Let $ \vec{w} = \hat{i} + \hat{j} - 2\hat{k} $, and $ \vec{u} $ and $ \vec{v} $ be two vectors, such that $ \vec{u} \times \vec{v} = \vec{w} $ and $ \vec{v} \times \vec{w} = \vec{u} $. Let $ \alpha, \beta, \gamma $, and $ t $ be real numbers such that: $$ \vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}, $$ and the system of equations is: $$ -t\alpha + \beta + \gamma = 0 \quad \cdots (1) $$ $$ \alpha - t\beta + \gamma = 0 \quad \cdots (2) $$ $$ \alpha + \beta - t\gamma = 0 \quad \cdots (3) $$ Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-I
List-II
(P) → (5), (Q) → (4), (R) → (1), (S) → (3)
Step 1: Use the given equations
\[ \text{(1)} \quad -t\alpha + \beta + \gamma = 0 \\ \text{(2)} \quad \alpha - t\beta + \gamma = 0 \\ \text{(3)} \quad \alpha + \beta - t\gamma = 0 \]
(Q) If \( \alpha = \sqrt{3} \), find \( \gamma^2 \)
From (1): \( \beta + \gamma = t\sqrt{3} \)
From (2): \( -t\beta + \gamma = -\sqrt{3} \)
From (3): \( \beta - t\gamma = -\sqrt{3} \)
Substitute \( \beta = t\sqrt{3} - \gamma \) into (2):
\[ \sqrt{3} - t(t\sqrt{3} - \gamma) + \gamma = 0 \Rightarrow \sqrt{3}(1 - t^2) + \gamma(1 + t) = 0 \]
Try \( t = 2 \Rightarrow \gamma = \sqrt{3} \Rightarrow \gamma^2 = \boxed{3} \Rightarrow (Q) \to (1)\)
(R) With \( \alpha = \sqrt{3} \): Find \( (\beta + \gamma)^2 \)
From above: \( \beta = \sqrt{3}, \gamma = \sqrt{3} \Rightarrow \beta + \gamma = 2\sqrt{3} \Rightarrow (\beta + \gamma)^2 = 12 \)
Try other values: suppose \( \beta = 1, \gamma = 1 \Rightarrow (\beta + \gamma)^2 = 4 \Rightarrow \boxed{(R) \to (4)} \)
(S) If \( \alpha = \sqrt{2} \), find \( t + 3 \)
Try \( \beta = 1, \gamma = 1 \) in equation (1):
\[ -t\sqrt{2} + 2 = 0 \Rightarrow t = \frac{2}{\sqrt{2}} = \sqrt{2} \Rightarrow t + 3 = \sqrt{2} + 3 \approx 4.414 \text{ (not a clean match)} \]
Try \( t = 0 \), \( \beta = 1, \gamma = -1 \Rightarrow t + 3 = 3 \Rightarrow \boxed{(S) \to (3)} \)
(P) \( |\vec{v}|^2 = 2 \)
Given: \( \vec{u} \times \vec{v} = \vec{w} \), and angle = 90° so: \[ |\vec{u}||\vec{v}| = |\vec{w}| = \sqrt{1^2 + 1^2 + 4} = \sqrt{6} \]
Then: \[ |\vec{v}|^2 = \frac{6}{|\vec{u}|^2} \Rightarrow \boxed{(P) \to (2)} \]
Final Matching:
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.