Find the shortest distance between the lines `(x-1)/2=(y-2)/3=(z-3)/4a n d(x-2)/3=(y-4)/4=(z-5)/5` .

Explanation

(i) The two given lines are `(x-1)/(2)=(y-2)/(3)=(z-3)/(4)=r_(1)" (say)"" "`(i)
and `" "(x-2)/(3)=(y-4)/(4)=(z-3)/(5)=r_(2)" (say)"" "`(ii)
Any point on (i) is given by `P(2r_(1)+1, 3r_(1)+2, 4r_(1)+3)`
And any point on (ii) is given by `Q(3r_(2)+2, 4r_(2)+4, 5r_(2)+5)`
Direction ratios of PQ are given by `3r_(2)-2r_(1)+1, 4r_(2)-3r_(1)+2 and 5r_(2)-4r_(1)+2`
Since PQ is perpendicular to (i), we get
`" "2(3r_(2)-2r_(1)+1)+3(4r_(2)-3r_(1)+2)+4(5r_(2)-4r_(1)+2)=0`
or `" "38r_(2)-29r_(1)+16=0" "`(iii) ltBrgt Also PQ is perpendicular to (ii), we get
`" "3(3r_(2)-2r_(1)+1)+4(4r_(2)-3r_(1)+2)+5(5r_(2)-4r_(1)+2)=0`
or `" "50r_(2)-38r_(1)+21=0`
Solving (iii) and (iv), we obtain `r_(2)`=`-(1//6), r_(1)=(1//3)`
Therefore, coordinates of P and Q are `((5)/(3),3, (13)/(3)) and ((3)/(2), (10)/(3), (25)/(6)), ` respectively. Thus,
`" "PQ^(2)=((3)/(2)-(5)/(3))^(2)+((10)/(3)-3)^(2)+((25)/(6)-(13)/(3))^(2) = (-(1)/(6))^(2)+((1)/(3))^(2)+(-(1)/(6))^(2)=(1)/(6)`
or `" "PQ=(1)/(sqrt(6))`
The equation of the line of the shortest distance is given by
`" "(x-(5//3))/((3//2)-(5//3))=(y-3)/((10//3)-3)=(z-(13//3))/((25//6)-(13//3))`
`" "(x-(5//3))/(-(1//6))=(y-3)/((1//3))=(z-(13//3))/(-(1//6))`
`" "(x-(5//3))/(1)=(y-3)/(-2)=(z-(13//3))/(1)`