JEE Advanced 2023 — Paper 1 — MATH-IV solution

JEE Advanced2023Paper 1MATH-IV

Let $ℓ_{1}$ and $ℓ_{2}$ be the lines $r_{1} = λ (\hat{i} + \hat{j} + \hat{k})$ and $r_{2} = (\hat{j} - \hat{k}) + μ (\hat{i} + \hat{k})$ , respectively. Let $X$ be the set of all the planes $H$ that contain the line $ℓ_{1}$ . For a plane $H$ , let $d (H)$ denote the smallest possible distance between the points of $ℓ_{2}$ and $H$ . Let $H_{0}$ be the plane in $X$ for which $d (H_{0})$ is the maximum value of $d (H)$ as $H$ varies over all planes in $X$ .

Match each entry in List-I to the correct entries in List-II.

List-I	List-II
(P) The value of $d (H_{0})$ is	(1) $3$
(Q) The distance of the point $(0, 1, 2)$ from $H_{0}$ is	(2) $\frac{1}{3}$
(R) The distance of origin from $H_{0}$ is	(3) 0
(S) The distance of origin from the point of intersection of planes $y = z$ , $x = 1$ and $H_{0}$ is	(4) $2$
	(5) $\frac{1}{2}$

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

Solution

Let $H : a x + b y + cz = 0$ (it passes through origin, on $ℓ_{1}$ ). Containing $ℓ_{1}$ with direction $(1, 1, 1)$ gives $a + b + c = 0$ .

$d (H)$ is maximised by making $ℓ_{2}$ parallel to $H$ (so the distance from $ℓ_{2}$ to $H$ is the constant perpendicular distance from any point of $ℓ_{2}$ ). Direction of $ℓ_{2}$ is $(1, 0, 1)$ , so $a + c = 0$ , which forces $b = 0$ .

Hence $H_{0}$ : $x - z = 0$ .

(P) Distance from any point of $ℓ_{2}$ , say $(0, 1, - 1)$ , to $H_{0}$ is $\frac{∣0 - ( - 1 ) ∣}{2} = \frac{1}{2}$ . $\to$ (5).

(Q) Distance of $(0, 1, 2)$ from $x - z = 0$ : $\frac{∣0 - 2∣}{2} = 2$ . $\to$ (4).

(R) Origin lies on $H_{0}$ , distance $= 0$ . $\to$ (3).

(S) Solve $y = z$ , $x = 1$ , $x = z$ simultaneously: $z = 1$ , $y = 1$ , $x = 1$ . Distance of $(1, 1, 1)$ from origin $= 3$ . $\to$ (1).

Mapping: P $\to$ 5, Q $\to$ 4, R $\to$ 3, S $\to$ 1. The correct option is (B).

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