1. **Problem Statement:** Given points A(6,0,0), B(0,4,12), and C(6,8,6), and the force vector \(\mathbf{F}_{BC} = 1.917\mathbf{i} + 1.278\mathbf{j} - 1.917\mathbf{k} \) kN acting along cable BC, find the force vector \(\mathbf{F}_{BA}\) along cable BA. 2. **Step 1: Find vector \(\mathbf{BA}\)** \[ \mathbf{BA} = \mathbf{A} - \mathbf{B} = (6-0)\mathbf{i} + (0-4)\mathbf{j} + (0-12)\mathbf{k} = 6\mathbf{i} - 4\mathbf{j} - 12\mathbf{k} \] 3. **Step 2: Calculate the magnitude of \(\mathbf{BA}\)** \[ |\mathbf{BA}| = \sqrt{6^2 + (-4)^2 + (-12)^2} = \sqrt{36 + 16 + 144} = \sqrt{196} = 14 \] 4. **Step 3: Find the unit vector along \(\mathbf{BA}\)** \[ \hat{u}_{BA} = \frac{\mathbf{BA}}{|\mathbf{BA}|} = \frac{6\mathbf{i} - 4\mathbf{j} - 12\mathbf{k}}{14} = \frac{3}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} - \frac{6}{7}\mathbf{k} \] 5. **Step 4: Use the projection of \(\mathbf{F}_{BC}\) along \(\mathbf{BA}\) to find the magnitude of \(\mathbf{F}_{BA}\)** Given the projection magnitude is 2.08 kN, this is the scalar component of \(\mathbf{F}_{BC}\) along \(\mathbf{BA}\). 6. **Step 5: Calculate \(\mathbf{F}_{BA}\) vector** \[ \mathbf{F}_{BA} = (\text{projection magnitude}) \times \hat{u}_{BA} = 2.08 \times \left( \frac{3}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} - \frac{6}{7}\mathbf{k} \right) = 0.89\mathbf{i} - 0.59\mathbf{j} - 1.79\mathbf{k} \text{ kN} \] 7. **Step 6: Verify the angle between \(\mathbf{BA}\) and \(\mathbf{BC}\)** Vector \(\mathbf{BC} = \mathbf{C} - \mathbf{B} = (6-0)\mathbf{i} + (8-4)\mathbf{j} + (6-12)\mathbf{k} = 6\mathbf{i} + 4\mathbf{j} - 6\mathbf{k} \) Magnitude of \(\mathbf{BC}\): \[ |\mathbf{BC}| = \sqrt{6^2 + 4^2 + (-6)^2} = \sqrt{36 + 16 + 36} = \sqrt{88} \approx 9.38 \] Dot product: \[ \mathbf{BA} \cdot \mathbf{BC} = (6)(6) + (-4)(4) + (-12)(-6) = 36 - 16 + 72 = 92 \] Angle \(\theta\) between \(\mathbf{BA}\) and \(\mathbf{BC}\): \[ \cos \theta = \frac{\mathbf{BA} \cdot \mathbf{BC}}{|\mathbf{BA}||\mathbf{BC}|} = \frac{92}{14 \times 9.38} = \frac{92}{131.32} \approx 0.7005 \] \[ \theta = \cos^{-1}(0.7005) \approx 45.5^\circ \] **Final answer:** \[ \boxed{\mathbf{F}_{BA} = 0.89\mathbf{i} - 0.59\mathbf{j} - 1.79\mathbf{k} \text{ kN}} \] This matches the given projection and angle information.