Verified Solution • 2025–26 Session moderate

QJEEM23Mad523: JEE Main 2023 Mathematics - Class 12 JEEM23Mad523 Solution: The shortest distance between the lines and is :

The shortest distance between the lines x−44=y+25=z+33\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3} and x−13=y−34=z−42\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2} is :

Multiple Choice Options

  • A 363 \sqrt{6}
  • B 626 \sqrt{2}
  • C 636 \sqrt{3}
  • D 262 \sqrt{6}

Step-by-Step Verified Solution

The given lines are

x−44=y+25=z+33\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3} and x−13=y−34=z−42\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}

amp; So, b⃗1=4i^+5j^+3k^amp;b⃗2=3i^+4j^+2k^amp;a⃗1=4i^−2j^−3k^amp;a⃗2=i^+3j^+4k^\begin{aligned}& \text { So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\& \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\& \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\&\vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k}\end{aligned}

amp;∴b⃗1×b⃗2=∣i^amp;j^amp;k^4amp;5amp;33amp;4amp;2∣amp;=(10−12)i^−(8−9)j^+(16−15)k^amp;=−2i^+j^+k^\begin{aligned}& \therefore \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\4 & 5 & 3 \\3 & 4 & 2\end{array}\right| \\\\& =(10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k} \\\\& =-2 \hat{i}+\hat{j}+\hat{k}\end{aligned}

amp; Shortest distance, d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣amp;=∣(3i^−5j^−7k^)⋅(−2i^+j^+k^)4+1+1∣amp;=∣−6−5−76∣=186=36 units \begin{aligned}& \text { Shortest distance, } d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\\\& =\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right| \\\\& =\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6} \text { units }\end{aligned}

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