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Mathematics · 3D Geometry

JEE Advanced 2023 — Paper 1 — Question 4

Let QQ be the cube with the set of vertices {(x1,x2,x3)∈R3:x1,x2,x3∈{0,1}}\left\{\left(x_{1}, x_{2}, x_{3}\right) \in R^{3}: x_{1}, x_{2}, x_{3} \in\{0,1\}\right\}. Let FF be the set of all

twelve lines containing the diagonals of the six faces of the cube Q . Let S be the set of all four lines containing the

main diagonals of the cube Q ; for instance, the line passing through the vertices (0,0,0)(0,0,0) and (1,1,1)(1,1,1) is in S .

For lines ℓ1\ell_{1} and ℓ2\ell_{2}, let d(ℓ1,ℓ2)\mathrm{d}\left(\ell_{1}, \ell_{2}\right) denote the shortest distance between them. Then the maximum value of

d(ℓ1,ℓ2)\mathrm{d}\left(\ell_{1}, \ell_{2}\right) as ℓ1\ell_{1} varies over F and ℓ2\ell_{2} varies over S , is

  1. Option A:

    16\frac{1}{\sqrt{6}}

    Correct
  2. Option B:

    18\frac{1}{\sqrt{8}}

  3. Option C:

    13\frac{1}{\sqrt{3}}

  4. Option D:

    112\frac{1}{\sqrt{12}}

Answer: A

Step-by-step solution

DR's of OG→=(1,1,1)\overrightarrow{\mathrm{OG}}=(1,1,1) DR's of AC→=(−1,1,0)\overrightarrow{\mathrm{AC}}=(-1,1,0) Equation of OG→=x1=y1=z1\overrightarrow{\mathrm{OG}}=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{1} Equation of AC→=x−1−1=y1=z0\overrightarrow{\mathrm{AC}}=\frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{0} OA→=i^\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}} Normal of OG→\overrightarrow{\mathrm{OG}} and AC→\overrightarrow{\mathrm{AC}} =∣i^j^k^111−110∣=(−i^−j^+2k^)=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 0\end{array}\right|=(-\hat{i}-\hat{j}+2 \hat{k}) S.D. =∣i^(−i^−j^+2k^)∣∣−i^−j^+2k^∣=16=\frac{|\hat{i}(-\hat{i}-\hat{j}+2 \hat{k})|}{|-\hat{i}-\hat{j}+2 \hat{k}|}=\frac{1}{\sqrt{6}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

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