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The shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is

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RPSC 2nd Grade Mathematics (Held on 30th June 2017) Official Paper
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  1. \(\frac{1}{6}\)
  2. \(\frac{1}{\sqrt6}\)
  3. \(\frac{1}{\sqrt3}\)
  4. \(\frac{1}{3}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{1}{\sqrt6}\)
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Detailed Solution

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Concept Used:

\( \frac{x-a_1}{x_1}=\frac{y-a_2}{y_1}=\frac{2-a_3}{z_1} \text { and } \frac{x-b_1}{x_2}=\frac{y-b_2}{y_2}=\frac{2-b_3}{z_2}\)

here, \( \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
\( \vec{n_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{n_2}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\)
 
then d= \( \frac{\left|\left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right) \cdot(\vec{a}-\vec{b})\right|}{\left|\overrightarrow{n_1} \times \overrightarrow{n_2}\right|}\) [where d has shortened the distance.

Calculation:

\(\vec{n_1}=2 \hat{i}+3 \hat{j}+4 \hat{k} \text { and } \vec{n_2}=3 \hat{i}+4 \hat{j}+5 \hat{k} \)

\( \vec{a}=1 \hat{i}-2\hat{j}+3\hat{k} \text { and } \vec{b} =2 \hat{i}+4 \hat{j}+5 \hat{k} \)
 
\( \Rightarrow \vec{n_1} \times \vec{n_2}=\left|\begin{array}{ccc} \hat{i}&\hat{j}&\hat{k}\\ 2 & 3 & 4\\ 3 & 4 & 5 \end{array}\right|=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)\)
\( \Rightarrow \vec{n_1} \times \vec{n_2}=- \hat{i}+2 \hat{j}- \hat{k} \)
 
\( \Rightarrow \left|\vec{n_1} \times \vec{n_2}\right|=\sqrt{(-1)^2+(2)^2+(-1)^2}= \sqrt{6} \)
\( \Rightarrow \left(\vec{n_1} \times \vec{n_2}\right) \cdot(\vec{a}-\vec{b})=(- \hat{i}+2 \hat{j}- \hat{k} ) \cdot(- \hat{i}-2 \hat{j}-2 \hat{k}) \)
\( \Rightarrow \left(\vec{n_1} \times \vec{n_2}\right) \cdot(\vec{a}-\vec{b})=1-4+2=-1\)
 
\( \Rightarrow d=\left|\frac{-1}{ \sqrt{6}}\right| = \frac{1}{ \sqrt{6}}\)
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