- #1
lizette
Hi! I'm new to the forums. I'm taking an introduction to physics class this semester and I've been having some difficulty with it. Oh, I also wanted to let you know that it's been a while since I've taken calculus or any other math class for that matter. But I need physics to graduate. Anywho ... the question that I have deals with vector components.
Two vectors of magnitudes a and b make an angle theta (which I'll represent as @) with each other when placed tail to tail. Prove, by taking components along two perpendicular anes, that
r = the square root of (a^2 + b^2 + 2abcos@)
gives the magnitude of the sum vector R (vector R = r with that arrow above it) of the two vectors.
Well this is what I have so far:
vector A = Axi + Ayi
vector B = Bxi + Byi
vector R = vector A + vector B
A^2 = Ax^2 + Ay^2
B^2 = Bx^2 + By^2
R^2 = A^2 + B^2
A dot B = A*B = ABcos@
I can see how r = square root of (A^2 + B^2) but where does the 2ABcos@ come in. I have a feeling that it deals with the A*B product, but I don't know how to fit it in.
Two vectors of magnitudes a and b make an angle theta (which I'll represent as @) with each other when placed tail to tail. Prove, by taking components along two perpendicular anes, that
r = the square root of (a^2 + b^2 + 2abcos@)
gives the magnitude of the sum vector R (vector R = r with that arrow above it) of the two vectors.
Well this is what I have so far:
vector A = Axi + Ayi
vector B = Bxi + Byi
vector R = vector A + vector B
A^2 = Ax^2 + Ay^2
B^2 = Bx^2 + By^2
R^2 = A^2 + B^2
A dot B = A*B = ABcos@
I can see how r = square root of (A^2 + B^2) but where does the 2ABcos@ come in. I have a feeling that it deals with the A*B product, but I don't know how to fit it in.