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

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(B) Given,$|\vec{a} \cdot \vec{b}|=|\vec{a} 	imes \vec{b}|$. Since $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos 	heta|$ and $|\vec{a} 	imes

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$\sec ^{-1}(-\sqrt{2})$

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(B) Given,$|\vec{a} \cdot \vec{b}|=|\vec{a} 	imes \vec{b}|$. Since $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos 	heta|$ and $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| |\sin 	heta|$,where $	heta$ is the angle between vectors $\vec{a}$ and $\vec{b}$,we have: $|\vec{a}||\vec{b}| |\cos 	heta| = |\vec{a}||\vec{b}| |\sin 	heta|$ $|\cos 	heta| = |\sin 	heta|$ $|	an 	heta| = 1$ This implies $	heta = \frac{\pi}{4}$ or $	heta = \frac{3\pi}{4}$. Given that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} Since $\cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} Note that $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$ because $\sec(\frac{3\pi}{4}) = -\sqrt{2}$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} < 0$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

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