We begin by considering two lines. Namely, f(t): x = 1 + 2t, y = 2 - 3t, z = 3 + 4t and h(s): x = -1 + 3s, y = 3-s, z = -5 + 5s. I now plot these two lines in 3 space in order to justify their skewness i.e. They do not intersect and are not parallel.

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I now introduce a new function D_l and this is defined as the distance between the two lines i.e. |(f(t)-h(s)|.  We now work with this equation to derive a general method for calculating the distance between two skew lines.

Before we begin, recall that |{(a,b,c)}| = \sqrt{a^2+b^2+c^2}

Now,

D_l = |{(1,2,3)-(-1,3,-5)+t(2,-3,4)-s(3,-1,5)}| \\= |{(2 + 2t - 3s, -1 -3t + s, 8 + 4t -5s)}|  \\= \sqrt{(2+2t-3s)^2+(-1 -3t+s)^2+(8+4t-5s)^2}\\=\sqrt{35s^2 - 58st - 94s + 29t^2 + 78t + 69}

I now bring in some Calculus. We use the fact that minimizing a function is the same as minimizing the square of that function (does not always hold but it holds here because we are dealing with a distance function that is non-negative and monotonic). Hence, we do the following:

(D_l)^2 =  35s^2 - 58st - 94s + 29t^2 + 78t + 69

We now take the partial derivatives (D_l)^2 with respect to s and t and we set it equal to zero. This is as follows.

\frac{\partial (D_l)^2}{\partial s} = 70s -58t -94 = 0

\frac{\partial (D_l)^2}{\partial t} = -58s+58t+78 = 0

Solving the system of linear equations we arrive at s=\frac{4}{3} and t=\frac{-1}{87}. We now substitute this back into our equation D_l and we arrive at a distance of \frac{16\sqrt{174}}{87} = 2.426.

Take note of the following: I left out a few details with respect to how rigorous you probably needed to be. What we really did is minimized the distance between the two lines but we did not check that this was a minimum. Please take care and remember to do this either using a Hessian matrix or the 2nd derivative test.

I hope that this is more intuitive than the method provided in Stewart.

 

 

 

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