If |vec{x}| = |vec{y}| = |vec{x} + vec{y}| = | vedclass

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(B) Given that $|\vec{x}| = 1$,$|\vec{y}| = 1$,and $|\vec{x} + \vec{y}| = 1$. Squaring the equation $|\vec{x} + \vec{y}| = 1$,we get $|\vec{x} + \vec{

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$\sqrt{3}$

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(B) Given that $|\vec{x}| = 1$,$|\vec{y}| = 1$,and $|\vec{x} + \vec{y}| = 1$. Squaring the equation $|\vec{x} + \vec{y}| = 1$,we get $|\vec{x} + \vec{y}|^2 = 1^2$. Using the property $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{x} \cdot \vec{y})$,we have $1^2 + 1^2 + 2(\vec{x} \cdot \vec{y}) = 1$. $2 + 2(\vec{x} \cdot \vec{y}) = 1$,which implies $2(\vec{x} \cdot \vec{y}) = -1$,so $\vec{x} \cdot \vec{y} = -\frac{1}{2}$. Now,we need to find $|\vec{x} - \vec{y}|$. $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$. Substituting the known values: $|\vec{x} - \vec{y}|^2 = 1^2 + 1^2 - 2(-\frac{1}{2}) = 1 + 1 + 1 = 3$. Therefore,$|\vec{x} - \vec{y}| = \sqrt{3}$.

If $|\vec{x}| = |\vec{y}| = |\vec{x} + \vec{y}| = 1$,then $|\vec{x} - \vec{y}| = $ . . . . . . .

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