Show that the two lines`(x-1)/2=(y-2)/3=(z-3)/4 nd (x-4)/5=(y-1)/z=z` interect. Find also the point of intersection of these lines.

Show that the two lines`(x-1)/2=(y-2)/3=(z-3)/4 nd (x-4)/5=(y-1)/z=z`

1.	Show that the two lines`(x-1)/2=(y-2)/3=(z-3)/4 nd (x-4)/5=(y-1)/z=z` interect. Find also the point of intersection of these lines.
Answer» We have `x_(1)=1,y_(1)=2,z_(1)=3` and `a_(1)=2,b_(1)=3,c_(1)=4` Also `x_(2)=4,y_(2)=1,z_(2)=0` and `a_(2)=5,b_(2)=2,c_(2)=1` If two lines intersect then shortest distance between them should be zero. `:.` Shortest distance between two given lines `=(\|(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2))\|)/(sqrt((b_(1)c_(2)-b_(2)c_(1))^(2)+(c_(1)a_(2)-c_(2)a_(1))^(2)+(a_(1)b_(2)-a_(2)b_(1))^(2))` `(\|(4-1,1-2,0-3),(2,3,4),(5,2,1)\|)/(sqrt((3.1-2.4)^(2)+(4.5-1.2)^(2)+(2.2-5.3)^(2))` `=(\|(3,-1,-3),(2,3,4),(5,2,1)\|)/(sqrt(25+34+121))` `=(3(3-8)+1(2-20)-3(4-15))/(sqrt(470))` `=(-15-18+33)/(sqrt(470))=0/(sqrt(470))=0` Therefore the given two lines are intersecting. For finding their point of intersection for the first line `(x-1)/2=(y-2)/3=(z-3)/4=lamda` `implies x=2lamda+1,y=3lamda+2` and `z=4lamda+3` Since the lines are intersecting. So, let us put these values in the equation of another line. Thus, `(2lamda+1-4)/5=(3lamda+2-1)/2=(4lamda+3)/1` `implies (2lamda-3)/5=(3lamda+1)/2=(4lamda+3)/1` `implies (2lamda-3)/5=(4lamda+3)/1` `implies 2lamda-3=20lamda+15` `implies 18lamda=-18=-1` So the required point of intersection is `x=2(-1)+1=-1` `y=3(-1)+2=-1` `z=4(-1)+3=-1` Thus, the ines intersect at `(-1,-1,-1)`.