Question 15

Since $MM^T = I$, the matrix $M$ is orthogonal. This implies that its rows and columns are orthonormal unit vectors.

Given the matrix $M$ and the vectors $\vec{u}, \vec{v}, \vec{w}$, we observe that $\vec{u}, \vec{v}, \vec{w}$ are the columns of $M$.

(P) Since the columns are unit vectors, $|\vec{w}|^2 = (-\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{3}})^2 + \mu^2 = 1 \Rightarrow \frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \Rightarrow \mu^2 = 1/6$.

Since the rows are also unit vectors, the sum of squares of elements in the third row is $\gamma^2 + \delta^2 + \mu^2 = 1$.

Thus, $\gamma^2 + \delta^2 = 1 - 1/6 = 5/6$. Matches with (5).

(Q) Given $x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$. Since $\vec{u}, \vec{v}, \vec{w}$ are orthonormal, taking the dot product with $\vec{u}$ gives:

$x(\vec{u} \cdot \vec{u}) + y(\vec{u} \cdot \vec{v}) + z(\vec{u} \cdot \vec{w}) = \vec{u} \cdot \hat{j} \Rightarrow x(1) + 0 + 0 = \text{y-component of } \vec{u} = 1/\sqrt{3}$. Matches with (4).

(R) $|\vec{u} \cdot (\vec{v} \times \vec{w})|$ is the absolute value of the scalar triple product of the columns, which is equal to $|\det(M)|$.

Since $MM^T = I$, $(\det M)^2 = 1$, so $|\det M| = 1$. Matches with (2).

(S) In an orthonormal system, $\vec{v} \times \vec{w} = \pm \vec{u}$.

Therefore, $\vec{u} \times (\vec{v} \times \vec{w}) = \vec{u} \times (\pm \vec{u}) = \vec{0}$.

The magnitude is $0$. Matches with (1).

Correct Option: A