Mathematics · 3D Geometry

JEE Advanced 2024 — Paper 1 — Question 16

Let γ∈R\gamma \in 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 entries in List-II.

List - IList - II
(P) γ \gamma equals(1) −i^−j^+k^-\hat{i}-\hat{j}+\hat{k}
(Q) A possible choice for n^\hat{n} is(2) 32\sqrt{\frac{3}{2}}
(R) OR→1 \overrightarrow{O R}_{1} equals(3) 1
(S) A possible value of OR1→⋅n^\overrightarrow{\mathrm{OR}_{1}} \cdot \hat{n} is(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}}

The correct option is:

  1. Option A:

    (P) →\rightarrow (3)(Q) →\rightarrow (4)(R)→(1)(R) \rightarrow(1) (S)→(2)(S) \rightarrow(2)

  2. Option B:

    (\mathrm{P}) \rightarrow(5)$$(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1)$$(S) \rightarrow(2)

  3. Option C:

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

    Correct
  4. Option D:

    (\mathrm{P}) \rightarrow(3)$$(Q) \rightarrow(1)$$(R) \rightarrow(4)$$(S) \rightarrow(5)

Answer: C

Step-by-step solution

x+111=y+212=z+293=α \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=\alpha x+163=y+112=z+44=β\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{4}=\beta ⇒α−11=3β−16⇒α−3β=−5\Rightarrow \alpha-11=3 \beta-16 \Rightarrow \alpha-3 \beta=-5 2α−21=2β−11⇒α−β=52 \alpha-21=2 \beta-11 \Rightarrow \alpha-\beta=5 α=10,β=5\alpha=10, \beta=5 So, γ=1\gamma=1 n^=∣i^j^k123321∣=−4i^+8j^−4k^\hat{n}=\left|\begin{array}{lll}\hat{i} & \hat{j} & k\\ 1 & 2 & 3\\ 3 & 2 & 1\end{array}\right|=-4 \hat{i}+8 \hat{j}-4 \hat{k} n^=(16i−26j^+16k^)\hat{n}=\left(\frac{1}{\sqrt{6}} i-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}\right) R1(−1,−1,1)\mathrm{R}_{1}(-1,-1,1) OR→=(−i^−j^+k^)\overrightarrow{\mathrm{OR}}=(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) OR→⋅n^=23\overrightarrow{\mathrm{OR}} \cdot \hat{\mathrm{n}}=\sqrt{\frac{2}{3}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Straight Lines in 3D Geometry