two vectors A and B have precisely equal magnitudes. For the magnitude of A +B to be larger than the magnitude of A-B by the factor n, what must be the angle between them?
Let's denote the magnitude of a vector V by |V|
Given:
|A + B| = n|A - B|
Squaring both sides:
|A + B|² = n²|A - B|²
Since V² = |V|²,
(A + ² = n²(A - ²
Performing the operation on vectors:
A² + 2AB + B² = n²(A² - 2AB + B²)
|A| = |B|==>A² = B²
2A² + 2AB = n²(2A² - 2AB)
2AB(1 + n²) = 2A²(n² - 1)
AB = 2A²(n² - 1)/2(1 + n²)
|A||B|cosα = A²(n² - 1)/(1 + n²)
cosα = |A|²(n² - 1)/|A|²(1 + n²)
cosα = (n² - 1)/(n² + 1)
α = arcos[(n² - 1)/(n² + 1)]
But A - B = 0
Comments
Let's denote the magnitude of a vector V by |V|
Given:
|A + B| = n|A - B|
Squaring both sides:
|A + B|² = n²|A - B|²
Since V² = |V|²,
(A +
² = n²(A - 
²
Performing the operation on vectors:
A² + 2AB + B² = n²(A² - 2AB + B²)
|A| = |B|==>A² = B²
2A² + 2AB = n²(2A² - 2AB)
2AB(1 + n²) = 2A²(n² - 1)
AB = 2A²(n² - 1)/2(1 + n²)
|A||B|cosα = A²(n² - 1)/(1 + n²)
cosα = |A|²(n² - 1)/|A|²(1 + n²)
cosα = (n² - 1)/(n² + 1)
α = arcos[(n² - 1)/(n² + 1)]
But A - B = 0