For any two vectors overrightarrow{A} and ove | vedclass

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(D) Given the condition: $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A} 	imes \overrightarrow{B}|$.Using the definitions of dot a

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

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(D) Given the condition: $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A} 	imes \overrightarrow{B}|$.Using the definitions of dot and cross products: $AB \cos 	heta = AB \sin 	heta$.Dividing both sides by $AB \cos 	heta$ (assuming $A, B 
eq 0$),we get $	an 	heta = 1$,which implies $	heta = 45^\circ$.The magnitude of the resultant vector $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$ is given by $|\overrightarrow{C}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Substituting $	heta = 45^\circ$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$:$|\overrightarrow{C}| = \sqrt{A^2 + B^2 + 2AB \left(\frac{1}{\sqrt{2}}ight)}$.$|\overrightarrow{C}| = \sqrt{A^2 + B^2 + \sqrt{2} AB}$.

For any two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$,if $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A} \times \overrightarrow{B}|$,the magnitude of $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$ is equal to

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