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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2023 · 1 February · Morning Shift · Mathematics · Three Dimensional Geometry

The shortest distance between the lines (x - 5)/(1) = (y - 2)/(2) =...

The shortest distance between the lines x−51=y−22=z−4−3\dfrac{x - 5}{1} = \dfrac{y - 2}{2} = \dfrac{z - 4}{- 3} and x+31=y+54=z−1−5\dfrac{x + 3}{1} = \dfrac{y + 5}{4} = \dfrac{z - 1}{- 5} is

Options

  1. A737\sqrt{3}
  2. B535\sqrt{3}
  3. C636\sqrt{3}
  4. D434\sqrt{3}

Correct answer

C. 636\sqrt{3}

Step-by-step solution

Shortest distance between two lines

x−x1a1=y−y1b1=z−z1c1\dfrac{x - {x}_{1}}{{a}_{1}} = \dfrac{y - {y}_{1}}{{b}_{1}} = \dfrac{z - {z}_{1}}{{c}_{1}} and x−x2a2=y−y2b2=z−z2c2\dfrac{x - {x}_{2}}{{a}_{2}} = \dfrac{y - {y}_{2}}{{b}_{2}} = \dfrac{z - {z}_{2}}{{c}_{2}} is given as

\dfrac{\left|\begin{cases}{x}_{1} - {x}_{2} & {y}_{1} - {y}_{2} & {z}_{1} - {z}_{2} \\ {a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2}\end{cases}\right|}{\sqrt{{\left({a}_{1}{b}_{2} - {a}_{2}{b}_{1}\right)}^{2} + {\left({b}_{1}{c}_{2} - {b}_{2}{c}_{1}\right)}^{2} + {\left({c}_{1}{a}_{2} - {c}_{2}{a}_{1}\right)}^{2}}}

= \dfrac{\left|\begin{cases}5 - \left(- 3\right) & 2 - \left(- 5\right) & 4 - 1 \\ 1 & 2 & - 3 \\ 1 & 4 & - 5\end{cases}\right|}{\sqrt{{\left(4 - 2\right)}^{2} + {\left(- 10 + 12\right)}^{2} + {\left(- 3 + 5\right)}^{2}}}

= \dfrac{\left|\begin{cases}8 & 7 & 3 \\ 1 & 2 & - 3 \\ 1 & 4 & - 5\end{cases}\right|}{\sqrt{{\left(2\right)}^{2} + {\left(2\right)}^{2} + {\left(2\right)}^{2}}}

=∣8(−10+12)−7(−5+3)+3(4−2)∣4+4+4= \dfrac{\left|8\left(- 10 + 12\right) - 7\left(- 5 + 3\right) + 3\left(4 - 2\right)\right|}{\sqrt{4 + 4 + 4}}

=16+14+612=3623= \dfrac{16 + 14 + 6}{\sqrt{12}} = \dfrac{36}{2\sqrt{3}}

=183=63= \dfrac{18}{\sqrt{3}} = 6\sqrt{3}

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