Question:

Let $\Delta = \left| \begin{matrix} \sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \theta \\ -\sin \theta \sin \varphi & \sin \theta \cos \varphi & 0 \end{matrix} \right|$. Then

Updated On: Apr 15, 2025
  • △ is independent of θ
  • △ is independent of ϕ
  • △ is a constant
  • (d△/dθ)θ=π /2 =0
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The Correct Option is B, D

Solution and Explanation

We are given the determinant of a 3x3 matrix, and we need to analyze the properties of \( \Delta \).

Step 1: Determinant of the Matrix
The matrix \( A \) is: \[ A = \begin{pmatrix} \sin \theta \cos \varphi & \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \cos \varphi & \cos \theta \sin \varphi & -\sin \theta \\ -\sin \theta \sin \varphi & \sin \theta \cos \varphi & 0 \end{pmatrix} \] We need to compute the determinant of this matrix. 

Step 2: Expanding the Determinant
Expanding \( \Delta \) along the third row: \[ \Delta = (-\sin \theta \sin \varphi) \cdot \left| \begin{matrix} \sin \theta \sin \varphi & \cos \theta \\ \cos \theta \sin \varphi & -\sin \theta \end{matrix} \right| + \sin \theta \cos \varphi \cdot \left| \begin{matrix} \sin \theta \cos \varphi & \cos \theta \\ \cos \theta \cos \varphi & -\sin \theta \end{matrix} \right| \] After simplifying this determinant expansion, we find that \( \Delta \) becomes independent of \( \varphi \), and the derivative with respect to \( \theta \) evaluated at \( \theta = \frac{\pi}{2} \) is zero. 

Step 3: Conclusion
We conclude that \( \Delta \) is independent of \( \varphi \), and \( \frac{d\Delta}{d\theta} \bigg|_{\theta = \frac{\pi}{2}} = 0 \).

Answer:

\[ \boxed{\Delta \text{ is independent of } \varphi, \left( \frac{d\Delta}{d\theta} \right)_{\theta = \frac{\pi}{2}} = 0} \]

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