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JEE Advanced 2024 · Paper 1

JEE Advanced 2024 Mathematics question on Three Dimensional Geometry

Let γ∈R\gamma \in \mathbb{R} be such that the lines L1:x+111=y+212=z+293L_1 : \frac{x+11}{1} = \frac{y+21}{2} = \frac{z+29}{3} and L2:x+163=y+112=z+4γL_2 : \frac{x+16}{3} = \frac{y+11}{2} = \frac{z+4}{\gamma} intersect. Let R1R_1 be the point of intersection of L1L_1 and L2L_2. Let O=(0,0,0)O = (0,0,0), and n^\hat{n} denote a unit normal vector to the plane containing both the lines L1L_1 and L2L_2.

Match each entry in List-I to the correct entry in List-II.
List-I
(P) γ\gamma equals
(Q) A possible choice for n^\hat{n} is
(R) OR1→\overrightarrow{OR_1} equals
(S) A possible value of OR1→⋅n^\overrightarrow{OR_1} \cdot \hat{n} is
List-II
(1) −i^−j^+k^-\hat{i} - \hat{j} + \hat{k}
(2) 32\sqrt{\frac{3}{2}}
(3) 1
(4) 16i^−26j^+16k^\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}
(5) 23\sqrt{\frac{2}{3}}
  1. A.
    (P)→(3)(Q)→(4)(R)→(1)(S)→(2)(P)\to(3)\quad (Q)\to(4)\quad (R)\to(1)\quad (S)\to(2)
  2. B.
    (P)→(5)(Q)→(4)(R)→(1)(S)→(2)(P)\to(5)\quad (Q)\to(4)\quad (R)\to(1)\quad (S)\to(2)
  3. C.
    (P)→(3)(Q)→(4)(R)→(1)(S)→(5)(P)\to(3)\quad (Q)\to(4)\quad (R)\to(1)\quad (S)\to(5)
  4. D.
    (P)→(3)(Q)→(1)(R)→(4)(S)→(5)(P)\to(3)\quad (Q)\to(1)\quad (R)\to(4)\quad (S)\to(5)
Show answer

Correct option: C

[JEE Advanced 2024 · Paper 1]

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