Let's solve both parts (a) and (b) step by step. --- ### (a) \(\mathbf{u} = (-2, 1)\), \(\mathbf{v} = (3, 1)\) #### **Step 1: Find the inner product \(\mathbf{u} \cdot \mathbf{v}\) and the norms** - **Inner product:** \[ \mathbf{u} \cdot \mathbf{v} = (-2)(3) + (1)(1) = -6 + 1 = -5 \] - **Norms:** \[ \|\mathbf{u}\| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \] \[ \|\mathbf{v}\| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \] **Explanation:** We calculate the dot product by multiplying corresponding components and summing. The norm is the square root of the sum of the squares of the components. --- #### **Step 2: Find the cosine of the angle** \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|} = \frac{-5}{\sqrt{5}\sqrt{10}} = \frac{-5}{\sqrt{50}} = \frac{-5}{5\sqrt{2}} = -\frac{1}{\sqrt{2}} \] **Explanation:** The cosine of the angle is given by the ratio of the dot product to the product of the norms. **Final answer (a):** \[ \boxed{\cos \theta = -\frac{1}{\sqrt{2}}} \] --- ### (b) \(\mathbf{u} = (1, 2, 3)\), \(\mathbf{v} = (4, 4, -4)\) #### **Step 1: Find the inner product \(\mathbf{u} \cdot \mathbf{v}\) and the norms** - **Inner product:** \[ \mathbf{u} \cdot \mathbf{v} = (1)(4) + (2)(4) + (3)(-4) = 4 + 8 - 12 = 0 \] - **Norms:** \[ \|\mathbf{u}\| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] \[ \|\mathbf{v}\| = \sqrt{4^2 + 4^2 + (-4)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3} \] **Explanation:** Calculate the dot product and the magnitudes as in part (a). --- #### **Step 2: Find the cosine of the angle** \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|} = \frac{0}{\sqrt{14} \cdot 4\sqrt{3}} = 0 \] **Explanation:** If the dot product is zero, the vectors are orthogonal, so the cosine of the angle is zero. **Final answer (b):** \[ \boxed{\cos \theta = 0} \]

**Final Answers:** **(a)** \(\boxed{\cos \theta = -\frac{1}{\sqrt{2}}}\) **(b)** \(\boxed{\cos \theta = 0}\)