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

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.