1. **Problem:** Show that the vectors $\vec{A} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}$, $\vec{B} = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$, and $\vec{C} = 3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}$ form the sides of a triangle. 2. **Formula and rule:** Three vectors form the sides of a triangle if the sum of the three vectors is zero, i.e., $$\vec{A} + \vec{B} + \vec{C} = \mathbf{0}$$ This means the vectors can be arranged head-to-tail to form a closed triangle. 3. **Intermediate work:** Calculate the sum: $$\vec{A} + \vec{B} + \vec{C} = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) + (\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}) + (3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k})$$ Combine like terms: $$= (2 + 1 + 3)\mathbf{i} + (-1 - 3 - 4)\mathbf{j} + (1 - 5 - 4)\mathbf{k}$$ $$= 6\mathbf{i} - 8\mathbf{j} - 8\mathbf{k}$$ 4. **Check if sum is zero:** $$6\mathbf{i} - 8\mathbf{j} - 8\mathbf{k} \neq \mathbf{0}$$ Since the sum is not zero, these three vectors as given do not directly form the sides of a triangle when placed head-to-tail in the given order. 5. **Alternative check:** To form a triangle, one vector must be the sum of the other two with opposite direction, i.e., $$\vec{C} = - (\vec{A} + \vec{B})$$ Calculate $-(\vec{A} + \vec{B})$: $$\vec{A} + \vec{B} = (2 + 1)\mathbf{i} + (-1 - 3)\mathbf{j} + (1 - 5)\mathbf{k} = 3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}$$ Then, $$-(\vec{A} + \vec{B}) = -3\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}$$ Compare with $\vec{C} = 3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}$, which is the negative of this vector. 6. **Conclusion:** Since $\vec{C} = \vec{A} + \vec{B}$ is false but $\vec{C} = -(\vec{A} + \vec{B})$ is false as well, the vectors do not form a triangle by simple addition. However, if we consider the vectors as sides of a triangle, the condition is that the sum of the vectors taken in order around the triangle is zero: $$\vec{A} + \vec{B} + \vec{C} = \mathbf{0}$$ Since the sum is not zero, the vectors as given do not form the sides of a triangle. **Final answer:** The vectors $2\mathbf{i} - \mathbf{j} + \mathbf{k}$, $\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$, and $3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}$ do not form the sides of a triangle because their vector sum is not zero.