Find the magnitude of the vector sum Vector A is 5.00 m long in a \( 0^{\circ} \) direction. Vector B is 6.00 m long in a \( 210^{\circ} \) directior

To find the magnitude of the vector sum of Vector A and Vector B, we can break each vector into its components and then sum those components. ### Step 1: Break down the vectors into components **Vector A:** - Magnitude: \( A = 5.00 \, \text{m} \) - Direction: \( 0^{\circ} \) (which is along the positive x-axis) The components of Vector A are: - \( A_x = A \cos(0^{\circ}) = 5.00 \cos(0) = 5.00 \, \text{m} \) - \( A_y = A \sin(0^{\circ}) = 5.00 \sin(0) = 0.00 \, \text{m} \) So, \( \mathbf{A} = (5.00, 0.00) \). **Vector B:** - Magnitude: \( B = 6.00 \, \text{m} \) - Direction: \( 210^{\circ} \) The components of Vector B are: - \( B_x = B \cos(210^{\circ}) = 6.00 \cos(210^{\circ}) = 6.00 \left(-\frac{\sqrt{3}}{2}\right) = -3.00\sqrt{3} \, \text{m} \) - \( B_y = B \sin(210^{\circ}) = 6.00 \sin(210^{\circ}) = 6.00 \left(-\frac{1}{2}\right) = -3.00 \, \text{m} \) So, \( \mathbf{B} = (-3.00\sqrt{3}, -3.00) \). ### Step 2: Sum the components Now we can sum the x and y components of the two vectors: **Total x-component:** \[ R_x = A_x + B_x = 5.00 + (-3.00\sqrt{3}) \] **Total y-component:** \[ R_y = A_y + B_y = 0.00 + (-3.00) = -3.00 \] ### Step 3: Calculate the magnitude of the resultant vector The magnitude of the resultant vector \( \mathbf{R} \) can be found using the Pythagorean theorem: \[ R = \sqrt{R_x^2 + R_y^2} \] Substituting the values: \[ R = \sqrt{(5.00 - 3.00\sqrt{3})^2 + (-3.00)^2} \] Calculating \( R_x \): \[ R_x = 5.00 - 3.00\sqrt{3} \approx 5.00 - 5.196 = -0.196 \, \text{m} \] Now substituting back into the magnitude equation: \[ R = \sqrt{(-0.196)^2 + (-3.00)^2} \] \[ R = \sqrt{0.038416 + 9.00} = \sqrt{9.038416} \approx 3.01 \, \text{m} \] ### Final Answer The magnitude of the vector sum \( \mathbf{R} \) is approximately \( 3.01 \, \text{m} \).