Step 1: Understanding the Concept:
When a vector rotates, its magnitude remains unchanged, but its direction (angle with the axes) changes. We can solve this by finding the initial angle of the vector and adjusting it by the rotation angle. Step 2: Key Formula or Approach:
1. Magnitude $|\vec{A}| = \sqrt{x^2 + y^2}$.
2. Initial angle $\theta = \tan^{-1}(y/x)$.
3. New vector $\vec{A'} = |A|\cos(\theta')\hat{i} + |A|\sin(\theta')\hat{j}$. Step 3: Detailed Explanation:
1. Find Magnitude:
\[ |\vec{A}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 \]
2. Find Initial Angle:
\[ \tan \theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ \text{ (with the positive x-axis)} \]
3. Apply Rotation: The vector is at 30° and rotates 30° clockwise.
\[ \text{New angle } \theta' = 30^\circ - 30^\circ = 0^\circ \]
4. Find New Vector: The vector now lies entirely along the positive x-axis.
\[ \vec{A'} = 2 \cos(0^\circ)\hat{i} + 2 \sin(0^\circ)\hat{j} = 2(1)\hat{i} + 0 = 2\hat{i} \] Step 4: Final Answer
The new vector is $2\hat{i}$.