Question 15

MATRIX MATCHHARD

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

$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$

Suppose that $MM^T = I$ ,

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

Let $\vec{u} = \alpha \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \gamma \hat{k}, \quad \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \beta \hat{j} + \delta \hat{k} \quad \text{and} \quad \vec{w} = -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \mu \hat{k}$

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

List - I

The value of $\gamma^2 + \delta^2$ is

If $x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$ for some real numbers $x, y, z$ , then the value of $x$ is

The value of $|\vec{u} \cdot (\vec{v} \times \vec{w})|$ is

The value of $|\vec{u} \times (\vec{v} \times \vec{w})|$ is

List-II

$0$

$1$

$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{3}}$

$\frac{5}{6}$

(A)

P-5, Q-4, R-2, S-1

(B)

P-4, Q-5, R-1, S-2

(C)

P-5, Q-3, R-2, S-1

(D)

P-5, Q-4, R-1, S-2

Detailed Solution

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

Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

Original Mocks & Regular Test Series
Real NTA-like Interface with Analytics
Many Free Tests Available