If |hat{a} cdot hat{b}| = frac{1}{2},then |ha | vedclass

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(C) Given that $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$.Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$.Thus,$|\cos

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$1$ or $\sqrt{3}$

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(C) Given that $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$.Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$.Thus,$|\cos 	heta| = \frac{1}{2}$,which implies $	heta = 60^\circ$ or $	heta = 120^\circ$.We know that $|\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b})$.Case $1$: If $\hat{a} \cdot \hat{b} = \frac{1}{2}$,then $|\hat{a} - \hat{b}|^2 = 1 + 1 - 2(\frac{1}{2}) = 1$,so $|\hat{a} - \hat{b}| = 1$.Case $2$: If $\hat{a} \cdot \hat{b} = -\frac{1}{2}$,then $|\hat{a} - \hat{b}|^2 = 1 + 1 - 2(-\frac{1}{2}) = 3$,so $|\hat{a} - \hat{b}| = \sqrt{3}$.Therefore,$|\hat{a} - \hat{b}|$ can be $1$ or $\sqrt{3}$.

If $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$,then $|\hat{a} - \hat{b}|$ may be $:-$

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