The shortest distance between the lines ${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$ and ${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$ is :

<p>${\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k$</p> <p>${\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k$</p> <p>$\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k$</p> <p>$\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k$</p> <p>S.D. $$ = {{\left| {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right|} \over {\left| {\overrightarrow a \times \overrightarrow b } \right|}}$$</p> <p>$$\left[ {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right] = \left| {\matrix{ 0 & { - 10} & { - 2} \cr 2 & { - 7} & 5 \cr 2 & 1 & { - 3} \cr } } \right|$$</p> <p>$\therefore$ $10( - 16) - 2(16) = - 192$</p> <p>$$\left| {\left[ {\matrix{ {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & {\matrix{ {\overrightarrow a } & {\overrightarrow b } \cr } } \cr } } \right]} \right| = 192$$</p> <p>$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 7} & 5 \cr 2 & 1 & { - 3} \cr } } \right| = 16\widehat i + 16\widehat j + 16\widehat k$$</p> <p>$\overrightarrow a \times \overrightarrow b = 16\sqrt 3$</p> <p>S.D. $= {{192} \over {16\sqrt 3 }} = 4\sqrt 3$</p>