Question

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 :

• A. 2
• B. \(\sqrt3\)
• C. \(\frac{1}{\sqrt3}\)
• D. 6

Answer 1

The Correct Option is A

To determine the magnitude of the x-component of the vector given the magnitude of the y-component and the angle it makes with the y-axis, we can use trigonometric concepts.

1. The vector makes an angle of \(30^\circ\) with the y-axis. Hence, the angle it makes with the x-axis is \(90^\circ - 30^\circ = 60^\circ\).
2. We are given that the magnitude of the y-component of the vector is \(2\sqrt{3}\).
3. From trigonometry, we know:
   • The y-component of the vector, \(V_y = V \cdot \cos(\theta)\)
   • The x-component of the vector, \(V_x = V \cdot \sin(\theta)\)
4. Let the magnitude of the vector be \(V\). Then, using the y-component formula:
   \(V \cdot \cos(30^\circ) = 2\sqrt{3}\)
5. Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we can write:
   \(V \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\)
6. Solving for \(V\), we get:
   \(V = \frac{2\sqrt{3} \times 2}{\sqrt{3}} = 4\)
7. Now, using the x-component formula:
   \(V_x = V \cdot \sin(30^\circ)\)
8. Since \(\sin(30^\circ) = \frac{1}{2}\), the x-component becomes:
   \(V_x = 4 \cdot \frac{1}{2} = 2\)

Thus, the magnitude of the x-component of the vector is 2, which matches the given correct answer.