If |vec A times vec B| = sqrt 3 vec A cdot ve | vedclass

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(D) Given: $|\vec A 	imes \vec B| = \sqrt 3 (\vec A \cdot \vec B)$Using the definitions of cross product and dot product: $AB \sin 	heta = \sqrt 3 A

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$({A^2} + {B^2} + AB)^{1/2}$

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(D) Given: $|\vec A 	imes \vec B| = \sqrt 3 (\vec A \cdot \vec B)$Using the definitions of cross product and dot product: $AB \sin 	heta = \sqrt 3 AB \cos 	heta$Dividing both sides by $AB \cos 	heta$ (assuming $A, B 
eq 0$): $	an 	heta = \sqrt 3$Thus,$	heta = 60^\circ$The magnitude of the resultant vector $|\vec A + \vec B|$ is given by: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$Substituting $	heta = 60^\circ$: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos 60^\circ}$Since $\cos 60^\circ = 1/2$: $|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB(1/2)}$Therefore,$|\vec A + \vec B| = (A^2 + B^2 + AB)^{1/2}$

If $|\vec A \times \vec B| = \sqrt 3 \vec A \cdot \vec B,$ then the value of $|\vec A + \vec B|$ is

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