Prove that the lines through `A(0,-1,-1)a n dB(4,5,1)`intersecrs the line through `C(3,9,2)a n dD(-4,4,4)dot`Also, find their point of intersection.
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We know that the cartesian equation of a line that passes through two points `(x_(1),y_(1),z_(1))` and `(x_(2),y_(2),z_(2))` is
`(x-x_(1))/(x_(2)-x_(1))=(y-y_(1))/(y_(2)-y_(1))=(z-z_(1))/(z_(2)-z_(1))`
Hence the cartesian equation of line passes through `A(0,-1,-1)` and `B(4,5,1)` is
`(x-0)/(4-0)=(y+1)/(5+1)=(z+1)/(1+1)`
`impliesx/4=(y+1)/6=(z+1)/2`……………..i
and cartesian equation of the line passes through `C(3,9,4)` and `D(-4,4,4)` is
`(x-3)/(-4-3)=(y-9)/(4-9)=(z-4)/(4-4)`
`implies (x-3)/(-7)=(y-9)/(-5)=(z-4)/0`………………ii
If the lines intersect, then shortest distance between both of them should be zero.
`:.` Shortest distance between the 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))` ltbgt `=(|(3-0,9+1,4+1),(4,6,2),(-7,-5,0)|)/(sqrt((6.0+10)^(2)+(-14-0)^(2)+(-20+42)^(2))`
`=(|(3,10,5),(4,6,2),(-7,-5,0)|)/(sqrt(100+196+484))`
`=(3(0+10)-10(14)+55(-20+42))/(sqrt(780))`
`=(30-140+110)/(sqrt(780))=0`
So, the given lines intersect.