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QJEEM23Maf14a: JEE Main 2023 Mathematics - Class 12 JEEM23Maf14a Solution: The shortest distance between the lines and is :

The shortest distance between the lines

x−51=y−22=z−4−3{{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}} and

x+31=y+54=z−1−5{{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}} is :

Multiple Choice Options

  • A 737\sqrt 3
  • B 535\sqrt 3
  • C 434\sqrt 3
  • D 636\sqrt 3

Step-by-Step Verified Solution

amp;L1:x−51=y−22=z−4−3amp;a1→=5i^+2j^+4k^amp;r1→=i^+2j^−3k^amp; L2:x+31=y+54=z−1−5amp;a2→=−3i^−5j^+k^amp;r2→=i^+4j^−5k^amp;r1→×r2→=∣i^amp;j^amp;k^1amp;2amp;−31amp;4amp;−5∣amp;=2i^+2j^+2k^\begin{aligned}& \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\& \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\& \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\\\& \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & -3 \\1 & 4 & -5\end{array}\right| \\\\& =2 \hat{i}+2 \hat{j}+2 \hat{k}\end{aligned}

amp; Shortest distance (d) =∣(r1→×r2→)⋅(a1→−a2→)∣r1→×r2→∣∣amp;=3623=63\begin{aligned}& \text { Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\\\& =\frac{36}{2 \sqrt{3}}=6 \sqrt{3}\end{aligned}

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