- #1

lizette

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.