For any two vectors vec{A} and vec{B},if vec{ | vedclass

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(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross products,we have $AB \cos 	heta = A

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$\sqrt{A^{2} + B^{2} + \sqrt{2} AB}$

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(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross products,we have $AB \cos 	heta = AB \sin 	heta$.Dividing both sides by $AB$ (assuming $A, B 
eq 0$),we get $\cos 	heta = \sin 	heta$,which implies $	an 	heta = 1$.Thus,$	heta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 	heta}$.Substituting $	heta = 45^{\circ}$,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 45^{\circ}}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB 	imes \frac{1}{\sqrt{2}}}$.Simplifying,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + \sqrt{2} AB}$.

For any two vectors $\vec{A}$ and $\vec{B}$,if $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$,the magnitude of $(\vec{A} + \vec{B})$ is: $(\tan \frac{\pi}{4} = 1, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}})$

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