For real numbers $\alpha ,\beta ,\gamma ,\delta$ and $\mu$, consider the matrix

$M=\left[\begin{array}{ccc}\alpha & \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{3}}& \beta & \frac{1}{\sqrt{3}}\\ \gamma & \delta & \mu \end{array}\right]$

Suppose that

$M{M}^T=I$

where $M^T$ is the transpose of the matrix $M$ and $I$ is the $3\times 3$ identity matrix. Let

$\overrightarrow{u}=\alpha \hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\gamma \hat{k}$

$\overrightarrow{v}=\frac{1}{\sqrt{2}}\hat{i}+\beta \hat{j}+\delta \hat{k}$

$\overrightarrow{w}=-\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\mu \hat{k}$

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I (P) The value of ${\gamma }^2+{\delta }^2$ (Q) If $x\overrightarrow{u}+y\overrightarrow{v}+z\overrightarrow{w}=\overrightarrow{j}$  for some real numbers x, y and z, then the value of x  is (R) The value of $|\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w}\left)\right|$ (S) The value of $|\overrightarrow{u}\times (\overrightarrow{v}\times \overrightarrow{w}\left)\right|$

List-II $\text{(1)}0$ $\text{(2)}1$ $\text{(3)}\frac{1}{\sqrt{2}}$ $\text{(4)}\frac{1}{\sqrt{3}}$ $\text{(5)}\frac{5}{6}$

The correct option is P → (5), Q → (4), R → (2), S → (1).

Step-by-step Solution:

We are given an orthogonal matrix $M$ of size $3\times 3$ such that $M{M}^T=I$. For an orthogonal matrix, both the rows and the columns form an orthonormal set of vectors, meaning:

1. The length (magnitude) of each row vector and each column vector is 1.

2. The dot product of any two distinct row vectors or column vectors is 0.

Also, $M^{T}M=I$ as well.

Notice that the given vectors $\overrightarrow{u}$, $\overrightarrow{v}$, and $\overrightarrow{w}$ are formed by the columns of the matrix $M$:

$\overrightarrow{u}={\left[\begin{array}{ccc}\alpha & \frac{1}{\sqrt{3}}& \gamma \end{array}\right]}^T$

$\overrightarrow{v}={\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \beta & \delta \end{array}\right]}^T$

$\overrightarrow{w}={\left[\begin{array}{ccc}-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}& \mu \end{array}\right]}^T$

Since the column vectors of an orthogonal matrix form an orthonormal basis:

• $\left|\overrightarrow{u}\right|=\left|\overrightarrow{v}\right|=\left|\overrightarrow{w}\right|=1$

• $\overrightarrow{u}\cdot \overrightarrow{v}=\overrightarrow{v}\cdot \overrightarrow{w}=\overrightarrow{w}\cdot \overrightarrow{u}=0$

Evaluation of Item (P):

Consider the third row of matrix $M$, which is $\left[\begin{array}{ccc}\gamma & \delta & \mu \end{array}\right]$. Since the sum of squares of elements of any row is 1:

${\gamma }^2+{\delta }^2+{\mu }^2=1$

Also, looking at column 3, the vector $\overrightarrow{w}$ has magnitude 1:

${(-\frac{1}{\sqrt{2}})}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2+{\mu }^2=1$

$\frac{1}{2}+\frac{1}{3}+{\mu }^2=1\Rightarrow \frac{5}{6}+{\mu }^2=1\Rightarrow {\mu }^2=1-\frac{5}{6}=\frac{1}{6}$

Substituting ${\mu }^2=\frac{1}{6}$ into ${\gamma }^2+{\delta }^2+{\mu }^2=1$:

${\gamma }^2+{\delta }^2+\frac{1}{6}=1\Rightarrow {\gamma }^2+{\delta }^2=\frac{5}{6}$

Thus, P → (5).

Evaluation of Item (Q):

We are given that $x\overrightarrow{u}+y\overrightarrow{v}+z\overrightarrow{w}=\hat{j}$.

Since $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ are mutually orthogonal unit vectors, taking the dot product of both sides with $\overrightarrow{u}$ gives:

$x(\overrightarrow{u}\cdot \overrightarrow{u})+y(\overrightarrow{v}\cdot \overrightarrow{u})+z(\overrightarrow{w}\cdot \overrightarrow{u})=\hat{j}\cdot \overrightarrow{u}$

Since $\overrightarrow{u}\cdot \overrightarrow{u}=1$ and $\overrightarrow{v}\cdot \overrightarrow{u}=\overrightarrow{w}\cdot \overrightarrow{u}=0$:

$x=\hat{j}\cdot \overrightarrow{u}$

From the definition of $\overrightarrow{u}=\alpha \hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\gamma \hat{k}$, the $\hat{j}$ component of $\overrightarrow{u}$ is $\frac{1}{\sqrt{3}}$.

Therefore, $x=\frac{1}{\sqrt{3}}$.

Thus, Q → (4).

Evaluation of Item (R):

The expression $|\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w}\left)\right|$ represents the absolute value of the scalar triple product of vectors $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$.

Since $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ are the columns of the matrix $M$:

$\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w})=det\left(M\right)$

For any orthogonal matrix $M$, $det\left(M\right)=\pm 1$.

Taking the absolute value:

$|\overrightarrow{u}\cdot (\overrightarrow{v}\times \overrightarrow{w}\left)\right|=|det(M\left)\right|=1$

Thus, R → (2).

Evaluation of Item (S):

Using the vector triple product expansion:

$\overrightarrow{u}\times (\overrightarrow{v}\times \overrightarrow{w})=(\overrightarrow{u}\cdot \overrightarrow{w})\overrightarrow{v}-(\overrightarrow{u}\cdot \overrightarrow{v})\overrightarrow{w}$

Since $\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ are mutually orthogonal:

Therefore:

$\overrightarrow{u}\times (\overrightarrow{v}\times \overrightarrow{w})=0\overrightarrow{v}-0\overrightarrow{w}=\overrightarrow{0}$

Taking the magnitude:

$|\overrightarrow{u}\times (\overrightarrow{v}\times \overrightarrow{w}\left)\right|=0$

Thus, S → (1).

Combining all results:

• P → (5)

• Q → (4)

• R → (2)

• S → (1)