Orthogonal Matrix Properties and Vector Matching List

Given the matrix $M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$

We are given that $M M^T = I$, which means $M$ is an orthogonal matrix. Thus, $M^T M = I$ also holds true.

Let the vectors $\vec{u}, \vec{v}, \vec{w}$ be defined as: $\vec{u} = \alpha \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \gamma \hat{k}$ $\vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \beta \hat{j} + \delta \hat{k}$ $\vec{w} = -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \mu \hat{k}$

Notice that $\vec{u}, \vec{v}, \vec{w}$ correspond to the first, second, and third columns of $M$, respectively. Since $M$ is orthogonal, its column vectors form an orthonormal basis for $\mathbb{R}^3$. Therefore, we have the following properties:

1. Unit vectors: $|\vec{u}| = |\vec{v}| = |\vec{w}| = 1$
2. Mutually orthogonal: $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{u} = 0$
3. Determinant property: $|\det M| = 1$

Similarly, the row vectors of $M$ are also orthonormal.

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Analysis of List-I Entries:

(P) To find the value of $\gamma^2 + \delta^2$:

Since the third row vector $(\gamma, \delta, \mu)$ has a norm of $1$: $\gamma^2 + \delta^2 + \mu^2 = 1 \implies \gamma^2 + \delta^2 = 1 - \mu^2$

Since $\vec{w}$ is a unit vector: $|\vec{w}|^2 = \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \mu^2 = 1$ $\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \implies \frac{5}{6} + \mu^2 = 1 \implies \mu^2 = \frac{1}{6}$

Substituting $\mu^2 = \frac{1}{6}$ into the equation above: $\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$

Thus, $(\text{P}) \rightarrow (5)$.

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(Q) To find the value of $x$ in $x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$:

Taking the dot product with $\vec{u}$ on both sides of the vector equation: $\vec{u} \cdot (x\vec{u} + y\vec{v} + z\vec{w}) = \vec{u} \cdot \hat{j}$ $x(\vec{u} \cdot \vec{u}) + y(\vec{u} \cdot \vec{v}) + z(\vec{u} \cdot \vec{w}) = \vec{u} \cdot \hat{j}$

Using the orthonormal properties ($\vec{u} \cdot \vec{u} = 1$ and $\vec{u} \cdot \vec{v} = \vec{u} \cdot \vec{w} = 0$): $x = \vec{u} \cdot \hat{j}$

From the definition of $\vec{u}$, its $\hat{j}$-component is $\frac{1}{\sqrt{3}}$. Therefore: $x = \frac{1}{\sqrt{3}}$

Thus, $(\text{Q}) \rightarrow (4)$.

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(R) To find the value of $|\vec{u} \cdot (\vec{v} \times \vec{w})|$:

The scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ is equal to the determinant of the matrix formed by $\vec{u}, \vec{v}, \vec{w}$ as its column vectors, which is $\det M$.

Since $M M^T = I$: $\det(M M^T) = \det(I) \implies (\det M)^2 = 1 \implies |\det M| = 1$

Therefore: $|\vec{u} \cdot (\vec{v} \times \vec{w})| = |\det M| = 1$

Thus, $(\text{R}) \rightarrow (2)$.

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(S) To find the value of $|\vec{u} \times (\vec{v} \times \vec{w})|$:

Using the vector triple product expansion formula: $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$

Since $\vec{u}, \vec{v}, \vec{w}$ are mutually orthogonal, $\vec{u} \cdot \vec{w} = 0$ and $\vec{u} \cdot \vec{v} = 0$. Hence: $\vec{u} \times (\vec{v} \times \vec{w}) = \vec{0}$

Therefore, its magnitude is: $|\vec{u} \times (\vec{v} \times \vec{w})| = 0$

Thus, $(\text{S}) \rightarrow (1)$.

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Conclusion:

The correct matching is: $(\text{P}) \rightarrow (5), \quad (\text{Q}) \rightarrow (4), \quad (\text{R}) \rightarrow (2), \quad (\text{S}) \rightarrow (1)$

This corresponds to Option A.