Posted by **Anonymous** on Sunday, February 3, 2013 at 11:21am.

find shortest distance between the line :

(x - 8)/ 3 = (y + 19)/ -16 = (z - 10)/ 7

and

(x - 15)/ 3 = (y - 29) /8 = (z - 5) / -5

- maths -
**Reiny**, Sunday, February 3, 2013 at 11:59am
A(8,-19,10) is a point on the first line and

B(15,29,5) is a point on the second line

vector AB = (7,48,-5)

the second line has direction v = (3,8,-5)

so the projection of vectorAB on u = AB·v/|v|

= (21 + 384 + 25)/√(9 + 64 + 25) = 430/√98

= 430/(4√2)

|AB| = √(49 + 2304 + 25) = √2378

so we can use Pythagoras

let the distance between the lines be h

h^2 + (430/√98)^2 = (√2378)^2

h^2+ 184900/98 = 2378

h^2 = 48144/98 = 24072/49

h = √24072 /7 = 2√6018 /7 = appr 22.1645

better check my arithmetic on that one, easy to make typing errors.

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