Question

A vector \(\sqrt{3}\hat{i} + \hat{j}\) rotates about its tail through an angle 30\(^\circ\) in clock wise direction then the new vector is

Hint:

Always check the magnitude first. The magnitude must remain 2. Options (c) and (d) have magnitudes 1 and can be eliminated immediately.

Answer 1

Step 1: Understanding the Question:
A vector is rotated in a 2D plane. Rotation changes the direction but leaves the magnitude of the vector unchanged.

Step 2: Key Formula or Approach:
1. Magnitude \(|\vec{A}| = \sqrt{x^2 + y^2}\).
2. Initial angle \(\theta = \tan^{-1}(y/x)\).

Step 3: Detailed Explanation:
Let the initial vector be \(\vec{A} = \sqrt{3}\hat{i} + \hat{j}\).
Magnitude: \(|\vec{A}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\).
Initial angle with positive x-axis: \(\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ\).
The vector is rotated 30\(^\circ\) clockwise.
New angle with x-axis = \(30^\circ - 30^\circ = 0^\circ\).
The new vector has magnitude 2 and is directed along the positive x-axis.
New Vector = \(2(\cos 0^\circ \hat{i} + \sin 0^\circ \hat{j}) = 2(1\hat{i} + 0\hat{j}) = 2\hat{i}\).

Step 4: Final Answer:
The new vector is \(2\hat{i}\).