Discussion
| List-I | List-II |
|---|---|
| (P) The value of \(\gamma^{2} + \delta^{2}\) is | (1) 0 |
| (Q) If \(x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}\) for some real numbers \(x, y\) and \(z\), then the value of \(x\) is | (2) 1 |
| (R) The value of \(|\vec{u} \cdot (\vec{v} \times \vec{w})|\) is | (3) \(\frac{1}{\sqrt{2}}\) |
| (S) The value of \(|\vec{u} \times (\vec{v} \times \vec{w})|\) is | (4) \(\frac{1}{\sqrt{3}}\) |
| (5) \(\frac{5}{6}\) |
Answer: A
The correct match is (P) → (5), (Q) → (4), (R) → (2), (S) → (1).
\(MM^{T} = I\) makes \(M\) orthogonal, so its rows are orthonormal and so are its columns, and \(\vec{u}, \vec{v}, \vec{w}\) are exactly the three columns.
Row 1 has unit length: \(\alpha^{2} + \frac{1}{2} + \frac{1}{2} = 1\), so \(\alpha = 0\). Rows 1 and 2 are perpendicular: \(\frac{\beta}{\sqrt{2}} - \frac{1}{\sqrt{6}} = 0\), so \(\beta = \frac{1}{\sqrt{3}}\). Then the unit-length columns give:
\[ \begin{aligned} \gamma^{2} &= 1 - 0 - \tfrac{1}{3} = \tfrac{2}{3} \\ \delta^{2} &= 1 - \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} \\ \gamma^{2} + \delta^{2} &= \tfrac{5}{6} \end{aligned} \]
(Q) In an orthonormal basis the coefficient of \(\vec{u}\) is the dot product with \(\vec{u}\): \(x = \vec{u} \cdot \hat{j} = \frac{1}{\sqrt{3}}\). (R) \(\vec{u} \cdot (\vec{v} \times \vec{w}) = \det M = \pm 1\), so its modulus is 1. (S) \(\vec{v} \times \vec{w}\) is a unit vector perpendicular to both, which is \(\pm\vec{u}\), so \(\vec{u} \times (\vec{v} \times \vec{w}) = \vec{0}\) and its modulus is 0.
The fourth option swaps R and S, the third gives \(x\) the value \(\frac{1}{\sqrt{2}}\), and the second misplaces P as well.
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