If hat{a}, hat{b}, and hat{c} are unit vector | vedclass

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(B) Let the angle between $\hat{a}$ and $\hat{c}$ be $	heta.$Given $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0}.$Rearranging the terms,we get $\hat

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$\frac{\pi}{3}$

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(B) Let the angle between $\hat{a}$ and $\hat{c}$ be $	heta.$Given $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0}.$Rearranging the terms,we get $\hat{a} + \hat{c} = \sqrt{3}\hat{b}.$Squaring both sides by taking the dot product with itself:$(\hat{a} + \hat{c}) \cdot (\hat{a} + \hat{c}) = (\sqrt{3}\hat{b}) \cdot (\sqrt{3}\hat{b}).$Expanding the dot product:$\hat{a} \cdot \hat{a} + \hat{a} \cdot \hat{c} + \hat{c} \cdot \hat{a} + \hat{c} \cdot \hat{c} = 3(\hat{b} \cdot \hat{b}).$Since $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,$\hat{a} \cdot \hat{a} = 1, \hat{b} \cdot \hat{b} = 1, \hat{c} \cdot \hat{c} = 1.$Also,$\hat{a} \cdot \hat{c} = |\hat{a}||\hat{c}| \cos 	heta = \cos 	heta.$Substituting these values:$1 + 2\cos 	heta + 1 = 3(1).$$2 + 2\cos 	heta = 3.$$2\cos 	heta = 1.$$\cos 	heta = \frac{1}{2}.$Therefore,$	heta = \frac{\pi}{3}.$

If $\hat{a}, \hat{b},$ and $\hat{c}$ are unit vectors satisfying $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0},$ then the angle between the vectors $\hat{a}$ and $\hat{c}$ is

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