#
If (a-a')^2+(b-b')^2+(c-c')^2=p

and (ab'-a'b)^2+(bc'-b'c)^2+(ca'-c'a)^2=q

then the perpendicular distance of the line

ax+by+cz=1,a'x+b'y+c'z=1 is

1
Comment has been deleted

Comment has been deleted

okk

madan how did u get (p/q)^0.5

finally got it

Log in to add a comment