To find the resultant vector using the numerical method, we need to break each vector into its components and then sum the components. ### Step 1: Break each vector into its components#### Vector \(\vec{A}\) Magnitude: \(A = 60\) Angle: \(\theta_1 = 45^\circ\) Components: \[ A_x = A \cos(\theta_1) = 60 \cos(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] \[ A_y = A \sin(\theta_1) = 60 \sin(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] #### Vector \(\vec{B}\) Magnitude: \(B = 60\) Angle: \(\theta_2 = 20^\circ\) (below the horizontal axis, so it's negative) Components: \[ B_x = B \cos(\theta_2) = 60 \cos(20^\circ) = 60 \times 0.9397 = 56.38 \] \[ B_y = B \sin(\theta_2) = 60 \sin(20^\circ) = 60 \times 0.3420 = 20.52 \] Since \(\theta_2\) is below the horizontal axis, \(B_y\) is negative: \[ B_y = -20.52 \] #### Vector \(\vec{C}\) Magnitude: \(C = 60\) Angle: \(180^\circ\) (directly to the left) Components: \[ C_x = C \cos(180^\circ) = 60 \cos(180^\circ) = 60 \times (-1) = -60 \] \[ C_y = C \sin(180^\circ) = 60 \sin(180^\circ) = 60 \times 0 = 0 \] ### Step 2: Sum the components#### Sum of \(x\)-components: \[ R_x = A_x + B_x + C_x \] \[ R_x = 42.43 + 56.38 - 60 \] \[ R_x = 38.81 \] #### Sum of \(y\)-components: \[ R_y = A_y + B_y + C_y \] \[ R_y = 42.43 - 20.52 + 0 \] \[ R_y = 21.91 \] ### Step 3: Find the magnitude and direction of the resultant vector#### Magnitude: \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(38.81)^2 + (21.91)^2} \] \[ R = \sqrt{1506.29 + 479.36} \] \[ R = \sqrt{1985.65} \] \[ R \approx 44.56 \] #### Direction (angle \(\theta\) with respect to the positive x-axis): \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \] \[ \theta = \tan^{-1}\left(\frac{21.91}{38.81}\right) \] \[ \theta \approx 29.4^\circ \] ### ResultThe resultant vector has a magnitude of approximately \(44.56\) and makes an angle of approximately \(29.4^\circ\) with the positive x-axis.

To find the resultant vector using the numerical method, we need to break each vector into its components and then sum the components. ### Step 1: Break each vector into its components#### Vector \(\vec{A}\) Magnitude: \(A = 60\) Angle: \(\theta_1 = 45^\circ\) Components: \[ A_x = A \cos(\theta_1) = 60 \cos(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] \[ A_y = A \sin(\theta_1) = 60 \sin(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] #### Vector \(\vec{B}\) Magnitude: \(B = 60\) Angle: \(\theta_2 = 20^\circ\) (below the horizontal axis, so it's negative) Components: \[ B_x = B \cos(\theta_2) = 60 \cos(20^\circ) = 60 \times 0.9397 = 56.38 \] \[ B_y = B \sin(\theta_2) = 60 \sin(20^\circ) = 60 \times 0.3420 = 20.52 \] Since \(\theta_2\) is below the horizontal axis, \(B_y\) is negative: \[ B_y = -20.52 \] #### Vector \(\vec{C}\) Magnitude: \(C = 60\) Angle: \(180^\circ\) (directly to the left) Components: \[ C_x = C \cos(180^\circ) = 60 \cos(180^\circ) = 60 \times (-1) = -60 \] \[ C_y = C \sin(180^\circ) = 60 \sin(180^\circ) = 60 \times 0 = 0 \] ### Step 2: Sum the components#### Sum of \(x\)-components: \[ R_x = A_x + B_x + C_x \] \[ R_x = 42.43 + 56.38 - 60 \] \[ R_x = 38.81 \] #### Sum of \(y\)-components: \[ R_y = A_y + B_y + C_y \] \[ R_y = 42.43 - 20.52 + 0 \] \[ R_y = 21.91 \] ### Step 3: Find the magnitude and direction of the resultant vector#### Magnitude: \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(38.81)^2 + (21.91)^2} \] \[ R = \sqrt{1506.29 + 479.36} \] \[ R = \sqrt{1985.65} \] \[ R \approx 44.56 \] #### Direction (angle \(\theta\) with respect to the positive x-axis): \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \] \[ \theta = \tan^{-1}\left(\frac{21.91}{38.81}\right) \] \[ \theta \approx 29.4^\circ \] ### ResultThe resultant vector has a magnitude of approximately \(44.56\) and makes an angle of approximately \(29.4^\circ\) with the positive x-axis.

When using the graphical method for vector addition, how is the resultant vector represented in the diagram?Group of answer choicesAs the longest vector in the diagramAs the vector closest to the x-axisAs the sum of vector magnitudesAs the diagonal connecting the tail of the first vector to the head of the last vector.

Find the resultant of the following vectors. Round to 1 d.p.30∠70°and10∠340°

Given the vectors M = 5i - 3j and  N  = -3i + 2k, what is the resultant?Group of answer choices-i – j2i - 3j + 2k-i – ki + 3j - 2k

How do you calculate the resultant force from a vector diagram?