Question

Let \(\hat{i}, \hat{j}, \hat{k}\) be the unit vectors in the x, y, and z directions, respectively. If the vector \(\hat{i} + \hat{j}\) is rotated about positive \(\hat{k}\) by \(135^\circ\), one gets

• A. \(-\hat{i}\)
• B. \(-\hat{j}\)
• C. \(\frac{-1}{\sqrt{2}}\hat{j}\)
• D. \(-\sqrt{2}\hat{i}\)

Hint:

To rotate a vector about the z-axis by an angle \(\theta\), use the rotation matrix and apply it to the vector components.

Answer 1

The Correct Option is D

Step 1: Rotation matrix for a vector.
The rotation of a vector \(\vec{V} = \hat{i} + \hat{j}\) by an angle \(\theta\) about the z-axis can be described using the rotation matrix: \[ \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}. \] Step 2: Apply the rotation to the vector.
Here, \(\theta = 135^\circ\), so the rotation matrix becomes: \[ \begin{bmatrix} \cos 135^\circ & -\sin 135^\circ \\ \sin 135^\circ & \cos 135^\circ \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}. \] Multiplying the rotation matrix by the vector \(\hat{i} + \hat{j}\), we get: \[ \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 1
1 \end{bmatrix} = \begin{bmatrix} -\sqrt{2} \\ 0 \end{bmatrix}. \] Thus, the rotated vector is \(-\sqrt{2} \hat{i}\), which corresponds to option (D).