hat{a}, hat{b},and hat{c} are three unit vect | vedclass

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(A) Given the vector triple product identity: $\hat{a} 	imes (\hat{b} 	imes \hat{c}) = (\hat{a} \cdot \hat{c}) \hat{b} - (\hat{a} \cdot \hat{b}) \ha

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$\frac{5 \pi}{6}$

Vedclass · Answer

(A) Given the vector triple product identity: $\hat{a} 	imes (\hat{b} 	imes \hat{c}) = (\hat{a} \cdot \hat{c}) \hat{b} - (\hat{a} \cdot \hat{b}) \hat{c}$.Comparing this with the given equation: $(\hat{a} \cdot \hat{c}) \hat{b} - (\hat{a} \cdot \hat{b}) \hat{c} = \frac{\sqrt{3}}{2} \hat{b} + \frac{\sqrt{3}}{2} \hat{c}$.Since $\hat{b}$ and $\hat{c}$ are not parallel,we can equate the coefficients of $\hat{b}$ and $\hat{c}$:$\hat{a} \cdot \hat{c} = \frac{\sqrt{3}}{2}$ and $-(\hat{a} \cdot \hat{b}) = \frac{\sqrt{3}}{2} \Rightarrow \hat{a} \cdot \hat{b} = -\frac{\sqrt{3}}{2}$.Let $	heta$ be the angle between $\hat{a}$ and $\hat{b}$. Since $\hat{a}$ and $\hat{b}$ are unit vectors,$\hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos 	heta = \cos 	heta$.Therefore,$\cos 	heta = -\frac{\sqrt{3}}{2}$.Since $\cos 	heta = -\frac{\sqrt{3}}{2}$,we have $	heta = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.

$\hat{a}, \hat{b}$,and $\hat{c}$ are three unit vectors such that $\hat{a} \times(\hat{b} \times \hat{c})=\frac{\sqrt{3}}{2}(\hat{b}+\hat{c})$. If $\hat{b}$ is not parallel to $\hat{c}$,then the angle between $\hat{a}$ and $\hat{b}$ is

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