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3-1.Vectors
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A vector in $x-y$ plane makes an angle of $30^{\circ}$ with $y$-axis The magnitude of $y$-component of vector is $2 \sqrt{3}$. The magnitude of $x$-component of the vector will be
A
$\frac{1}{\sqrt{3}}$
B
$6$
C
$\sqrt{3}$
D
$2$
(JEE MAIN-2023)
Solution
$A _y= A \cos 30^{\circ}=2 \sqrt{3}$
$\Rightarrow A \frac{\sqrt{3}}{2}=2 \sqrt{3}$
$\Rightarrow A =4$
Now $A_x=A \sin 30^{\circ}=4 \times \frac{1}{2}=2$
Standard 11
Physics
