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Magnitude of shortest distance between lines

Short answer

Line 1: A(0,-1,2), d1=(4,3,2). Line 2: direction d2=(1,1,1) from cross product, point B(-24/13,-21/13,0). n=d1×d2=(1,-2,1). Distance = |AB·n|/√6 = (34/13)/√6 = 17√6/39 ≈ 1.07 units.

Find the magnitude of the line of the shortest distance between the lines x/4 = (y+1)/3 = (z−2)/2, 5x−2y−3z+6 = 0, x−3y+2z−3 = 0. [Second line as printed: the pair of planes 5x−2y−3z+6 = 0 and x−3y+2z−3 = 0; scan unclear]

Reading. The second line is the intersection of the planes 5x − 2y − 3z + 6 = 0 and x − 3y + 2z − 3 = 0, and I take that as printed. The first line is x/4 = (y + 1)/3 = (z − 2)/2.

Line 1

Through A(0, −1, 2), direction d₁ = (4, 3, 2).

Line 2: direction and a point

Direction = cross product of the plane normals:

d₂ = (5, −2, −3) × (1, −3, 2)
   = ((−2)(2) − (−3)(−3), (−3)(1) − (5)(2), (5)(−3) − (−2)(1))
   = (−13, −13, −13)  ∝ (1, 1, 1)

Point: put z = 0: 5x − 2y = −6 and x − 3y = 3. Solving: x = −24/13, y = −21/13. So B(−24/13, −21/13, 0).

Common perpendicular direction

n = d₁ × d₂ = (4, 3, 2) × (1, 1, 1) = (3 − 2, 2 − 4, 4 − 3) = (1, −2, 1),   |n| = √6

Shortest distance

AB = B − A = (−24/13, −21/13 + 1, −2) = (−24/13, −8/13, −2).

AB · n = −24/13 + 16/13 − 2 = −8/13 − 26/13 = −34/13

S.D. = |AB · n| / |n| = (34/13)/√6 = 34/(13√6) = 17√6/39 ≈ 1.07

Answer: the magnitude of the shortest distance is 17√6/39 ≈ 1.07 units.

✓ Answer checked by Claude Opus 5.5 (high effort)

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