1. **State the problem:** We need to find the vector expression for the tension $\mathbf{T}$ in cable AB with magnitude 2.4 kN, acting on member AD, and also find the magnitude of the projection of $\mathbf{T}$ along line AC. 2. **Find the unit vector along AB:** Given vector from A to B is $\mathbf{r}_{AB} = 2\mathbf{i} + 1\mathbf{j} - 5\mathbf{k}$. Magnitude of $\mathbf{r}_{AB}$ is: $$\sqrt{2^2 + 1^2 + (-5)^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$$ Unit vector along AB is: $$\mathbf{n}_{AB} = \frac{2\mathbf{i} + 1\mathbf{j} - 5\mathbf{k}}{\sqrt{30}}$$ 3. **Calculate tension vector $\mathbf{T}$:** Tension magnitude $T = 2.4$ kN. So, $$\mathbf{T} = T \mathbf{n}_{AB} = 2.4 \times \frac{2\mathbf{i} + 1\mathbf{j} - 5\mathbf{k}}{\sqrt{30}} = \frac{2.4}{\sqrt{30}} (2\mathbf{i} + 1\mathbf{j} - 5\mathbf{k})$$ Calculate scalar: $$\frac{2.4}{\sqrt{30}} \approx 0.438$$ Therefore, $$\mathbf{T} = 0.876\mathbf{i} + 0.438\mathbf{j} - 2.19\mathbf{k} \text{ kN}$$ 4. **Find unit vector along AC:** Vector from A to C is $\mathbf{r}_{AC} = 2\mathbf{i} - 2\mathbf{j} - 5\mathbf{k}$. Magnitude: $$\sqrt{2^2 + (-2)^2 + (-5)^2} = \sqrt{4 + 4 + 25} = \sqrt{33}$$ Unit vector along AC: $$\mathbf{n}_{AC} = \frac{2\mathbf{i} - 2\mathbf{j} - 5\mathbf{k}}{\sqrt{33}}$$ 5. **Calculate projection of $\mathbf{T}$ along AC:** Projection magnitude is the dot product: $$T_{AC} = \mathbf{T} \cdot \mathbf{n}_{AC} = (0.876)(\frac{2}{\sqrt{33}}) + (0.438)(\frac{-2}{\sqrt{33}}) + (-2.19)(\frac{-5}{\sqrt{33}})$$ Calculate each term: $$= \frac{1.752}{\sqrt{33}} - \frac{0.876}{\sqrt{33}} + \frac{10.95}{\sqrt{33}} = \frac{1.752 - 0.876 + 10.95}{\sqrt{33}} = \frac{11.826}{\sqrt{33}}$$ Calculate numeric value: $$\sqrt{33} \approx 5.7446$$ So, $$T_{AC} \approx \frac{11.826}{5.7446} = 2.06 \text{ kN}$$ **Final answers:** $$\mathbf{T} = 0.876\mathbf{i} + 0.438\mathbf{j} - 2.19\mathbf{k} \text{ kN}$$ $$T_{AC} = 2.06 \text{ kN}$$