GATE PH 2007 • Q24 MCQ • 1 Mark

Section 1: Mathematical Physics

Linear Algebra & Vector Calculus

Subtopic: Tensors & Coordinate Transformations

Question Statement:

Consider a vector $\vec{p} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ in the coordinate system $(\hat{i}, \hat{j}, \hat{k})$. The axes are rotated anti-clockwise about the $Y$-axis by an angle of $60^\circ$. The vector $\vec{p}$ in the rotated coordinate system $(\hat{i}', \hat{j}', \hat{k}')$ is

Options:

  • (A) $(1 - \sqrt{3})\hat{i}' + 3\hat{j}' + (1 + \sqrt{3})\hat{k}'$
  • (B) $(1 + \sqrt{3})\hat{i}' + 3\hat{j}' + (1 - \sqrt{3})\hat{k}'$
  • (C) $(1 - \sqrt{3})\hat{i}' + (3 + \sqrt{3})\hat{j}' + 2\hat{k}'$
  • (D) $(1 - \sqrt{3})\hat{i}' + (3 - \sqrt{3})\hat{j}' + 2\hat{k}'$

Official Answer Key:

A

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2007 Q.24)** **1. Coordinate Transformation Principles**: - **Field & Concept**: *Section 1: Mathematical Physics* (`Tensors & Coordinate Transformations`) - When coordinate axes are rotated anti-clockwise by an angle $\theta = 60^\circ$ about the $Y$-axis, the transformation of components is given by the rotation matrix $R_y(\theta)$: $\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ **2. Detailed Step-by-Step Evaluation**: - With $\theta = 60^\circ$, $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $x' = x \cos 60^\circ - z \sin 60^\circ = 2\left(\frac{1}{2}\right) - 2\left(\frac{\sqrt{3}}{2}\right) = 1 - \sqrt{3}$ $y' = y = 3$ $z' = x \sin 60^\circ + z \cos 60^\circ = 2\left(\frac{\sqrt{3}}{2}\right) + 2\left(\frac{1}{2}\right) = \sqrt{3} + 1 = 1 + \sqrt{3}$ - Therefore, the vector in the rotated frame is: $\vec{p}' = (1 - \sqrt{3})\hat{i}' + 3\hat{j}' + (1 + \sqrt{3})\hat{k}'$ **3. Official Verification & Conclusion**: - **Official Master Answer**: **Option (A)**.

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