Add Vectors with R² = A² + B² + 2AB cosθ and Read the Special Cases

For vectors of magnitudes A and B at an angle θ, R² = A² + B² + 2 A B cosθ and the resultant makes an angle α with A where tan α = B sinθ / (A + B cosθ). The difference is |A - B|² = A² + B² - 2 A B cosθ. The resultant is largest (A + B) at 0° and smallest (|A - B|) at 180°. A vector of length l turned through θ about its tail changes by 2 l sin(θ/2); the tip of a seconds hand of length 1 cm moves through 90° in 15 s, so its velocity (magnitude 2π/60 cm/s) changes by sqrt(2) times that, (π/30) sqrt(2) cm/s.

If |A| = |B|, the resultant bisects the angle between them, so a resultant making θ/2 with A means |A| = |B|.

If |A + B| = |A - B| the vectors are perpendicular, and if (A + B) is perpendicular to (A - B) their magnitudes are equal. If |A| + |B| = |A + B| the vectors are parallel. Two equal vectors whose resultant has the same magnitude as either make an angle of 120°.

The resultant of (A + B) and (A - B) is 2A, which makes an angle of 0° with A.

Three forces can be in equilibrium only if each is between the difference and the sum of the other two, so (3, 5, 6) N can balance but (3, 5, 1), (3, 5, 9) and (3, 5, 15) cannot; with equal limits, (10, 10, 20) can balance as a degenerate triangle. For four forces P, 2P, 3P, 4P along the sides of a square in order, opposite sides combine to (3P - P) = 2P and (4P - 2P) = 2P at right angles, so the resultant is 2 sqrt(2) P. If two forces sum to 16 N and their 8 N resultant is perpendicular to the smaller force, the larger force is the hypotenuse: (16 - F)² = F² + 64 gives F = 6 N and 10 N. The angle between two paths from velocity components is the difference of their arctangents: components (1, sqrt 3) at 60° and (2, 2) at 45° differ by 15°.