Two vectors A and B have precisely equal magnitudes. For the magnitude of A + B to be 62 times greater than the magnitude of A - B, what must be the angle between them? I think I'm on the right track... But I can't seem to get this question. I've been stuck on it for a while now.
angle in between = cos of theta= dot product of first vector and second vector DIVIDED by magnitude of the first and second vector. Hope that helps.
LMFAO sorry for the late re For the magnitude of A + B to be 62 times greater than the magnitude of A - B... means |A + B| = 62 * |A - B| not 62(A+B) = A-B... lols didnt see this post till today and it takes a while to figure it out. i did it in a weird way. if you draw arbitrary vectors A and B with B on positive x-axis and call the angle between Θ. using geometry you'll notice |A + B| = [|B| + |A|cosΘ]² + [|A|sinΘ]² and |A - B| = [|A|cosΘ - |B|]² + [|A|sinΘ]² magnitude of A + B is 62 times the magnitude of A - B: |A + B| = 62 * |A - B| [|B| + |A|cosΘ]² + [|A|sinΘ]² = 62 * [ [|A|cosΘ - |B|]² + [|A|sinΘ]² ] expand and you should get this: [|B|² + 2|A||B|cosΘ + |A|²cos²Θ + |A|²sin²Θ] = 62 [|A|²cos²Θ - 2|A||B|cosΘ + |B|² + |A|²sin²Θ] |A|²cos²Θ + |A|²sin²Θ = |A|²(cos²Θ + sin²Θ) = |A|² [|B|² + 2|A||B|cosΘ + |A|²] = 62 [|A|² - 2|A||B|cosΘ + |B|²] since A and B have the the same magnitude, |A| = |B| [|B|² + 2|B||B|cosΘ + |B|²] = 62 [|B|² - 2|B||B|cosΘ + |B|²] the whole thing is now simplified to: [2|B|² + 2|B|²cosΘ] = 62 [2|B|² - 2|B|²cosΘ] further simplified to: (1 + cosΘ) = 62 (1 - cosΘ) do a little algebra: 1 + cosΘ = 62 - 62cosΘ 63cosΘ = 61 cosΘ = 61/63 Θ ≈ 14.48° I think this is the answer.
