Question

find the shortest distance between the lines x-3/1=y-5/-2=z-7/1 and x+1/7=y+1/-6=z+1/1

Answer 1

So...I'm having trouble reading your equations.

Answer 2

Equations of lines are
\[\frac{ x-3 }{1 }=\frac{ y-5 }{ -2 }=\frac{ z-7 }{1 }=r1 (say) ...(1)\]
and \[\frac{ x+1 }{ 7 }=\frac{ y+1 }{ -6}=\frac{ z+1 }{1 }=r2 (say) ....(2) \]
Any point P on (1) is (r1+3,-2r1+5,r1+7)
Any point Q on (2) is (7r2-1,-6r2-1,r2-1)
Let P and Q be the shortest points,
D,r's of PQ are <x2-x1,y2-y1,z2-z1>

Answer 3

D.r's are <1,-2,1>
D,r's of (2) are <7,-6,1 >

Answer 4

use the property that shortest distance is perpendicular to both lines
and find two equations in r1,r2 and solve to find r1,r2
then find P and Q
\[then PQ=\sqrt{\left( x2-x1 \right)^{2}+\left( y2-y1 \right)^{2}+\left( z2-z1 \right)^{2 }}\]

Answer 5

why is there 1 on bottom? that is weird to me.

Answer 6

did you just do that because it was easy for you to see certain things about the problem?

Answer 7

3,5,7 are the coordinates of any points on line (1) and 1,-2 and 1 are direction ratios

Answer 8

@surjithayer can u plz ans. my next ques.

Answer 9

eq of a line is \[\frac{ x-x1 }{ a }=\frac{ y-y1 }{ b }=\frac{ z-z1 }{ c }\]
Here a,b,c are d.r's