For non-zero vectors vec{a}, vec{b}, vec{c},t | vedclass

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(C) Given $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$. Since $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a} 	imes

Vedclass · Accepted Answer

$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

Vedclass · Answer

(C) Given $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$. Since $|(\vec{a} 	imes \vec{b}) \cdot \vec{c}| = |\vec{a} 	imes \vec{b}| |\vec{c}| |\cos 	heta|$,where $	heta$ is the angle between $(\vec{a} 	imes \vec{b})$ and $\vec{c}$,we have $|\vec{a} 	imes \vec{b}| |\vec{c}| |\cos 	heta| = |\vec{a}||\vec{b}||\vec{c}|$. Substituting $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin \phi$,where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$,we get $|\vec{a}||\vec{b}| \sin \phi |\vec{c}| |\cos 	heta| = |\vec{a}||\vec{b}||\vec{c}|$. This implies $\sin \phi = 1$ and $|\cos 	heta| = 1$. Thus,$\phi = 90^{\circ}$ (meaning $\vec{a} \perp \vec{b}$) and $	heta = 0^{\circ}$ or $180^{\circ}$ (meaning $\vec{c}$ is parallel to the normal of the plane containing $\vec{a}$ and $\vec{b}$). Since $\vec{c}$ is parallel to $\vec{a} 	imes \vec{b}$,it follows that $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \vec{b}$. Therefore,$\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,and $\vec{c} \cdot \vec{a} = 0$.

For non-zero vectors $\vec{a}, \vec{b}, \vec{c}$,the condition $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$ holds if and only if:

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