Given: \[ |\vec{a}| = \alpha, \quad |\vec{b}| = \alpha, \quad \text{and} \quad |\vec{a} + \vec{b}| = \alpha \] Step 1: Using Magnitude Formula \[ |\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab \cos\theta} \] Substituting the given values: \[ \alpha = \sqrt{\alpha^2 + \alpha^2 + 2(\alpha)(\alpha) \cos\theta} \] \[ \alpha^2 = 2\alpha^2 + 2\alpha^2 \cos\theta \] \[ \alpha^2 - 2\alpha^2 = 2\alpha^2 \cos\theta \] \[ - \alpha^2 = 2\alpha^2 \cos\theta \] \[ \cos\theta = -\frac{1}{2} \] Step 2: Finding \( |\vec{a} - \vec{b}| \) \[ |\vec{a} - \vec{b}| = \sqrt{a^2 + b^2 - 2ab \cos\theta} \] Substituting values: \[ |\vec{a} - \vec{b}| = \sqrt{\alpha^2 + \alpha^2 - 2(\alpha)(\alpha) \cos\theta} \] Since \( \cos\theta = -\frac{1}{2} \), we get: \[ |\vec{a} - \vec{b}| = \sqrt{2\alpha^2 - 2\alpha^2\left(-\frac{1}{2}\right)} \] \[ |\vec{a} - \vec{b}| = \sqrt{2\alpha^2 + \alpha^2} \] \[ |\vec{a} - \vec{b}| = \sqrt{3\alpha^2} \] \[ |\vec{a} - \vec{b}| = \sqrt{3} \alpha \]