Vector Equation of a Line and Shortest Distance
In three-dimensional space, line L₁ passes through A(1, 2, −1) with direction d₁ = (2, −1, 3). Line L₂ passes through B(4, 0, 2) with direction d₂ = (1, 1, −2). A student is asked to determine whether the lines intersect, are parallel, or are skew, and to find the shortest distance between them if they are skew. The student computes d₁ · d₂ = 2 − 1 − 6 = −5, concludes the lines are not parallel, and then states that because the dot product is not zero the lines must intersect. The marking guide rewards a complete vector argument and correct use of the scalar triple product.
1. Show that the lines are skew and find the shortest distance between them.
Read the worked answer
The direction vectors are not scalar multiples (2, −1, 3) ≠ λ(1, 1, −2), so the lines are not parallel.
Test for intersection: solve A + s d₁ = B + t d₂. 1 + 2s = 4 + t … (i) 2 − s = t … (ii) −1 + 3s = 2 − 2t … (iii)
From (ii), t = 2 − s. Substituting into (i): 1 + 2s = 4 + 2 − s, so 3s = 5, s = 5/3. Then t = 1/3. Check (iii): LHS = −1 + 5 = 4; RHS = 2 − 2/3 = 4/3. Not equal, so no solution: the lines are skew.
The shortest distance is |(B − A) · (d₁ × d₂)| / |d₁ × d₂|. Now B − A = (3, −2, 3). Compute d₁ × d₂ = |i j k; 2 −1 3; 1 1 −2| = i[(−1)(−2) − 3(1)] − j[2(−2) − 3(1)] + k[2(1) − (−1)(1)] = i(2 − 3) − j(−4 − 3) + k(2 + 1) = (−1, 7, 3). |d₁ × d₂| = √(1 + 49 + 9) = √59. Then (B − A) · (−1, 7, 3) = −3 − 14 + 9 = −8. Distance = 8/√59 ≈ 1.04 units.
2. Explain the error in the student's reasoning about the dot product and state what the dot product of direction vectors does tell you.
Read the worked answer
The student confused a necessary condition with a sufficient one. d₁ · d₂ = 0 would mean the directions are perpendicular, which says nothing about whether the lines meet; perpendicular lines in space need not intersect. Conversely, a nonzero dot product does not imply intersection. The dot product of direction vectors only measures the angle between the directions, not any positional relationship.
Whether two lines intersect depends on whether the position equation A + s d₁ = B + t d₂ has a consistent solution for s and t. Non-parallel lines either intersect at a single point or are skew. The correct test is to solve the three coordinate equations and check for consistency, as done above, where the third equation failed.
A useful diagnostic is the scalar triple product (B − A) · (d₁ × d₂): if it is zero the lines are coplanar and, being non-parallel, must intersect; if nonzero they are skew. Here it equals −8 ≠ 0, confirming skewness.