1. **Stating the problem:** We are given points A, B, C, D with coordinates and tensions in cables. We want to find the tension vector $\mathbf{T}$ in cable AB and the tension $\mathbf{T}_{AC}$ in cable AC. 2. **Given data:** - $\overrightarrow{OB} = 3.7\mathbf{i} + 2.1\mathbf{j} + 0\mathbf{k}$ m - $\overrightarrow{OA} = 0\mathbf{i} + 1.3\mathbf{j} + 5.4\mathbf{k}$ m - $\overrightarrow{AD} = 0\mathbf{i} + 0\mathbf{j} + 1.3\mathbf{k}$ m (height difference from A to D) - $\mathbf{T} = (T_x \mathbf{i} + T_y \mathbf{j} + T_z \mathbf{k})$ kN tension in cable AB - $\mathbf{T}_{AC} = T_{AC,x} \mathbf{i}$ kN tension in cable AC 3. **Find vector $\overrightarrow{AB}$:** $$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3.7 - 0)\mathbf{i} + (2.1 - 1.3)\mathbf{j} + (0 - 5.4)\mathbf{k} = 3.7\mathbf{i} + 0.8\mathbf{j} - 5.4\mathbf{k}$$ 4. **Calculate magnitude of $\overrightarrow{AB}$:** $$|\overrightarrow{AB}| = \sqrt{3.7^2 + 0.8^2 + (-5.4)^2} = \sqrt{13.69 + 0.64 + 29.16} = \sqrt{43.49} \approx 6.595$$ 5. **Unit vector along AB:** $$\hat{u}_{AB} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{3.7}{6.595}\mathbf{i} + \frac{0.8}{6.595}\mathbf{j} + \frac{-5.4}{6.595}\mathbf{k} \approx 0.561\mathbf{i} + 0.121\mathbf{j} - 0.819\mathbf{k}$$ 6. **Express tension vector $\mathbf{T}$ in terms of magnitude $T$:** $$\mathbf{T} = T \hat{u}_{AB} = T(0.561\mathbf{i} + 0.121\mathbf{j} - 0.819\mathbf{k})$$ 7. **Given $\mathbf{T}_{AC} = T_{AC,x} \mathbf{i}$, tension along x-axis only.** 8. **Summary:** - $\mathbf{T} = T(0.561\mathbf{i} + 0.121\mathbf{j} - 0.819\mathbf{k})$ kN - $\mathbf{T}_{AC} = T_{AC,x} \mathbf{i}$ kN Since no numerical tension magnitudes are given, this is the vector form of tensions based on geometry.