Vector vec{A} of magnitude 5 sqrt{3} units,an | vedclass

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(C) The magnitude of the vector product (cross product) of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $|\vec{A} 	imes \vec{B}| = |\

Vedclass · Accepted Answer

$25 \sqrt{3}$ units

Vedclass · Answer

(C) The magnitude of the vector product (cross product) of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $|\vec{A} 	imes \vec{B}| = |\vec{A}| |\vec{B}| \sin 	heta$ Given: $|\vec{A}| = 5 \sqrt{3}$ units $|\vec{B}| = 10$ units $	heta = 30^{\circ}$ Substituting these values into the formula: $|\vec{A} 	imes \vec{B}| = (5 \sqrt{3}) 	imes (10) 	imes \sin 30^{\circ}$ $|\vec{A} 	imes \vec{B}| = 50 \sqrt{3} 	imes \frac{1}{2}$ $|\vec{A} 	imes \vec{B}| = 25 \sqrt{3}$ units Therefore,the correct option is $C$.

Vector $\vec{A}$ of magnitude $5 \sqrt{3}$ units,another vector $\vec{B}$ of magnitude $10$ units are inclined to each other at an angle of $30^{\circ}$. The magnitude of the vector product of the two vectors is $\left[\sin 30^{\circ}=\frac{1}{2}\right]$

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