Question Number 96089 by bebo last updated on 29/May/20
Answered by mr W last updated on 30/May/20
Commented by mr W last updated on 30/May/20
$$\boldsymbol{{METHOD}}\:\boldsymbol{{I}} \\ $$$${point}\:{A}\:{on}\:{line}\:\mathrm{1}: \\ $$$${x}=\mathrm{3}+\mathrm{3}{s} \\ $$$${y}=\mathrm{8}−{s} \\ $$$${z}=\mathrm{3}+{s} \\ $$$${point}\:{B}\:{on}\:{line}\:\mathrm{2}: \\ $$$${x}=−\mathrm{3}−\mathrm{3}{t} \\ $$$${y}=−\mathrm{7}+\mathrm{2}{t} \\ $$$${z}=\mathrm{6}+\mathrm{4}{t} \\ $$$${square}\:{of}\:{distance}\:{between}\:{A}\:\&\:{B}: \\ $$$${D}={d}^{\mathrm{2}} =\left(\mathrm{3}+\mathrm{3}{s}+\mathrm{3}+\mathrm{3}{t}\right)^{\mathrm{2}} +\left(\mathrm{8}−{s}+\mathrm{7}−\mathrm{2}{t}\right)^{\mathrm{2}} +\left(\mathrm{3}+{s}−\mathrm{6}−\mathrm{4}{t}\right)^{\mathrm{2}} \\ $$$${D}=\mathrm{9}\left({s}+{t}+\mathrm{2}\right)^{\mathrm{2}} +\left({s}+\mathrm{2}{t}−\mathrm{15}\right)^{\mathrm{2}} +\left({s}−\mathrm{4}{t}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${such}\:{that}\:{d}\:{is}\:{minimum}: \\ $$$$\frac{\partial{D}}{\partial{s}}=\mathrm{18}\left({s}+{t}+\mathrm{2}\right)+\mathrm{2}\left({s}+\mathrm{2}{t}−\mathrm{15}\right)+\mathrm{2}\left({s}−\mathrm{4}{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{11}{s}+\mathrm{7}{t}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial{D}}{\partial{t}}=\mathrm{18}\left({s}+{t}+\mathrm{2}\right)+\mathrm{4}\left({s}+\mathrm{2}{t}−\mathrm{15}\right)−\mathrm{8}\left({s}−\mathrm{4}{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{7}{s}+\mathrm{29}{t}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}: \\ $$$${s}=\mathrm{0},\:{t}=\mathrm{0} \\ $$$$\Rightarrow{point}\:{on}\:{line}\:\mathrm{1}\:{is}\:{A}\left(\mathrm{3},\mathrm{8},\mathrm{3}\right) \\ $$$$\Rightarrow{point}\:{on}\:{line}\:\mathrm{2}\:{is}\:{B}\left(−\mathrm{3},−\mathrm{7},\mathrm{6}\right) \\ $$$$ \\ $$$${the}\:{distance}\:{between}\:{them}\:{is}\:{the} \\ $$$${shortest}\:{distance}\:{between}\:{both}\:{lines}: \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{3}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{7}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{6}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{30}} \\ $$$$ \\ $$$${eqn}.\:{of}\:{the}\:{shortest}\:{distance}\:{is}\:{AB}: \\ $$$$\frac{{x}−\mathrm{3}}{\mathrm{6}}=\frac{{y}+\mathrm{7}}{\mathrm{15}}=\frac{{z}−\mathrm{6}}{−\mathrm{3}} \\ $$$${or} \\ $$$$\frac{{x}−\mathrm{3}}{\mathrm{2}}=\frac{{y}+\mathrm{7}}{\mathrm{5}}=\frac{{z}−\mathrm{6}}{−\mathrm{1}} \\ $$
Commented by mr W last updated on 30/May/20
$$\boldsymbol{{METHOD}}\:\boldsymbol{{II}} \\ $$$$\boldsymbol{{l}}_{\mathrm{1}} =\left(\mathrm{3},−\mathrm{1},\mathrm{1}\right) \\ $$$$\boldsymbol{{l}}_{\mathrm{2}} =\left(−\mathrm{3},\mathrm{2},\mathrm{4}\right) \\ $$$$\boldsymbol{{l}}_{\mathrm{3}} ={l}_{\mathrm{1}} ×{l}_{\mathrm{2}} =\left(−\mathrm{6},−\mathrm{15},\mathrm{3}\right) \\ $$$${P}\left(\mathrm{3},\mathrm{8},\mathrm{3}\right),\:{Q}\left(−\mathrm{3},−\mathrm{7},\mathrm{6}\right) \\ $$$$\boldsymbol{{PQ}}=\left(\mathrm{6},\mathrm{15},−\mathrm{3}\right) \\ $$$${d}=\frac{\boldsymbol{{PQ}}\centerdot\boldsymbol{{l}}_{\mathrm{3}} }{\mid\boldsymbol{{l}}_{\mathrm{3}} \mid} \\ $$$$=\frac{\left(\mathrm{6},\mathrm{15},−\mathrm{3}\right)\centerdot\left(−\mathrm{6},−\mathrm{15},\mathrm{3}\right)}{\:\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }}=\mathrm{3}\sqrt{\mathrm{30}} \\ $$
Answered by Sourav mridha last updated on 30/May/20