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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2024 · Paper 1 · Mathematics · Three Dimensional Geometry

Let γ ∈ ℝ be such that the lines L_1: (x+11)/(1)= y+21 2 = z+29 3...

Let γ∈R\gamma \in \mathbb{R} be such that the lines L1:x+111=y+212=z+293\mathrm{L}_1: \frac{x+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{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 R1\mathrm{R}_1 be the point of intersection of L1\mathrm{L}_1 and L2\mathrm{L}_2. Let O=(0,0,0)\mathrm{O}=(0,0,0), and n^\hat{\mathrm{n}} denote a unit normal vector to the plane containing both the lines L1\mathrm{L}_1 and L2\mathrm{L}_2. Match each entry in List-I to the correct entry in List-II. Let γ ∈ ℝ be such that the lines L_1: (x+11)/(1)= y+21 2 = z+29 3 and L_2: (x+16)/(3)=(y+11)/(2)=(z+4)/(γ) intersect. The correct option is

Options

  1. A(P)→(3)(Q)→(4)(R)→(1)(S)→(2)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)
  2. B(P)→(5)(Q)→(4)(R)→(1)(S)→(2)(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)
  3. C(P)→(3)(Q)→(4)(R)→(1)(S)→(5)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)
  4. D(P)→(3)(Q)→(1)(R)→(4)(S)→(5)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(1) \quad(\mathrm{R}) \rightarrow(4) \quad(\mathrm{S}) \rightarrow(5)

Correct answer

C. (P)→(3)(Q)→(4)(R)→(1)(S)→(5)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)

Step-by-step solution

L1:x+111=y+212=z+293=aL2:x+163=y+112=z+4γ=b\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{z}+29}{3}=\mathrm{a} \\ & \mathrm{L}_2: \frac{\mathrm{x}+16}{3}=\frac{\mathrm{y}+11}{2}=\frac{\mathrm{z}+4}{\gamma}=\mathrm{b}\end{aligned}x=a−11=3b−16⇒a−3b=−5...(1)y=2a−21=2b−11⇒2a−2b=10...(2)z=3a−29=br−4⇒3a−bγ=25...(3)\begin{aligned} & x=a-11=3 b-16 \Rightarrow a-3 b=-5...(1) \\ & y=2 a-21=2 b-11 \Rightarrow 2 a-2 b=10...(2) \\ & z=3 a-29=b r-4 \Rightarrow 3 a-b \gamma=25...(3)\end{aligned}from (1) & (2)\mathrm{a}=10, \mathrm{~b}=5Nowfrom(3)Now from (3)3(10)-5 \gamma=25 \quad \therefore \gamma=1 \begin{aligned} & \mathrm{R}_1 \equiv(-1,-1,1) \\ & \mathrm{OR}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{n}}=\left|\begin{array}{lrr}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right|=-4 \hat{\mathrm{i}}-(-8) \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{n}}=-4 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}=-4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \hat{\mathrm{n}}= \pm \frac{4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{4 \sqrt{6}}= \pm \frac{(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{6}} \\ & \overrightarrow{\mathrm{OR}_1 \cdot \hat{\mathrm{n}}= \pm(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\left(\frac{\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}}\end{aligned}
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