A vector of magnitude 14 lies in the xy- plan | vedclass

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(A) Let the vector be $\vec{A}$ with magnitude $|\vec{A}| = 14$. Given that the vector makes an angle $	heta = 60^\circ$ with the $x-$ axis. The comp

Vedclass · Accepted Answer

$7, 7\sqrt{3}$

Vedclass · Answer

(A) Let the vector be $\vec{A}$ with magnitude $|\vec{A}| = 14$. Given that the vector makes an angle $	heta = 60^\circ$ with the $x-$ axis. The component of the vector along the $x-$ axis is given by $A_x = |\vec{A}| \cos 	heta = 14 \cos 60^\circ$. Since $\cos 60^\circ = 1/2$,we have $A_x = 14 	imes (1/2) = 7$. The component of the vector along the $y-$ axis is given by $A_y = |\vec{A}| \sin 	heta = 14 \sin 60^\circ$. Since $\sin 60^\circ = \sqrt{3}/2$,we have $A_y = 14 	imes (\sqrt{3}/2) = 7\sqrt{3}$. Thus,the components are $7$ and $7\sqrt{3}$.

$A$ vector of magnitude $14$ lies in the $xy-$ plane and makes an angle of $60^\circ$ with the $x-$ axis. The components of the vector in the direction of the $x-$ axis and $y-$ axis are:

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