Calculate the resultant force R by summing the given forces F1, F2, and F3.

Here's the solution to question 1.

Part (i): Show that the system of forces is not in equilibrium.

Step 1: Calculate the resultant force $\mathbf{R}$ by summing the given forces $\mathbf{F}_1$, $\mathbf{F}_2$, and $\mathbf{F}_3$.

$$\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3$$ $$\mathbf{R} = (\mathbf{i} + \mathbf{j} + 3\mathbf{k}) + (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + (-\mathbf{i} - \mathbf{j} - \mathbf{k})$$ $$\mathbf{R} = (1+2-1)\mathbf{i} + (1+1-1)\mathbf{j} + (3+2-1)\mathbf{k}$$ $$\mathbf{R} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$$

Since the resultant force $\mathbf{R} \neq \mathbf{0}$, the system of forces is not in equilibrium.

Part (ii): A fourth force $\mathbf{F}_4$ is added to the system at the point $\mathbf{r}_4$ with position vector $\mathbf{r}_4 = \mathbf{i} - \mathbf{j} + \mathbf{k}$ to form a couple. Find:

a) The force $\mathbf{F}_4$.

Step 2: For the system to form a couple, the net force must be zero. Therefore, $\mathbf{F}_4$ must be equal and opposite to the resultant force of the first three forces.

$$\mathbf{F}_4 = -\mathbf{R}$$ $$\mathbf{F}_4 = -(2\mathbf{i} + \mathbf{j} + 4\mathbf{k})$$ $$\boxed{\mathbf{F}_4 = -2\mathbf{i} - \mathbf{j} - 4\mathbf{k}}$$

b) The moment of the couple.

Step 3: Calculate the moment of each force about the origin using the formula $\mathbf{M} = \mathbf{r} \times \mathbf{F}$.

$$\mathbf{M}_1 = \mathbf{r}_1 \times \mathbf{F}_1 = (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + \mathbf{j} + 3\mathbf{k})$$ $$$$