maths
posted by Anonymous .
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

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.