1. **Problem statement:** We have two vectors \( \vec{a} \) and \( \vec{b} \) each with magnitude 8 units. \( \vec{a} \) makes a 45° angle with the positive x-axis, and \( \vec{b} \) is directed along the negative x-axis. We need to find: a. The magnitude and direction of \( \vec{a} + \vec{b} \) b. The magnitude and direction of \( \vec{a} - \vec{b} \) 2. **Formulas and rules:** - Vector components: For a vector with magnitude \( r \) and angle \( \theta \), components are \( r\cos\theta \) (x-component) and \( r\sin\theta \) (y-component). - Magnitude of vector \( \vec{v} = (v_x, v_y) \) is \( \sqrt{v_x^2 + v_y^2} \). - Direction (angle) of vector \( \vec{v} \) is \( \tan^{-1}(\frac{v_y}{v_x}) \). 3. **Find components of \( \vec{a} \):** \( a_x = 8 \cos 45^\circ = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \) \( a_y = 8 \sin 45^\circ = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \) 4. **Find components of \( \vec{b} \):** Since \( \vec{b} \) is along negative x-axis with magnitude 8: \( b_x = -8 \), \( b_y = 0 \) 5. **Calculate \( \vec{a} + \vec{b} \):** \( (a_x + b_x, a_y + b_y) = (4\sqrt{2} - 8, 4\sqrt{2} + 0) = (4\sqrt{2} - 8, 4\sqrt{2}) \) 6. **Magnitude of \( \vec{a} + \vec{b} \):** $$\sqrt{(4\sqrt{2} - 8)^2 + (4\sqrt{2})^2}$$ Simplify: \( (4\sqrt{2} - 8)^2 = (4\sqrt{2})^2 - 2 \times 4\sqrt{2} \times 8 + 8^2 = 32 - 64\sqrt{2} + 64 \) \( (4\sqrt{2})^2 = 32 \) Sum: \( 32 - 64\sqrt{2} + 64 + 32 = 128 - 64\sqrt{2} \) So magnitude: $$\sqrt{128 - 64\sqrt{2}}$$ 7. **Direction of \( \vec{a} + \vec{b} \):** $$\theta = \tan^{-1}\left( \frac{4\sqrt{2}}{4\sqrt{2} - 8} \right)$$ 8. **Calculate \( \vec{a} - \vec{b} \):** \( (a_x - b_x, a_y - b_y) = (4\sqrt{2} - (-8), 4\sqrt{2} - 0) = (4\sqrt{2} + 8, 4\sqrt{2}) \) 9. **Magnitude of \( \vec{a} - \vec{b} \):** $$\sqrt{(4\sqrt{2} + 8)^2 + (4\sqrt{2})^2}$$ Simplify: \( (4\sqrt{2} + 8)^2 = 32 + 64\sqrt{2} + 64 = 96 + 64\sqrt{2} \) \( (4\sqrt{2})^2 = 32 \) Sum: \( 96 + 64\sqrt{2} + 32 = 128 + 64\sqrt{2} \) So magnitude: $$\sqrt{128 + 64\sqrt{2}}$$ 10. **Direction of \( \vec{a} - \vec{b} \):** $$\theta = \tan^{-1}\left( \frac{4\sqrt{2}}{4\sqrt{2} + 8} \right)$$ **Final answers:** - \( |\vec{a} + \vec{b}| = \sqrt{128 - 64\sqrt{2}} \), direction \( = \tan^{-1}\left( \frac{4\sqrt{2}}{4\sqrt{2} - 8} \right) \) - \( |\vec{a} - \vec{b}| = \sqrt{128 + 64\sqrt{2}} \), direction \( = \tan^{-1}\left( \frac{4\sqrt{2}}{4\sqrt{2} + 8} \right) \)