Vector Algebra

Concept:The condition $MM^T = I$ means the rows of $M$ form an orthonormal set of vectors (unit length and mutually perpendicular).Explanation:From $MM^T = I$, the rows $\vec{u}, \vec{v}, \vec{w}$ (as given) are orthonormal.Each row vector has magnitude $1$.For $\vec{u}$: $\alpha^2 + \frac13 + \gamma^2 = 1 \implies \alpha^2+\gamma^2 = \frac23$.For $\vec{v}$: $\frac12 + \beta^2 + \delta^2 = 1 \implies \beta^2+\delta^2 = \frac12$.For $\vec{w}$: $\frac12 + \frac13 + \mu^2 = 1 \implies \mu^2 = \frac16$.The third row $(\gamma,\delta,\mu)$ also has unit length: $\gamma^2+\delta^2+\mu^2 = 1$.Substitute $\mu^2$: $\gamma^2+\delta^2 = 1-\frac16 = \frac56$. So (P) matches $\frac56$ → (5).For (Q): $x\vec{u}+y\vec{v}+z\vec{w} = \hat{j}$. Dot both sides with $\vec{u}$: $x(\vec{u}\cdot\vec{u}) + y(\vec{v}\cdot\vec{u}) + z(\vec{w}\cdot\vec{u}) = \vec{u}\cdot\hat{j}$.Since $\vec{u},\vec{v},\vec{w}$ are orthonormal, $\vec{u}\cdot\vec{v}=0$, $\vec{u}\cdot\vec{w}=0$, and $\vec{u}\cdot\vec{u}=1$. Also $\vec{u}\cdot\hat{j} = \frac{1}{\sqrt3}$. Thus $x = \frac{1}{\sqrt3}$ → (4).For (R): The scalar triple product $|\vec{u}\cdot(\vec{v}\times\vec{w})|$ equals the absolute determinant of the matrix formed by the orthonormal rows, which is $1$. So (R) → (2).For (S): Using identity $\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$.Since $\vec{u}\cdot\vec{v}=0$ and $\vec{u}\cdot\vec{w}=0$, the result is $\vec{0}$, so its magnitude is $0$. Thus (S) → (1).Answer:(P) → (5), (Q) → (4), (R) → (2), (S) → (1). The correct option is A.