The shortest distance between the lines
${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :
Solution
<p>${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$</p>
<p>${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$</p>
<p>Now, $$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 3 & { - 1} \cr
2 & 1 & 3 \cr
} } \right| = 10\widehat i - 8\widehat j - 4\widehat k$$</p>
<p>and $${\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k$$</p>
<p>$\therefore$ $$S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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