question goes like this: A dog in an open field...

Thanks Greg!

Ahh of course! That's solved the mystery! Thanks a lot DrClaude :)

Hello, I have a question about why I can't determine the angle between two vectors using their cross product. Say there are two vectors in the XY-plane that we want to find the angle between: A = -2.00i + 6.00j B = 2.00i - 3.00j The method to do this would be to work out the scalar product of...

Hello, I'm an animator working in the field of videogames. However before taking a U-turn in my career, I got an MPhys degree in Physics with Astrophysics from the University of Manchester. Lately I've wanted to get my brain back into Physics mode and recapture some of the knowledge that has...