A vector in x-y plane makes an angle of 30º with y-axis. The magnitude of y-component of vector is \(2\sqrt3\). The magnitude of x -component of the vector will be :
- 2
- \(\sqrt3\)
- \(\frac{1}{\sqrt3}\)
- 6
The Correct Option is A
Solution and Explanation
Understanding the Problem
- A vector makes a 30° angle with the y-axis.
- The magnitude of the y-component (\(A_y\)) is \(2\sqrt{3}\).
- We need to find the magnitude of the x-component (\(A_x\)).
Solution
1. Relating \(A_y\) to the Vector's Magnitude (A):
Since the angle is with the y-axis, we use cosine:
\( A_y = A \cos(30^\circ) \)
Given \(A_y = 2\sqrt{3}\), we have:
\( 2\sqrt{3} = A \cos(30^\circ) \)
2. Solving for the Vector's Magnitude (A):
We know \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\).
Substituting:
\( 2\sqrt{3} = A \left( \frac{\sqrt{3}}{2} \right) \)
Solving for A:
\( A = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4 \)
3. Finding the x-Component (\(A_x\)):
Since the angle is with the y-axis, the x-component uses sine:
\( A_x = A \sin(30^\circ) \)
We found \(A = 4\), and \(\sin(30^\circ) = \frac{1}{2}\).
Substituting:
\( A_x = 4 \left( \frac{1}{2} \right) = 2 \)
Final Answer
The magnitude of the x-component of the vector is 2.
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