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

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(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given that their sum is a unit vector,$|\vec{a} + \vec{b}|

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$\sqrt{3}$ units

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(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given that their sum is a unit vector,$|\vec{a} + \vec{b}| = 1$. 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 $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$. This simplifies to $2 + 2(\vec{a} \cdot \vec{b}) = 1$,which gives $2(\vec{a} \cdot \vec{b}) = -1$. 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,$|\vec{a} - \vec{b}|^2 = 1 + 1 - (-1) = 1 + 1 + 1 = 3$. Therefore,$|\vec{a} - \vec{b}| = \sqrt{3}$ units.

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

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