Physics (Vectors)

Question

A Vector 𝑨⃗ is 2.80 cm long and is 60.0 degrees above the x-axis in the first quadrant. Vector 𝑩⃗⃗ is 1.90 cm long and is 60.0 degrees below the x-axis in the fourth quadrant. Use components to find the magnitude and direction of 𝑨⃗ - 𝑩⃗.

Answer 1

A_x = 2.80cos60° = 1.40 cm; A_y = 2.80sin60° = 2.42 cm

B_x = 1.90cos(-60°) = 0.95 cm, B_y = 1.90sin(-60°) = - 1.65 cm

So, (A - B)_x = A_x - B_x = 1.40 - 0.95 = 0.45 cm

(A - B)_y = A_y - B_y = 2.42 - (- 1.65) = 4.07 cm.

Then magnitude of (A - B) = (0.45² + 4.07²)^1/2 = 4.095 cm.

Because (A - B)_x > 0 and (A - B)_y > 0 this vector lokated in 1st quadrant.

Now tan∝ = 4.07/0.45 = 9.0444, so ∝ tan^-1(9.0444) = 83.69°,

∝ is angle above x-axis