1. **Problem statement:** (i) Given vectors \(\vec{a}, \vec{b}, \vec{c}\) such that \(\vec{a} + \vec{b} + \vec{c} = 0\) and magnitudes \(|\vec{a}|=2\), \(|\vec{b}|=3\), \(|\vec{c}|=4\), find the angle between \(\vec{a}\) and \(\vec{b}\). (ii) Given \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\), find the angle between \(\vec{a}\) and \(\vec{b}\). 2. **Formula and rules:** - Dot product formula: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). - Magnitude squared: \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\). - For vectors \(\vec{u}, \vec{v}\), \(|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + 2\vec{u} \cdot \vec{v} + |\vec{v}|^2\). --- ### Part (i): 3. From \(\vec{a} + \vec{b} + \vec{c} = 0\), rearranged as \(\vec{c} = - (\vec{a} + \vec{b})\). 4. Taking magnitudes squared: $$|\vec{c}|^2 = |-(\vec{a} + \vec{b})|^2 = |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$$ 5. Substitute known magnitudes: $$4^2 = 2^2 + 2\vec{a} \cdot \vec{b} + 3^2$$ $$16 = 4 + 2\vec{a} \cdot \vec{b} + 9$$ 6. Simplify: $$16 = 13 + 2\vec{a} \cdot \vec{b} \implies 2\vec{a} \cdot \vec{b} = 3 \implies \vec{a} \cdot \vec{b} = \frac{3}{2}$$ 7. Using dot product formula: $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 2 \times 3 \times \cos\theta = 6\cos\theta$$ 8. Equate and solve for \(\cos\theta\): $$6\cos\theta = \frac{3}{2} \implies \cos\theta = \frac{3}{2} \times \frac{1}{6} = \frac{1}{4}$$ 9. Therefore, the angle between \(\vec{a}\) and \(\vec{b}\) is: $$\theta = \cos^{-1}\left(\frac{1}{4}\right)$$ --- ### Part (ii): 10. Given \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\), square both sides: $$|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$$ 11. Expand using dot product: $$|\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{b}|^2$$ 12. Simplify: $$2\vec{a} \cdot \vec{b} = -2\vec{a} \cdot \vec{b} \implies 4\vec{a} \cdot \vec{b} = 0 \implies \vec{a} \cdot \vec{b} = 0$$ 13. Since \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0\), and assuming neither vector is zero, we get: $$\cos\theta = 0 \implies \theta = 90^\circ$$ --- **Final answers:** - (i) Angle between \(\vec{a}\) and \(\vec{b}\) is \(\cos^{-1}\left(\frac{1}{4}\right)\). - (ii) Angle between \(\vec{a}\) and \(\vec{b}\) is \(90^\circ\).