(D) Let the vector be $\vec{A}$. The $y$-component is given by $A_y = A \cos 30^{\circ}$.Given $A_y = 2 \sqrt{3}$,we have:$A \cos 30^{\circ} = 2 \sqrt

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(D) Let the vector be $\vec{A}$. The $y$-component is given by $A_y = A \cos 30^{\circ}$.Given $A_y = 2 \sqrt{3}$,we have:$A \cos 30^{\circ} = 2 \sqrt{3}$$A \left( \frac{\sqrt{3}}{2} \right) = 2 \sqrt{3}$$A = 4$Now,the $x$-component is given by $A_x = A \sin 30^{\circ}$.$A_x = 4 \times \sin 30^{\circ} = 4 \times \frac{1}{2} = 2$.Thus,the magnitude of the $x$-component is $2$.

Class 11 Physics 3-1.Vectors Resolution of a Vector into Rectangular Components

Medium

$A$ vector in the $x-y$ plane makes an angle of $30^{\circ}$ with the $y$-axis. The magnitude of the $y$-component of the vector is $2 \sqrt{3}$. The magnitude of the $x$-component of the vector will be:

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A
$\frac{1}{\sqrt{3}}$
B
$6$
C
$\sqrt{3}$
D
$2$

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3-1.Vectors

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Resolution of a Vector into Rectangular Components

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$A$ vector in the $x-y$ plane makes an angle of $30^{\circ}$ with the $y$-axis. The magnitude of the $y$-component of the vector is $2 \sqrt{3}$. The magnitude of the $x$-component of the vector will be:

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