If vec{A} and vec{B} are two vectors satisfyi | vedclass

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(A) Given the relation $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross product: $AB \cos 	heta = AB \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the relation $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}|$.Using the definitions of dot and cross product: $AB \cos 	heta = AB \sin 	heta$.Dividing both sides by $AB$ (assuming $A, B 
eq 0$),we get $\cos 	heta = \sin 	heta$,which implies $	an 	heta = 1$,so $	heta = 45^{\circ}$.The magnitude of the vector difference $|\vec{A} - \vec{B}|$ is given by the formula $\sqrt{A^{2} + B^{2} - 2AB \cos 	heta}$.Substituting $	heta = 45^{\circ}$,we get $\sqrt{A^{2} + B^{2} - 2AB \cos 45^{\circ}}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $\sqrt{A^{2} + B^{2} - 2AB \cdot \frac{1}{\sqrt{2}}} = \sqrt{A^{2} + B^{2} - \sqrt{2}AB}$.

Column $-I$	Column $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$,then the value of $|\vec{A} - \vec{B}|$ will be:

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