JEE Advanced 2024 — Paper 1 — MATH-IV solution

JEE Advanced2024Paper 1MATH-IV

Let $γ \in R$ be such that the lines $L_{1} : \frac{x + 11}{1} = \frac{y + 21}{2} = \frac{z + 29}{3}$ and $L_{2} : \frac{x + 16}{3} = \frac{y + 11}{2} = \frac{z + 4}{γ}$ intersect. Let $R_{1}$ be the point of intersection of $L_{1}$ and $L_{2}$ . Let $O = (0, 0, 0)$ , and $\overset{n}{^}$ denote a unit normal vector to the plane containing both the lines $L_{1}$ and $L_{2}$ .

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 $\overset{n}{^}$ is	(2) $\frac{3}{2}$
(R) $O R_{1}$ equals	(3) 1
(S) A possible value of $O R_{1} \cdot \overset{n}{^}$ is	(4) $\frac{1}{6} \hat{i} - \frac{2}{6} \hat{j} + \frac{1}{6} \hat{k}$
	(5) $\frac{2}{3}$

The correct option is

A
(P) $\to$ (3); (Q) $\to$ (4); (R) $\to$ (1); (S) $\to$ (2)
B
(P) $\to$ (5); (Q) $\to$ (4); (R) $\to$ (1); (S) $\to$ (2)
C
(P) $\to$ (3); (Q) $\to$ (4); (R) $\to$ (1); (S) $\to$ (5)
D
(P) $\to$ (3); (Q) $\to$ (1); (R) $\to$ (4); (S) $\to$ (5)

Solution

Direction vectors: $b_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $b_{2} = 3 \hat{i} + 2 \hat{j} + γ \hat{k}$ . Points on the lines: $A = (- 11, - 21, - 29)$ on $L_{1}$ , $B = (- 16, - 11, - 4)$ on $L_{2}$ , so $A B = - 5 \hat{i} + 10 \hat{j} + 25 \hat{k} = 5 (- \hat{i} + 2 \hat{j} + 5 \hat{k})$ .

The plane normal is parallel to $A B \times b_{1}$ . Computing:

$A B \times b_{1} = (- \hat{i} + 2 \hat{j} + 5 \hat{k}) \times (\hat{i} + 2 \hat{j} + 3 \hat{k}) \propto (6 - 10) \hat{i} + (5 + 3) \hat{j} + (- 2 - 2) \hat{k} = - 4 \hat{i} + 8 \hat{j} - 4 \hat{k}$

So a normal is proportional to $\hat{i} - 2 \hat{j} + \hat{k}$ . Unit form: $\overset{n}{^} = \frac{1}{6} \hat{i} - \frac{2}{6} \hat{j} + \frac{1}{6} \hat{k}$ . So (Q) $\to$ (4).

For $b_{2}$ to lie in the plane: $b_{2} \cdot \overset{n}{^} = 0$ . $(3) (1) + (2) (- 2) + γ (1) = 0 ⟹ γ = 1$ . So (P) $\to$ (3).

For intersection, parameterize $L_{1}$ : $(- 11 + λ, - 21 + 2 λ, - 29 + 3 λ)$ and $L_{2}$ : $(- 16 + 3 μ, - 11 + 2 μ, - 4 + μ)$ . From $x$ and $y$ equations, $λ = 10$ , $μ = 5$ . Then $O R_{1} = (- 1, - 1, 1) = - \hat{i} - \hat{j} + \hat{k}$ . So (R) $\to$ (1).

$O R_{1} \cdot \overset{n}{^} = \frac{- 1 + 2 + 1}{6} = \frac{2}{6} = \frac{2}{3}$ . So (S) $\to$ (5).

$∴ Option (C)$

← Back to full paper

Practice more MATH-IV

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →