Krushnaraj Pandya said:
Homework Statement
how to write the vector equation of the line of shortest distance between two skew lines in the shortest and most efficient way?
(The exact lines given in a particular problem in my book can be referenced- L1=(3i+8j+3k)+λ(3i-j+k) and L2=(-3i-7j+6k)+μ(-3i+2j+4k) )
2. Relevant methods
The vector we want would be perpendicular to both the lines, so we can use dot product=0 for both lines which gives us 2 equations in l,m and n where vector=li+mj+nk. We can find distance between the lines from a routine formula then equate the modulus of the vector sqrt(l^2+m^2+n^2) to that distance
The Attempt at a Solution
combined above
Instead of attempting fancy and sophisticated geometric constructions, the easiest way is just to compute the distance between two points (one on each line) and then minimize it. It is, in fact, easier to minimize the
square of the distance; that is an equivalent problem. Do you see why?
A point on L1 has the form ##\vec{x}_1 = (3+3s, 8-s,3+s)## and a point on L2 has the form ##\vec{x}_2 = (-3-3t, -7+2t, 6+4t)##, where I write ##s## and ##t## instead of your ##\lambda## and ##\mu.## The squared distance between the points is
$$S = \sum_{i=1}^3 [x_1(i)-x_2(i)]^2 = (6+3s+3t)^2 +(15-s-2t)^2 + (-3 +s + 4t)^2.$$
We can expand this out to get
$$S = 11 s^2 + 14 s t + 29 t^2 + 270 \hspace{4ex}(1)$$ The contour curves (the ##s-t## curves of constant values for ##S##) are ellipses with center at ##(s,t) = (0,0)##, and the values of ##S## get larger as we go away from this center. Thus, the least value of ##S## is ##S = 270,## obtained at ##s = t = 0##. The two points are ##\vec{x}_{1,best} = (3,8,3)## on L1 and ##\vec{x}_{2,best} = (-3,-7,6)## on L2.
Note: normally, minimization problems like this would be handled using calculus, but because of the special structure of this type of problem it can be handled using plain algebra alone, essentially by rewriting a quadratic of the form
$$S = a s^2 + b st + c t^2 + ps + qt + r$$ to first get rid of the linear terms ##ps## and ##qt##, thus getting a simple "pure" quadratic similar to (1) above (but with different numbers). Then find the center of the ellipse of constant ##S##. No calculus is needed at all.