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:

If: \[ MM^T=I \] then rows of \(M\) are orthonormal vectors.

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: \[ \boxed{\mathrm{(A)}} \]