The modulus of the vector product of two vect | vedclass

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Question

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$ and the angle between them be $	heta$.According to the problem,the magnitude of the vector product

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$ and the angle between them be $	heta$.According to the problem,the magnitude of the vector product is $\frac{1}{\sqrt{3}}$ times the scalar product:$|\vec{A} 	imes \vec{B}| = \frac{1}{\sqrt{3}} (\vec{A} \cdot \vec{B})$$AB \sin 	heta = \frac{1}{\sqrt{3}} AB \cos 	heta$Dividing both sides by $AB \cos 	heta$ (assuming $A, B 
eq 0$ and $	heta 
eq 90^{\circ}$):$	an 	heta = \frac{1}{\sqrt{3}}$Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.Converting degrees to radians: $	heta = 30^{\circ} 	imes \frac{\pi}{180^{\circ}} = \frac{\pi}{6}$.

The modulus of the vector product of two vectors is $\frac{1}{\sqrt{3}}$ times their scalar product. The angle between the vectors is

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