The resultant of two vectors having magnitude | vedclass

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(D) Let the magnitudes of the two vectors be $A = 2$ and $B = 3$. The resultant $R$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta} = 1$.Squaring

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(D) Let the magnitudes of the two vectors be $A = 2$ and $B = 3$. The resultant $R$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta} = 1$.Squaring both sides,we get $2^2 + 3^2 + 2(2)(3) \cos 	heta = 1^2$.$4 + 9 + 12 \cos 	heta = 1$.$13 + 12 \cos 	heta = 1$.$12 \cos 	heta = -12$,which gives $\cos 	heta = -1$.Therefore,$	heta = 180^\circ$.The cross product of two vectors is given by $|\vec{A} 	imes \vec{B}| = AB \sin 	heta$.Since $	heta = 180^\circ$,$\sin 180^\circ = 0$.Thus,$|\vec{A} 	imes \vec{B}| = 2 	imes 3 	imes 0 = 0$.

The resultant of two vectors having magnitudes $2$ and $3$ is $1$. What is their cross product?

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