Verified Solution • 2025–26 Session moderate

QJEEM23Mabb87: JEE Main 2023 Mathematics - Class 12 JEEM23Mabb87 Solution: The shortest distance between the lines and is :

The shortest distance between the lines x+21=y−2=z−52{{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2} and x−41=y−12=z+30{{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0} is :

Multiple Choice Options

  • A 8
  • B 7
  • C 6
  • D 9

Step-by-Step Verified Solution

Given, the lines are

x+21amp;=y−2=z−52    ..........(i) and x−41amp;=y−12=z+30    ..........(ii)\begin{aligned}\frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\\text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii)\end{aligned}

Formula for shortest distance between two skew-lines,

SDamp;=∣∣x2−x1amp;y2−y1amp;z2−z1a1amp;b1amp;c1a2amp;b2amp;c2∣∣i^amp;j^amp;k^a1amp;b1amp;c1a2amp;b2amp;c2∣∣=∣∣6amp;1amp;−81amp;−2amp;21amp;2amp;0∣∣i^amp;j^amp;k^1amp;−2amp;21amp;2amp;0∣∣amp;=∣6(−4)−1(−2)−8(4)∣−4i^+2j^+4k^∣∣amp;=∣−24+2−3236∣amp;=∣−546∣=∣−9∣=9\begin{aligned}S D & =\left|\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\a_1 & b_1 & c_1 \\a_2 & b_2 & c_2\end{array}\right|}{\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\a_1 & b_1 & c_1 \\a_2 & b_2 & c_2\end{array}\right|}\right|=\left|\frac{\left|\begin{array}{ccc}6 & 1 & -8 \\1 & -2 & 2 \\1 & 2 & 0\end{array}\right|}{\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & -2 & 2 \\1 & 2 & 0\end{array}\right|}\right| \\\\& =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}|}\right| \\\\& =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\\\& =\left|\frac{-54}{6}\right|=|-9|=9\end{aligned}

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