VECTORS are \(\rm \vec {a}\) and \(\rm \vec {B}\)Magnitude of sum of \(\rm \vec {a}\) and \(\rm \vec {b}\):\(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ}\)Magnitude of difference of \(\rm \vec {a}\) and \(\rm \vec {b}\):\(\rm \left|\vec a-\vec b\right| = \sqrt{a^2+b^2-2ab\cosθ}\)where a, b are magnitude of vectors \(\rm \vec a \text{ and } \vec b\); and θ is angle between them. Calculation:Given:\(\rm \left|\vec a\right|=\alpha\text{, }\left|\vec b\right|=\alpha\text{ and }\left|\vec a+\vec b\right|=\alpha\)As we know,\(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ}\)⇒ \(\rm \alpha = \sqrt{\alpha^2+\alpha^2+2(\alpha)(\alpha)\cosθ}\)⇒ \(\rm \alpha^2 = 2\alpha^2+2\alpha^2\cosθ\)⇒ \(\rm -1=2\cosθ\)⇒ \(\boldsymbol{\rm \cosθ=-\frac{1}{2}}\)Now, \(\rm \left|\vec a-\vec b\right| = \sqrt{a^2+b^2-2ab\cosθ}\)⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{\alpha^2+\alpha^2-2(\alpha)(\alpha)\cosθ}\)\(∵ \COS θ = -\frac{1}{2}\)⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{2\alpha^2-2\alpha^2 (\frac{-1}{2})}\)⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{2\alpha^2+\alpha^2}\)⇒ \(\boldsymbol{\rm \left|\vec a-\vec b\right| = \sqrt{3}\alpha}\)