InterviewSolution
Saved Bookmarks
| 1. |
Find the length and the equations of the line of shortest distance between the lines `(x-8)/(3)=(y+9)/(-16) =(z-10)/(7) " and " (x-15)/(3)=(y-29)/(8)=(z-5)/(-5)` |
|
Answer» The equations of the given lines are `(x-8)/(3)=(y+9)/(-16) =(z-10)/(7)=lambda`(say) `(x-15)/(3)=(y-29)/(8) =(z-5)/(-5)=mu `(say) Any point on the line (i) is `P(3lambda +8,-16lambda -9,7lambda +10)` Any point on the line(ii) is `Q( 3mu+ 15,8mu +29 ,- 5mu +5)` The direction rations of PQ are `(3mu -3lambda +7, 8mu+ 16lambda +38 , -5mu -7lambda -5)` Now PQ will be the shortest distance between (i) and (ii) only when PQ is perpendicular to each one of (i) and (ii) `:.{underset(3(3mu -3lambda+7) +8(8mu+16lambda+38)-5(-5mu-7lambda-5)=0)underset(" and ")(3(3mu -3lambda +7) -16(8mu +16lambda+38)+7(-5mu-7lambda-5)=0)` `rArr {underset(154lambda +98mu +350=0 )(314lambda +154mu+622 =0)rArr {underset(77lambda +49mu +175 =0)(157lambda +77mu +311=0)` On multiplying (iii) by 7 and (iv) by 11 and subtracting , we get `(1099 lambda -847 lambda) +(2177- 1925) =0 rArr 252 lambda=-252 rArr lambda =-1` Putting `lambda=-1 ` in (iv) we get `49 mu +(175 -77) =0 rArr 49 mu +98 =0 rArr 49mu =-98 rArr mu =-2` Now `lambda =-1` gives P (5,7,3) and `mu =-2 ` gives Q(9,13,15) `:.SD =|PQ| =sqrt((9-5)^(2)+(13-7)^(2) +(15 -3)^(2))` `=sqrt(16+36 +144)=sqrt(196) =14` units Equations of the line of shortest distance is the equation of PQ given by `(x-5)/(9-5) =(y-7)/(13-7) =(z-3)/(15-3)` `rArr (x-5)/(4)=(y-7)/(6)=(z-3)/(12)rArr (x-5)/(2)=(y-7)/(3)=(z-3)/(6)` Hence the equations of the line of shortest distance is `(x-5)/(2)=(y-7)/(3)=(z-3)/(6)` |
|