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

Get the slope/tangent vectors for each line:

dP/dt = ⟨-2, -4, 5⟩

dQ/dt = ⟨3, 5, 4⟩

Take their cross product - I'll call it v :

v = ⟨-2, -4, 5⟩×⟨3, 5, 4⟩ = ⟨-41, 23, 2⟩

Normalize this vector by multiplying it by 1/||v|| :

n = v/||v|| = ⟨-41, 23, 2⟩/(3√246)

Then the distance d between P and Q is

d = |n • (⟨-2, -4, 5⟩ - ⟨4, 5, -4⟩)|

d = 1/(3√246) |⟨-41, 23, 2⟩ • ⟨-6, -9, 9⟩|

d = 1/(3√246) |246 - 207 + 18|

d = 57/(3√246)

d = 19/√246