1. **Problem statement:** We need to find the vector expression for the tension $\mathbf{T}$ exerted by the cable on the fixed bracket at point A. 2. **Given data:** - Tension magnitude $|\mathbf{T}| = 2$ kN = 2000 N - Coordinates of points: - $A = (0.3, 1.2, 0)$ m (assuming $z=0$ at A since not given) - $B = (0, 0.4, 0.5)$ m 3. **Formula and approach:** The tension vector $\mathbf{T}$ points from A to B along the cable. $$\mathbf{T} = T \cdot \hat{u}$$ where $\hat{u}$ is the unit vector from A to B: $$\hat{u} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}$$ 4. **Calculate vector $\overrightarrow{AB}$:** $$\overrightarrow{AB} = B - A = (0 - 0.3, 0.4 - 1.2, 0.5 - 0) = (-0.3, -0.8, 0.5)$$ 5. **Calculate magnitude of $\overrightarrow{AB}$:** $$|\overrightarrow{AB}| = \sqrt{(-0.3)^2 + (-0.8)^2 + 0.5^2} = \sqrt{0.09 + 0.64 + 0.25} = \sqrt{0.98} \approx 0.99$$ 6. **Calculate unit vector $\hat{u}$:** $$\hat{u} = \frac{1}{0.99}(-0.3, -0.8, 0.5) \approx (-0.303, -0.808, 0.505)$$ 7. **Calculate tension vector $\mathbf{T}$:** $$\mathbf{T} = 2000 \times (-0.303, -0.808, 0.505) = (-606, -1616, 1010) \text{ N}$$ **Final answer:** $$\boxed{\mathbf{T} = -606\mathbf{i} - 1616\mathbf{j} + 1010\mathbf{k} \text{ N}}$$ This vector represents the tension force exerted by the cable at point A, directed from A towards B with magnitude 2000 N.