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प्रश्न
Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
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उत्तर
\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
Since the first line passes through the point (-1,-1,-1) and has direction ratios proportional to 7, -6 , 1 , its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[\text{ Here }, \]
\[ \overrightarrow{a_1} = - \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = 7 \hat{i} - 6 \hat{j} + \hat{k} \]
Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1, -2,1.Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[\text{ Here }, \]
\[ \overrightarrow{a_2} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} - 2 \hat{j} + \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 7 & - 6 & 1 \\ 1 & - 2 & 1\end{vmatrix}\]
\[ = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 8 \right)^2}\]
\[ = \sqrt{16 + 36 + 64}\]
\[ = \sqrt{116}\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \right) . \left( - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \right)\]
\[ = - 16 - 36 - 64\]
\[ = - 116\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 116}{\sqrt{116}} \right|\]
\[ = \sqrt{116}\]
\[ = 2\sqrt{29}\]
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