Resolving a vector intersection ambiguity in 3D
Two straight pipes are modelled as lines in three-dimensional space. Pipe A passes through the point A(1, 2, 3) and has direction vector d_A = (2, −1, 1). Pipe B passes through the point B(4, 0, 5) and has direction vector d_B = (−1, 2, 0). An engineer reports that the pipes 'cross', and plans a junction at the crossing point. A second engineer is not convinced, arguing that in three dimensions two lines with different directions usually do not meet at all - they may be skew. You must decide whether the pipes actually intersect, whether they are parallel, or whether they are skew, and then determine the shortest distance between the two lines if they do not meet.
1. Determine whether the two lines intersect, are parallel, or are skew. Justify your conclusion with an explicit calculation.
Read the worked answer
Write line A as r_A(s) = (1, 2, 3) + s(2, −1, 1) and line B as r_B(t) = (4, 0, 5) + t(−1, 2, 0).
First check directions: d_A = (2, −1, 1) and d_B = (−1, 2, 0) are not scalar multiples (2/(−1) ≠ −1/2), so the lines are not parallel.
Set r_A(s) = r_B(t):
1 + 2s = 4 − t … (i) 2 − s = 0 + 2t … (ii) 3 + s = 5 + 0t … (iii)
From (iii): s = 2. Substitute into (ii): 2 − 2 = 2t, so t = 0. Check (i): 1 + 2(2) = 5 and 4 − 0 = 4. Since 5 ≠ 4, the system is inconsistent. The lines do not intersect.
Since they are neither parallel nor intersecting, they are skew. The second engineer is correct: the pipes do not cross, and no junction point exists.
2. Find the shortest distance between the two skew lines.
Read the worked answer
The shortest distance between skew lines is the projection of the vector joining any point on one line to any point on the other onto the common normal n = d_A × d_B.
Compute n = d_A × d_B = |i j k; 2 −1 1; −1 2 0|.
i-component: (−1)(0) − (1)(2) = −2. j-component: −[(2)(0) − (1)(−1)] = −(1) = −1. k-component: (2)(2) − (−1)(−1) = 4 − 1 = 3.
So n = (−2, −1, 3).
Vector between the known points: AB = B − A = (4 − 1, 0 − 2, 5 − 3) = (3, −2, 2).
Distance = |AB · n| / |n| = |(3)(−2) + (−2)(−1) + (2)(3)| / √(4 + 1 + 9) = |−6 + 2 + 6| / √14 = 2/√14 ≈ 0.535.
The shortest distance is 2/√14 ≈ 0.53 units. This is the closest the two pipes come; any junction would require a connecting spool of at least this length.