Find the distance between the skew lines P(t)=(2,0,2)+t⟨−2,−4,5⟩ and Q(t)=(4,5,−4)+t⟨3,5,4⟩.

Question

Find the distance between the skew lines P(t)=(2,0,2)+t⟨−2,−4,5⟩ and Q(t)=(4,5,−4)+t⟨3,5,4⟩.

Hint: Take the cross product of the slope vectors of P and Q to find a vector normal to both of these lines.

Answer 1

The problem is to minimize the distance between the two lines.

Let A ( 2-2t ,-4t ,2+5t ) be a point on the first line and

B ( 4+3s, 5+5s, -4+4s ) a point on the second line.

Then their distance D (s, t) = { (2+3s+2t)² + ( 5 +5s+4t)² +(6+5t-4s)² }^1/2 ( Equation I )

To minimize D suffices to minimize d (s,t) = (2+3s+2t)² + ( 5 +5s+4t)² +(6+5t-4s)²

∂d/∂s =14 + 100s +12t

∂d/∂t = 108 +12 s + 90t

Solving the system of equations ∂d/∂s =0 and ∂d/∂t = 0 we get ( s , t ) = ( 1/246 , -443/369 ).

Substituting the values of s and t back in to Equation I we get the smallest distance between the lines

being 7 √246 / 246.