If vec{a} and vec{b} are two unit vectors inc | vedclass

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.The angle between them is $	heta = 60^{\circ}$.For the

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$| \vec{a} + \vec{b} | > 1$

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.The angle between them is $	heta = 60^{\circ}$.For the sum of vectors,the magnitude is given by $|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}| |\vec{b}| \cos 	heta}$.Substituting the values: $|\vec{a} + \vec{b}| = \sqrt{1^2 + 1^2 + 2(1)(1) \cos 60^{\circ}} = \sqrt{1 + 1 + 2(0.5)} = \sqrt{3} \approx 1.732$.Since $1.732 > 1$,it follows that $|\vec{a} + \vec{b}| > 1$.For the difference of vectors,the magnitude is given by $|\vec{a} - \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}| |\vec{b}| \cos 	heta}$.Substituting the values: $|\vec{a} - \vec{b}| = \sqrt{1^2 + 1^2 - 2(1)(1) \cos 60^{\circ}} = \sqrt{1 + 1 - 2(0.5)} = \sqrt{1} = 1$.Thus,the correct statement is $|\vec{a} + \vec{b}| > 1$.

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle of $60^{\circ}$ to each other,then:

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