If bar{a} and bar{b} are two vectors such tha | vedclass

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(B) Given $|\bar{a}| = |\bar{b}| = \sqrt{6}$ and $\bar{a} \cdot \bar{b} = -1$. We know that $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(	heta)$

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$35$

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(B) Given $|\bar{a}| = |\bar{b}| = \sqrt{6}$ and $\bar{a} \cdot \bar{b} = -1$. We know that $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(	heta)$. Substituting the values: $-1 = (\sqrt{6})(\sqrt{6}) \cos(	heta) = 6 \cos(	heta)$. So, $\cos(	heta) = -\frac{1}{6}$. Since $\sin^2(	heta) = 1 - \cos^2(	heta)$, we have $\sin^2(	heta) = 1 - (-\frac{1}{6})^2 = 1 - \frac{1}{36} = \frac{35}{36}$. Thus, $\sin(	heta) = \frac{\sqrt{35}}{6}$. Now, $|\bar{a} 	imes \bar{b}| = |\bar{a}| |\bar{b}| \sin(	heta) = (\sqrt{6})(\sqrt{6}) \sin(	heta) = 6 \sin(	heta)$. Therefore, $|\bar{a} 	imes \bar{b}| \sin(	heta) = 6 \sin(	heta) \cdot \sin(	heta) = 6 \sin^2(	heta) = 6 	imes \frac{35}{36} = \frac{35}{6}$.

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}|=|\bar{b}|=\sqrt{6}$ and $\bar{a} \cdot \bar{b}=-1$, then find the value of $|\bar{a} \times \bar{b}| \sin(\theta)$, where $\theta$ is the angle between $\bar{a}$ and $\bar{b}$.

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