• # question_answer Find the shortest distance between lines $\frac{x-3}{1}=\frac{y-5}{-\,2}=\frac{z-7}{1}$ and       $\frac{x+1}{7}=\frac{y+1}{-\,6}=\frac{z+1}{1}.$

Given equations of lines are                         $\frac{x-3}{1}=\frac{y-5}{-\,2}=\frac{z-7}{1}$              ?(i)             and       $\frac{x+1}{7}=\frac{y+1}{-\,6}=\frac{z+1}{1}$                    ?(ii) On comparing above equations with one point form of equation of line which is             $\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c},$ we get ${{a}_{1}}=1,$ ${{b}_{1}}=-\,2,$ ${{c}_{1}}=1,$ ${{x}_{1}}=3,$             ${{y}_{1}}=5,$ ${{z}_{1}}=7$ and       ${{a}_{2}}=7,$ ${{b}_{2}}=-\,6,$             ${{c}_{2}}=1,$ ${{x}_{2}}=-\,1,$             ${{y}_{2}}=-\,1,$ ${{z}_{2}}=-\,1$ We know that the shortest distance between two lines is given by $\therefore$  $=\left| \frac{-\,4(-2\,+6)+6(1-7)-8(-\,6+14)}{\sqrt{{{(4)}^{2}}+{{(6)}^{2}}+{{(8)}^{2}}}} \right|$ $=\left| \frac{-\,4(4)+6(-6)-8(8)}{\sqrt{16+36+64}} \right|$ $=\left| \frac{-\,16-36-64}{\sqrt{116}} \right|$ $=\left| \frac{-\,116}{\sqrt{116}} \right|=\frac{116}{\sqrt{116}}$ $=\frac{{{(\sqrt{116})}^{2}}}{\sqrt{116}}=\sqrt{116}$ Hence, the required shortest distance is $\sqrt{116}$units.
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