Q33 / 51 · JEE Advanced 2023 (Paper 1) · Mathematics

Let ℓ1 and ℓ2 be the lines r→1=λi^+j^+k^ and r→2=j^-k^+μi^+k^, respectively, Let X be the set of all the planes H that contain the line ℓ1. For a plane H, let dH denote the smallest possible distance between the points of ℓ2 and H. Let H0 be a plane in X for which dH0 is the maximum value of dH 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 dH0 is 1 3
Q The distance of the point 0, 1, 2 from H0 is 2 13
R The distance of origin from H0 is 3 0
S The distance of origin from the point of
intersection of planes y=z, x=1 and H0 is
4 2
    5 12

The correct option is

  • A
    P→2 Q→4 R→5 S→1
  • Correct answer
    B
    P→5 Q→4 R→3 S→1
  • C
    P→2 Q→1 R→3 S→2
  • D
    P→5 Q→1 R→4 S→2

Solution

Given,

H0 will be the plane containing the line ℓ1 and parallel to ℓ2.

So, the normal vector of plane parallel to ℓ1 and ℓ2 is given by,

i^j^k^111101=j^1-j^1-1+k^-1=i^-k^

Hence, the equation of plane H0  will be,

H0 : x-z=C which passes through origin,

So,C=0

∴ H0 : x-z=0

Now solving,

P dH0=1 distance of point 0, 1, -1 from H.

d=0--12=12 ∴ P→5

Q d=0-22=2 ∴ Q→4

R d=02=0 ∴ R→3

S Point of intersection will be of given planesy=z, x=1 & x-z=0  will be, 1, 1, 1 

Hence, distance d=1+1+1=3 ∴ S→1

∴ Option (B) is correct.

Hence, the correct answer is option: 'B'.

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