If the sum of two unit vectors is a unit vect | vedclass

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(B) Let the two unit vectors be $\vec{a}$ and $\vec{b}$,such that $|\vec{a}| = 1$ and $|\vec{b}| = 1$.Given that their sum is a unit vector,we have $|

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$\sqrt{3}$

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(B) Let the two unit vectors be $\vec{a}$ and $\vec{b}$,such that $|\vec{a}| = 1$ and $|\vec{b}| = 1$.Given that their sum is a unit vector,we have $|\vec{a} + \vec{b}| = 1$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 1^2$.Using the identity $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$,we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$,which implies $2(\vec{a} \cdot \vec{b}) = -1$,so $\vec{a} \cdot \vec{b} = -\frac{1}{2}$.Now,we need to find the magnitude of their difference,$|\vec{a} - \vec{b}|$.We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.Substituting the known values,$|\vec{a} - \vec{b}|^2 = 1 + 1 - 2(-\frac{1}{2}) = 1 + 1 + 1 = 3$.Therefore,$|\vec{a} - \vec{b}| = \sqrt{3}$.

If the sum of two unit vectors is a unit vector,then the magnitude of their difference is

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