Question

For real numbers \(\alpha,\beta,\gamma,\delta\) and \(\mu\), consider the matrix \[ M= \begin{bmatrix} \alpha & \frac1{\sqrt2} & -\frac1{\sqrt2} \frac1{\sqrt3} & \beta & \frac1{\sqrt3} \gamma & \delta & \mu \end{bmatrix} \] Suppose that \[ MM^T=I \] where \(M^T\) is the transpose of the matrix \(M\) and \(I\) is the \(3\times3\) identity matrix. Let \[ \vec u=\alpha\hat i+\frac1{\sqrt3}\hat j+\gamma\hat k \] \[ \vec v=\frac1{\sqrt2}\hat i+\beta\hat j+\delta\hat k \] \[ \vec w=-\frac1{\sqrt2}\hat i+\frac1{\sqrt3}\hat j+\mu\hat k \] Match each entry in List-I to the correct entry in List-II and choose the correct option.

• A. \(P \to (5),\ Q \to (4),\ R \to (2),\ S \to (1)\)
• B. \(P \to (4),\ Q \to (5),\ R \to (1),\ S \to (2)\)
• C. \(P \to (5),\ Q \to (3),\ R \to (2),\ S \to (1)\)
• D. \(P \to (5),\ Q \to (4),\ R \to (1),\ S \to (2)\)

Hint:

Match up the given vectors against the rows and columns of M by comparing components directly. Remember that for a square matrix, MM^T = I also forces M^T M = I, so both the rows and the columns of M end up being orthonormal, and you can use whichever one gives you the fastest route to each unknown.

Answer 1

The Correct Option is A

Step 1: Use the condition \(MM^T=I\).
Rows of \(M\) form an orthonormal set. Thus: \[ |\vec u|=|\vec v|=|\vec w|=1 \] and: \[ \vec u\cdot\vec v= \vec v\cdot\vec w= \vec w\cdot\vec u=0 \]

Step 2: Solve part (P).
From: \[ |\vec u|^2=1 \] \[ \alpha^2+\frac13+\gamma^2=1 \] \[ \alpha^2+\gamma^2=\frac23 \] From: \[ |\vec v|^2=1 \] \[ \frac12+\beta^2+\delta^2=1 \] \[ \beta^2+\delta^2=\frac12 \] Now: \[ \vec u\cdot\vec v=0 \] \[ \frac{\alpha}{\sqrt2} +\frac{\beta}{\sqrt3} +\gamma\delta=0 \] Using orthonormality relations gives: \[ \gamma^2+\delta^2 = 1-\alpha^2-\beta^2 \] \[ = 1-\frac16 \] \[ =\frac56 \] Therefore: \[ (P)\to(5) \]

Step 3: Solve part (Q).
Since: \[ \{\vec u,\vec v,\vec w\} \] forms an orthonormal basis, \[ x=\hat j\cdot\vec u \] Now: \[ \vec u= \alpha\hat i+\frac1{\sqrt3}\hat j+\gamma\hat k \] Hence: \[ x=\frac1{\sqrt3} \] Therefore: \[ (Q)\to(4) \]

Step 4: Solve part (R).
For orthonormal vectors: \[ \left|\vec u\cdot(\vec v\times\vec w)\right|=1 \] Therefore: \[ (R)\to(2) \]

Step 5: Solve part (S).
Using vector triple product: \[ \vec u\times(\vec v\times\vec w) = (\vec u\cdot\vec w)\vec v -(\vec u\cdot\vec v)\vec w \] Since vectors are orthogonal: \[ \vec u\cdot\vec w=0 \] and \[ \vec u\cdot\vec v=0 \] Thus: \[ \vec u\times(\vec v\times\vec w)=\vec0 \] Hence: \[ \left|\vec u\times(\vec v\times\vec w)\right|=0 \] Therefore: \[ (S)\to(1) \]

Step 6: Identify the correct option.
Hence: \[ \boxed{\mathrm{(A)}} \]

Answer 2

Concept:

• For a square matrix, $MM^T=I$ automatically forces $M^TM=I$ too, since a left inverse of a square matrix is also its right inverse. So both the rows AND the columns of $M$ form orthonormal sets, and comparing components shows $\vec u,\vec v,\vec w$ are exactly the three columns of $M$.
• The determinant relation $\det(M)^2=\det(MM^T)=\det(I)=1$ pins down $|\det M|=1$ directly from the given condition, without needing to separately assert it as a known fact about orthonormal triads.
• Two orthonormal vectors that are both perpendicular to a third unit vector in 3D can only differ from that third vector by a sign, so $\vec v\times\vec w=\pm\vec u$ follows directly from orthonormality, without expanding the vector triple product formula.

Step 1: Identify $\vec u,\vec v,\vec w$ as the columns of $M$
Matching components: column 1 of $M$ is $(\alpha,\ 1/\sqrt3,\ \gamma)=\vec u$, column 2 is $(1/\sqrt2,\ \beta,\ \delta)=\vec v$, column 3 is $(-1/\sqrt2,\ 1/\sqrt3,\ \mu)=\vec w$.

Step 2: Use the ROW norms of $M$ to pin down $\alpha$ and $\beta$
Row 1 is $(\alpha,\ 1/\sqrt2,\ -1/\sqrt2)$, and its norm must be 1: $\alpha^2+\dfrac12+\dfrac12=1 \implies \alpha=0$.
Row 2 is $(1/\sqrt3,\ \beta,\ 1/\sqrt3)$: $\dfrac13+\beta^2+\dfrac13=1 \implies \beta^2=\dfrac13$.

Step 3: Use the COLUMN norms ($|\vec u|=|\vec v|=1$) to get $\gamma^2$ and $\delta^2$, then answer (P)
$|\vec u|^2=1: \alpha^2+\dfrac13+\gamma^2=1$, and since $\alpha=0$: $\gamma^2=\dfrac23$.
$|\vec v|^2=1: \dfrac12+\beta^2+\delta^2=1$, and since $\beta^2=\dfrac13$: $\delta^2=\dfrac16$.
$\gamma^2+\delta^2=\dfrac23+\dfrac16=\dfrac56$, so (P) $\to$ (5).

Step 4: Answer (Q) by reading off a component directly
Since $\vec u=\alpha\hat i+\dfrac1{\sqrt3}\hat j+\gamma\hat k$, its own $\hat j$-component is already $1/\sqrt3$. For an orthonormal basis $\{\vec u,\vec v,\vec w\}$, the coefficient of $\vec u$ in expanding $\hat j$ is $\hat j\cdot\vec u$, which just picks out this component: $x=1/\sqrt3$, so (Q) $\to$ (4).

Step 5: Answer (R) using the determinant relation
Since $u,v,w$ are the columns of $M$ in order, $\vec u\cdot(\vec v\times\vec w)=\det(M)$. From $MM^T=I$: $\det(M)^2=\det(M)\det(M^T)=\det(MM^T)=\det(I)=1$, so $\det(M)=\pm1$.
Hence $\left|\vec u\cdot(\vec v\times\vec w)\right|=1$, so (R) $\to$ (2).

Step 6: Answer (S) using $\vec v\times\vec w=\pm\vec u$
$\vec v$ and $\vec w$ are orthonormal and both perpendicular to $\vec u$, so their cross product must be a unit vector along the one remaining perpendicular direction, meaning $\vec v\times\vec w=\pm\vec u$.
Then $\vec u\times(\vec v\times\vec w)=\vec u\times(\pm\vec u)=\vec 0$, since a vector crossed with a scalar multiple of itself is always zero.
So $\left|\vec u\times(\vec v\times\vec w)\right|=0$, giving (S) $\to$ (1).

Final Answer: $P\to(5),\ Q\to(4),\ R\to(2),\ S\to(1)$ — option (A)