Mathematics · 3D Geometry

JEE Main 2024 — 9 April, Shift 1 — Question 15

The shortest distance between the line x−34=y+7−11=z−15\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5} and x−53=y−9−6=z+21\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1} is :

  1. Option A:

    187563\frac{187}{\sqrt{563}}

    Correct
  2. Option B:

    178563\frac{178}{\sqrt{563}}

  3. Option C:

    185563\frac{185}{\sqrt{563}}

  4. Option D:

    179563\frac{179}{\sqrt{563}}

Answer: A

Step-by-step solution

figure

n→=p→×q→\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}

n⃗=∣i j k 4−1153−61∣=19i +11j +9k \vec{n}=\left| \begin{matrix}\overset{}{\mathop{i}}\, & \overset{}{\mathop{j}}\, & \overset{}{\mathop{k}}\, \\4 & -11 & 5 \\3 & -6 & 1 \\\end{matrix} \right|=19\overset{}{\mathop{i}}\,+11\overset{}{\mathop{j}}\,+9\overset{}{\mathop{k}}\,

S.d. == projection of AB→\overrightarrow{\mathrm{AB}} on n⃗\vec{n}

=∣AB→⋅n⃗∣n⃗∣∣=∣(2i^+16j^−3k^)⋅(19i^+11j^+9k^)361+121+81∣=\left|\frac{\overrightarrow{A B} \cdot \vec{n}}{|\vec{n}|}\right|=\left|\frac{(2 \hat{i}+16 \hat{j}-3 \hat{k}) \cdot(19 \hat{i}+11 \hat{j}+9 \hat{k})}{\sqrt{361+121+81}}\right|

=38+176−27563=\frac{38+176-27}{\sqrt{563}}

S.d. =187563=\frac{187}{\sqrt{563}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them