vec A and vec B are two vectors and theta is | vedclass

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(A) Given the equation: $|\vec A 	imes \vec B| = \sqrt{3}(\vec A \cdot \vec B)$.We know that the magnitude of the cross product is $|\vec A 	imes \v

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Given the equation: $|\vec A 	imes \vec B| = \sqrt{3}(\vec A \cdot \vec B)$.We know that the magnitude of the cross product is $|\vec A 	imes \vec B| = AB \sin 	heta$ and the dot product is $\vec A \cdot \vec B = AB \cos 	heta$.Substituting these into the given equation:$AB \sin 	heta = \sqrt{3} AB \cos 	heta$.Dividing both sides by $AB \cos 	heta$ (assuming $A, B 
eq 0$ and $\cos 	heta 
eq 0$):$\frac{\sin 	heta}{\cos 	heta} = \sqrt{3}$.$	an 	heta = \sqrt{3}$.Therefore,$	heta = 	an^{-1}(\sqrt{3}) = 60^\circ$.

$\vec A$ and $\vec B$ are two vectors and $\theta$ is the angle between them. If $|\vec A \times \vec B| = \sqrt{3}(\vec A \cdot \vec B)$,the value of $\theta$ is ......... $^\circ$.

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