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Question
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
Options
\[\sqrt{30}\]
\[2\sqrt{30}\]
\[5\sqrt{30}\]
\[3\sqrt{30}\]
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Solution
\[3\sqrt{30}\]
We have ,
\[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} . . . (1) \]
\[\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} . . . (2)\]
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3,-1 , 1 .
Its vector equation is ,
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} , \text{ where }\overrightarrow{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \text{ and } \overrightarrow{b_1} = 3 \hat{i} - \hat{j} + \hat{k} .\]
Also, line (2) passes through the point ( -3 , -7 , 6 ) and has direction ratios proportional to -3, 2 4 .
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} , \text{ where } \overrightarrow{a_2} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \text{ and } \overrightarrow{b_2} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} .\]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4\end{vmatrix}\]
\[ = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 6 \right)^2 + \left( - 15 \right)^2 + 3^2}\]
\[ = \sqrt{36 + 225 + 9}\]
\[ = \sqrt{270}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right) . \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right)\]
\[ = 36 + 225 + 9\]
\[ = 270\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{270}{\sqrt{270}} \right|\]
\[ = \sqrt{270}\]
\[ = 3\sqrt{30}\]
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