question_answer

Show that the lines \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] and \[\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{-\,2}\] are perpendicular.                       

Answer:

Given equations of lines are  \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] and \[\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{-\,2}.\] Then, DR's of given two lines are respectively \[{{a}_{1}}=2,\] \[{{b}_{1}}=3,\] \[{{c}_{1}}=4\] and \[{{a}_{2}}=1,\] \[{{b}_{2}}=2,\] \[{{c}_{2}}=-\,2\] Now, consider \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}\]             \[=2(1)+3(2)+4(-\,2)=2+6-8=8-8=0\]             \[\because \]       \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] Hence, the given lines are perpendicular.      Hence proved.