JEE Advanced 2023 · Paper 1 · Mathematics Q16

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MathematicsList match+3 / -1Question 16 of 51

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 the 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.

The correct option is:

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
$\sqrt3$
2
$\dfrac{1}{\sqrt3}$
3
$0$
4
$\sqrt2$
5
$\dfrac{1}{\sqrt2}$

P → 2; Q → 4; R → 5; S → 1

P → 5; Q → 4; R → 3; S → 1

P → 2; Q → 1; R → 3; S → 2

P → 5; Q → 1; R → 4; S → 2

Answer and worked solution

Official answer: B

Choosing $H_0$ . A plane containing $\ell_1$ that is not parallel to $\ell_2$ meets $\ell_2$ , so $d(H)=0$ . The maximum is reached when $H$ is parallel to $\ell_2$ , with normal
$(1,1,1)\times(1,0,1)=(1,0,-1),\qquad H_0:\ x-z=0.$

(P) Use the point $(0,1,-1)$ of $\ell_2$ : $d(H_0)=\frac{|0+1|}{\sqrt2}=\frac1{\sqrt2}$ → 5.

(Q) For $(0,1,2)$ : $\frac{|0-2|}{\sqrt2}=\sqrt2$ → 4.

(R) $H_0$ passes through the origin, so the distance is 0 → 3.

(S) Solving $y=z$ , $x=1$ , $x=z$ gives $(1,1,1)$ , at distance $\sqrt3$ → 1.

P→5, Q→4, R→3, S→1.

Topics:Equation of a plane Angles and distances Skew lines

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JEE Advanced 2023 · Paper 1, Mathematics question 16