JEE Advanced 2023 Paper 1 · Q16 · 3D Planes

Question

Let $\ell_1$ and $\ell_2$ be the lines $\vec{r}_1 = \lambda(\hat{i}+\hat{j}+\hat{k})$ and $\vec{r}_2 = (\hat{j}-\hat{k}) + \mu(\hat{i}+\hat{k})$, respectively. Let $X$ be the set of all planes $H$ that contain the line $\ell_1$. For a plane $H$, let $d(H)$ denote the smallest possible distance between the points of $\ell_2$ and $H$. Let $H_0$ be a 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:
(P) The value of $d(H_0)$ is
(Q) The distance of the point $(0, 1, 2)$ from $H_0$ is
(R) The distance of origin from $H_0$ is
(S) The distance of origin from the point of intersection of planes $y = z$, $x = 1$ and $H_0$ is
List-II: (1) $\sqrt{3}$, (2) $\dfrac{1}{\sqrt{3}}$, (3) $0$, (4) $\sqrt{2}$, (5) $\dfrac{1}{\sqrt{2}}$

SolveKar AI · Accepted Answer

$\ell_1$ direction $(1,1,1)$, $\ell_2$ direction $(1,0,1)$. The plane through $\ell_1$ that maximizes distance from $\ell_2$ is parallel to $\ell_2$, with normal $\vec{n} = (1,1,1)	imes(1,0,1) = (1, 0, -1)$. Plane $H_0: x - z = 0$. (P) $d(H_0) =$ distance of $(0,1,-1)\in\ell_2$ from $x-z=0$: $|0-(-1)|/\sqrt{2} = 1/\sqrt{2}$ — (5). (Q) $|0 - 2|/\sqrt{2} = \sqrt{2}$ — (4). (R) origin: $0$ — (3). (S) intersection of $y=z, x=1, x-z=0$: $x=1, z=1, y=1$, point $(1,1,1)$, distance $\sqrt{3}$ — (1).

JEE Advanced 2023 Paper 1 · Q16 · 3D Planes

Solution