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(B) Given that $|\overrightarrow A 	imes \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$.Using the definitions of cross product a

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given that $|\overrightarrow A 	imes \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$.Using the definitions of cross product and dot product:$|\overrightarrow A | |\overrightarrow B | \sin 	heta = |\overrightarrow A | |\overrightarrow B | \cos 	heta$.Dividing both sides by $|\overrightarrow A | |\overrightarrow B | \cos 	heta$ (assuming $\overrightarrow A, \overrightarrow B 
eq 0$):$\frac{\sin 	heta}{\cos 	heta} = 1$.$	an 	heta = 1$.Since $	an 45^\circ = 1$,the angle $	heta = 45^\circ$.

If $|\overrightarrow A \times \overrightarrow B | = |\overrightarrow A \cdot \overrightarrow B |$,then the angle between $\overrightarrow A$ and $\overrightarrow B$ will be ........ $^o$.

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