Question

Let $\overrightarrow{w} = \hat{\imath} + \hat{\jmath} - 2\hat{k}$, and $\overrightarrow{u}$ and $\overrightarrow{v}$ be two vectors, such that $\overrightarrow{u} \times \overrightarrow{v} = \overrightarrow{w}$ and $\overrightarrow{v} \times \overrightarrow{w} = \overrightarrow{u}$. Let $\alpha, \beta, \gamma$, and $t$ be real numbers such that

$\overrightarrow{u} = \alpha\hat{\imath} + \beta\hat{\jmath} + \gamma\hat{k}$, $-t\alpha + \beta + \gamma = 0$, $\alpha - t\beta + \gamma = 0$, and $\alpha + \beta - t\gamma = 0$.

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

List-I List-II

(P) $|\overrightarrow{v}|^2$ is equal to (1) 0

(Q) If $\alpha = \sqrt{3}$, then $\gamma^2$ is equal to (2) 1

(3) 2

(R) If $\alpha = \sqrt{3}$, then $(\beta + \gamma)^2$ is equal to (4) 3

(S) If $\alpha = \sqrt{2}$, then $t + 3$ is equal to (5) 5

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

Answer 1

Answer: (A)

A

Answer 2

First, let's analyze the given vector equations: $\overrightarrow{u} \times \overrightarrow{v} = \overrightarrow{w}$ and $\overrightarrow{v} \times \overrightarrow{w} = \overrightarrow{u}$.

From the properties of the cross product, $\overrightarrow{w}$ is orthogonal to both $\overrightarrow{u}$ and $\overrightarrow{v}$. Also, $\overrightarrow{u}$ is orthogonal to both $\overrightarrow{v}$ and $\overrightarrow{w}$. This implies that $\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}$ are mutually orthogonal vectors.

The magnitude of $\overrightarrow{w}$ is $|\overrightarrow{w}| = |\hat{\imath} + \hat{\jmath} - 2\hat{k}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.

Taking magnitudes of the given vector equations: $|\overrightarrow{u} \times \overrightarrow{v}| = |\overrightarrow{w}|$ and $|\overrightarrow{v} \times \overrightarrow{w}| = |\overrightarrow{u}|$. Since $\overrightarrow{u}$ and $\overrightarrow{v}$ are orthogonal, $|\overrightarrow{u} \times \overrightarrow{v}| = |\overrightarrow{u}||\overrightarrow{v}|\sin(90^\circ) = |\overrightarrow{u}||\overrightarrow{v}|$. Since $\overrightarrow{v}$ and $\overrightarrow{w}$ are orthogonal, $|\overrightarrow{v} \times \overrightarrow{w}| = |\overrightarrow{v}||\overrightarrow{w}|\sin(90^\circ) = |\overrightarrow{v}||\overrightarrow{w}|$. So, we have: $|\overrightarrow{u}||\overrightarrow{v}| = |\overrightarrow{w}| = \sqrt{6}$ $|\overrightarrow{v}||\overrightarrow{w}| = |\overrightarrow{u}|$ Substitute $|\overrightarrow{w}| = \sqrt{6}$ into the second equation: $\sqrt{6}|\overrightarrow{v}| = |\overrightarrow{u}|$. Substitute this into the first equation: $(\sqrt{6}|\overrightarrow{v}|)|\overrightarrow{v}| = \sqrt{6} \implies \sqrt{6}|\overrightarrow{v}|^2 = \sqrt{6}$. This gives $|\overrightarrow{v}|^2 = 1$.

(P) $|\overrightarrow{v}|^2$ is equal to 1. So (P) $\rightarrow$ (2).

We are given $\overrightarrow{u} = \alpha\hat{\imath} + \beta\hat{\jmath} + \gamma\hat{k}$ and a system of linear equations for $\alpha, \beta, \gamma$:

1. $-t\alpha + \beta + \gamma = 0$
2. $\alpha - t\beta + \gamma = 0$
3. $\alpha + \beta - t\gamma = 0$ This is a homogeneous system of linear equations. Since $\overrightarrow{u} \neq \overrightarrow{0}$ (because $|\overrightarrow{u}| = \sqrt{6} \neq 0$), the system must have a non-trivial solution. This requires the determinant of the coefficient matrix to be zero: $\begin{vmatrix} -t & 1 & 1 \\ 1 & -t & 1 \\ 1 & 1 & -t \end{vmatrix} = 0$ $-t(t^2 - 1) - 1(-t - 1) + 1(1 + t) = 0$ $-t^3 + t + t + 1 + 1 + t = 0$ $-t^3 + 3t + 2 = 0 \implies t^3 - 3t - 2 = 0$. Factoring the cubic: $(t+1)^2(t-2) = 0$. The possible values for $t$ are $t=-1$ or $t=2$.

Case 1: $t=-1$. The system becomes: $\alpha + \beta + \gamma = 0$ $\alpha + \beta + \gamma = 0$ $\alpha + \beta + \gamma = 0$ The condition is $\alpha + \beta + \gamma = 0$. We also know that $\overrightarrow{u}$ is orthogonal to $\overrightarrow{w} = \hat{\imath} + \hat{\jmath} - 2\hat{k}$. $\overrightarrow{u} \cdot \overrightarrow{w} = (\alpha\hat{\imath} + \beta\hat{\jmath} + \gamma\hat{k}) \cdot (\hat{\imath} + \hat{\jmath} - 2\hat{k}) = \alpha + \beta - 2\gamma = 0$. We have the system: I) $\alpha + \beta + \gamma = 0$ II) $\alpha + \beta - 2\gamma = 0$ Subtracting II from I gives $3\gamma = 0 \implies \gamma = 0$. Substituting $\gamma=0$ into I gives $\alpha + \beta = 0 \implies \beta = -\alpha$. So, for $t=-1$, $\beta = -\alpha$ and $\gamma = 0$. The condition $|\overrightarrow{u}|^2 = 6$ implies $\alpha^2 + \beta^2 + \gamma^2 = 6$. $\alpha^2 + (-\alpha)^2 + 0^2 = 6 \implies 2\alpha^2 = 6 \implies \alpha^2 = 3$.

Case 2: $t=2$. The system becomes: $-2\alpha + \beta + \gamma = 0$ $\alpha - 2\beta + \gamma = 0$ $\alpha + \beta - 2\gamma = 0$ From the first two equations: $(-2\alpha + \beta + \gamma) - (\alpha - 2\beta + \gamma) = 0 \implies -3\alpha + 3\beta = 0 \implies \alpha = \beta$. From the second and third equations: $(\alpha - 2\beta + \gamma) - (\alpha + \beta - 2\gamma) = 0 \implies -3\beta + 3\gamma = 0 \implies \beta = \gamma$. So, for $t=2$, $\alpha = \beta = \gamma$. The condition $|\overrightarrow{u}|^2 = 6$ implies $\alpha^2 + \beta^2 + \gamma^2 = 6$. $\alpha^2 + \alpha^2 + \alpha^2 = 6 \implies 3\alpha^2 = 6 \implies \alpha^2 = 2$. This is consistent with $\overrightarrow{u} \cdot \overrightarrow{w} = \alpha + \beta - 2\gamma = \alpha + \alpha - 2\alpha = 0$.

Now we evaluate the expressions in List-I based on these findings.

(P) $|\overrightarrow{v}|^2$: We found $|\overrightarrow{v}|^2 = 1$. Matches (2).

(Q) If $\alpha = \sqrt{3}$, then $\gamma^2$ is equal to. If $\alpha = \sqrt{3}$, then $\alpha^2 = 3$. This occurs when $t=-1$. In this case, $\gamma = 0$. So $\gamma^2 = 0^2 = 0$. Matches (1).

(R) If $\alpha = \sqrt{3}$, then $(\beta + \gamma)^2$ is equal to. If $\alpha = \sqrt{3}$, then this is the case $t=-1$, where $\beta = -\alpha$ and $\gamma = 0$. $\beta = -\sqrt{3}$ and $\gamma = 0$. $(\beta + \gamma)^2 = (-\sqrt{3} + 0)^2 = (-\sqrt{3})^2 = 3$. Matches (4).

(S) If $\alpha = \sqrt{2}$, then $t + 3$ is equal to. If $\alpha = \sqrt{2}$, then $\alpha^2 = 2$. This occurs when $t=2$. So $t=2$. $t + 3 = 2 + 3 = 5$. Matches (5).

Summary of matches: (P) $\rightarrow$ (2) (Q) $\rightarrow$ (1) (R) $\rightarrow$ (4) (S) $\rightarrow$ (5)

Let's check the options provided. (A) (P) $\rightarrow$ (2) (Q) $\rightarrow$ (1) (R) $\rightarrow$ (4) (S) $\rightarrow$ (5) - This matches our results. (B) (P) $\rightarrow$ (2) (Q) $\rightarrow$ (4) (R) $\rightarrow$ (3) (S) $\rightarrow$ (5) - Incorrect. (C) (P) $\rightarrow$ (2) (Q) $\rightarrow$ (1) (R) $\rightarrow$ (4) (S) $\rightarrow$ (3) - Incorrect. (D) (P) $\rightarrow$ (5) (Q) $\rightarrow$ (4) (R) $\rightarrow$ (1) (S) $\rightarrow$ (3) - Incorrect.

The correct option is (A).