Question

Let y ∈ R be such that the lines \(L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}\) and \(L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}\) intersect. Let R₁ be the point of intersection of L₁ and L₂. Let O = (0, 0 ,0), and \(\hat{n}\) denote a unit normal vector to the plane containing both the lines L₁ and L₂.
Match each entry in List-I to the correct entry in List-II.

List - I		List - II
(P)	γ equals	(1)	\(-\hat{i}-\hat{j}+\hat{k}\)
(Q)	A possible choice for \(\hat{n}\) is	(2)	\(\sqrt{\frac{3}{2}}\)
(R)	\(\overrightarrow{OR_1}\) equals	(3)	1
(S)	A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is	(4)	\(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\)
		(5)	\(\sqrt{\frac{2}{3}}\)

The correct option is

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

Answer 1

The Correct Option is C

For the given problem, we have two lines \(L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}\) and \(L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}\). These lines intersect at a point \(R_1\). We need to determine the matched items from List-I to List-II based on the conditions provided.

1. Finding Intersection: Set the parameter \(t\) for \(L_1\) as \(x = t - 11\), \(y = 2t - 21\), \(z = 3t - 29\) and parameter \(s\) for \(L_2\) as \(x = 3s - 16\), \(y = 2s - 11\), \(z = \gamma s - 4\). Equate to satisfy all three coordinates simultaneously for intersection.
2. Solving Equations: Solve \(x\) and \(y\) equations:
   • \(t - 11 = 3s - 16 \implies t = 3s - 5\)
   • \(2t - 21 = 2s - 11 \implies t = s + 5\)
3. Calculate \(z\):
   • From \(L_1\), \(z = 3(10) - 29 = 1\)
   • From \(L_2\), substitute \(s = 5\): \(\gamma(5) - 4 = 1 \implies \gamma = 1\)
4. Match Lists: Evaluate \(\overrightarrow{OR_1}\), unit normal \(\hat{n}\), and calculate corresponding values:
   • \(\overrightarrow{OR_1} = (t - 11, 2t - 21, 3t - 29) = (-1,-1,1)\)
   • Unit normal \(\hat{n}\) possibilities:
     • Cross product of direction vectors \(\overrightarrow{d_1} = (1,2,3)\) and \(\overrightarrow{d_2} = (3,2,\gamma)\) gives a normal vector \(\overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = (-4\hat{i} + 8\hat{j} - 4\hat{k})\)
     • Normalize \(\hat{n} = \frac{1}{\sqrt{6}}(-2, 4, -2) = \frac{1}{\sqrt{6}}(-1, 2, -1)\)
   • Calculate \(\overrightarrow{OR_1}.\hat{n}\):\(\frac{1}{\sqrt{6}}((-1)\cdot(-1) + (-1)\cdot2 + (1)\cdot(-1)) = \frac{-2}{\sqrt{6}}\)

(P)	γ equals	3
(Q)	A possible choice for \(\hat{n}\)	\(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\)
(R)	\(\overrightarrow{OR_1}\) equals	\(-\hat{i}-\hat{j}+\hat{k}\)
(S)	A possible value of \(\overrightarrow{OR_1}.\hat{n}\)	\(\sqrt{\frac{2}{3}}\)

Thus, the correct matching is (P) → (3), (Q) → (4), (R) → (1), (S) → (5).

Answer 2

We are given two lines \(L_1\) and \(L_2\) with parameters, and we want to find \(\gamma\), the point of intersection \(R_1\), a unit normal vector \(\hat{n}\) to the plane containing both lines, and the scalar product \(\overrightarrow{OR_1} \cdot \hat{n}\).

---

Step 1: Parametric form of lines

\[ L_1: x = -11 + t, \quad y = -21 + 2t, \quad z = -29 + 3t \] \[ L_2: x = -16 + 3s, \quad y = -11 + 2s, \quad z = -4 + \gamma s \] ---

Step 2: Find \(t, s\) such that lines intersect

\[ -11 + t = -16 + 3s \implies t = -5 + 3s \] \[ -21 + 2t = -11 + 2s \implies 2t - 2s = 10 \implies t - s = 5 \] Substitute \(t\) from first into second: \[ (-5 + 3s) - s = 5 \implies 2s = 10 \implies s = 5 \] \[ t = -5 + 3 \times 5 = 10 \] ---

Step 3: Use \(z\)-coordinate to find \(\gamma\)

\[ -29 + 3t = -4 + \gamma s \implies -29 + 30 = -4 + 5 \gamma \implies 1 = -4 + 5 \gamma \implies \gamma = 1 \] ---

Step 4: Find coordinates of \(R_1\)

\[ x = -11 + 10 = -1, \quad y = -21 + 20 = -1, \quad z = -29 + 30 = 1 \] So, \[ \overrightarrow{OR_1} = (-1, -1, 1) \] ---

Step 5: Direction vectors of \(L_1\) and \(L_2\)

\[ \vec{d}_1 = (1, 2, 3), \quad \vec{d}_2 = (3, 2, 1) \] ---

Step 6: Find normal vector to the plane

\[ \vec{n} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \\ \end{vmatrix} = (-4, 8, -4) \] ---

Step 7: Unit normal vector \(\hat{n}\)

\[ |\vec{n}| = \sqrt{(-4)^2 + 8^2 + (-4)^2} = \sqrt{16 + 64 + 16} = \sqrt{96} = 4 \sqrt{6} \] \[ \hat{n} = \frac{1}{4 \sqrt{6}} (-4, 8, -4) = \frac{1}{\sqrt{6}} (-1, 2, -1) \] ---

Step 8: Calculate \(\overrightarrow{OR_1} \cdot \hat{n}\)

\[ (-1, -1, 1) \cdot \frac{1}{\sqrt{6}} (-1, 2, -1) = \frac{1}{\sqrt{6}} (1 - 2 - 1) = \frac{-2}{\sqrt{6}} = -\frac{2}{\sqrt{6}} \] ---

Matching List-I to List-II:

\[ (P) \to (3) \quad \gamma = 1 \] \[ (Q) \to (4) \quad \hat{n} = \frac{1}{\sqrt{6}} (-\hat{i} + 2 \hat{j} - \hat{k}) \] \[ (R) \to (1) \quad \overrightarrow{OR_1} = -\hat{i} - \hat{j} + \hat{k} \] \[ (S) \to (5) \quad \overrightarrow{OR_1} \cdot \hat{n} = \sqrt{\frac{2}{3}} \quad \text{(absolute value)} \] ---

Final answer:

\[ \boxed{ (P) \to (3), \quad (Q) \to (4), \quad (R) \to (1), \quad (S) \to (5) } \]