1. **Problem statement:** Given forces $F_1 = 90$ N, $F_2 = 150$ N, and $F_3 = 200$ N with specified directions, express each force in Cartesian vector form and find the resultant force vector. 2. **Force directions and angles:** - $F_3$ is along the positive $z$-axis. - $F_2$ makes a $60^\circ$ angle with the horizontal $y$-axis. - $F_1$ makes angles $45^\circ$ and $60^\circ$ in the $xy$-plane (we interpret this as $F_1$ making $45^\circ$ with $x$-axis and $60^\circ$ with $y$-axis, consistent with the problem description). 3. **Expressing forces in vector form:** - $F_3$ along $z$-axis: $$\vec{F}_3 = 0\hat{i} + 0\hat{j} + 200\hat{k}$$ - $F_2$ makes $60^\circ$ with $y$-axis, so its components are: - $x$-component: $0$ (assuming it lies in $yz$-plane) - $y$-component: $150 \cos 60^\circ = 150 \times 0.5 = 75$ - $z$-component: $150 \sin 60^\circ = 150 \times \frac{\sqrt{3}}{2} \approx 129.9$ So, $$\vec{F}_2 = 0\hat{i} + 75\hat{j} + 129.9\hat{k}$$ - $F_1$ makes $45^\circ$ with $x$-axis and $60^\circ$ with $y$-axis in $xy$-plane, so: - $x$-component: $90 \cos 45^\circ = 90 \times \frac{\sqrt{2}}{2} \approx 63.64$ - $y$-component: $90 \cos 60^\circ = 90 \times 0.5 = 45$ - $z$-component: $0$ (force lies in $xy$-plane) So, $$\vec{F}_1 = 63.64\hat{i} + 45\hat{j} + 0\hat{k}$$ 4. **Calculate resultant force:** Sum components: - $x$-component: $63.64 + 0 + 0 = 63.64$ - $y$-component: $45 + 75 + 0 = 120$ - $z$-component: $0 + 129.9 + 200 = 329.9$ Resultant force vector: $$\vec{F}_R = 63.64\hat{i} + 120\hat{j} + 329.9\hat{k}$$ 5. **Magnitude of resultant force:** $$|\vec{F}_R| = \sqrt{63.64^2 + 120^2 + 329.9^2} \approx \sqrt{4050 + 14400 + 108840} = \sqrt{127290} \approx 356.8$$ N **Final answer:** $$\vec{F}_R = 63.64\hat{i} + 120\hat{j} + 329.9\hat{k} \quad \text{with magnitude} \quad 356.8$$ N