1. **State the problem:** Find the magnitude and coordinate direction angles of the resultant force \( \mathbf{F}_R = \mathbf{F}_1 + \mathbf{F}_2 \). 2. **Write the given forces:** \( \mathbf{F}_1 = 0\mathbf{i} + 60\mathbf{j} + 80\mathbf{k} \) lb \( \mathbf{F}_2 = 50\mathbf{i} - 100\mathbf{j} + 100\mathbf{k} \) lb 3. **Calculate the resultant force components:** \[ \mathbf{F}_R = (0+50)\mathbf{i} + (60 - 100)\mathbf{j} + (80 + 100)\mathbf{k} = 50\mathbf{i} - 40\mathbf{j} + 180\mathbf{k} \] 4. **Calculate the magnitude of the resultant force:** \[ |\mathbf{F}_R| = \sqrt{50^{2} + (-40)^{2} + 180^{2}} = \sqrt{2500 + 1600 + 32400} = \sqrt{36500} \] \[ |\mathbf{F}_R| = 190.79 \text{ lb (approx)} \] 5. **Calculate the coordinate direction angles \( \alpha, \beta, \gamma \):** \[ \alpha = \cos^{-1} \left( \frac{F_{Rx}}{|\mathbf{F}_R|} \right) = \cos^{-1} \left( \frac{50}{190.79} \right) = \cos^{-1} (0.262) = 75.0^\circ \] \[ \beta = \cos^{-1} \left( \frac{F_{Ry}}{|\mathbf{F}_R|} \right) = \cos^{-1} \left( \frac{-40}{190.79} \right) = \cos^{-1} (-0.21) = 102.2^\circ \] \[ \gamma = \cos^{-1} \left( \frac{F_{Rz}}{|\mathbf{F}_R|} \right) = \cos^{-1} \left( \frac{180}{190.79} \right) = \cos^{-1} (0.943) = 19.5^\circ \] **Final answers:** - Magnitude of resultant force \( |\mathbf{F}_R| \approx 190.8 \text{ lb} \) - Direction angles: \( \alpha = 75.0^\circ, \beta = 102.2^\circ, \gamma = 19.5^\circ \)