Shortest Distance Between Skew and Parallel Lines

Convert both lines to vector form if needed. Identify a₁, a₂, b₁, and b₂. Compute b₁ × b₂. If it is zero, use the parallel-lines formula. If it is non-zero, compute a₂ - a₁, then the scalar triple product (a₂ - a₁) · (b₁ × b₂). Take the modulus and divide by |b₁ × b₂|. State the distance with correct units if given.

A 3D diagram should show two non-intersecting, non-parallel lines, their direction vectors, and the common perpendicular segment connecting them.

Using |a₂ - a₁| as the shortest distance without checking perpendicularity to both line directions.

The shortest distance between two skew lines is the length of the common perpendicular between them. Skew lines are non-parallel, non-intersecting lines in three-dimensional space.

The shortest distance between two skew lines is the length of the common perpendicular between them. Skew lines are non-parallel, non-intersecting lines in three-dimensional space. For skew lines r = a₁ + λb₁ and r = a₂ + μb₂, shortest distance = |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|, with condition b₁ × b₂ ≠ 0. For parallel lines, use distance = |(a₂ - a₁) × b| / |b|, where b is the common direction vector. For r = i + 2j + 3k + λ(i - j + k) and r = 2i - j + k + μ(2i + j - k), use a₁ = (1,2,3), a₂ = (2,-1,1), b₁ = (1,-1,1), and b₂ = (2,1,-1) in the skew-lines distance formula.

For lines r = a₁ + λb₁ and r = a₂ + μb₂, the vector a₂ - a₁ joins one point on the first line to one point on the second line. The vector b₁ × b₂ gives a direction perpendicular to both lines. Projecting a₂ - a₁ on this perpendicular direction gives the shortest distance. For skew lines r = a₁ + λb₁ and r = a₂ + μb₂, shortest distance = |(a₂ - a₁) · (b₁ × b₂)| / |b₁ × b₂|, with condition b₁ × b₂ ≠ 0. For parallel lines, use distance = |(a₂ - a₁) × b| / |b|, where b is the common direction vector. Find the shortest distance between r = (1,0,2) + λ(1,2,2) and r = (0,1,1) + μ(2,1,3). Here a₁ = (1,0,2), a₂ = (0,1,1), b₁ = (1,2,2), b₂ = (2,1,3). a₂ - a₁ = (-1,1,-1). b₁ × b₂ = |i j k; 1 2 2; 2 1 3| = i(6 - 2) - j(3 - 4) + k(1 - 4) = (4,1,-3). Dot product: (-1,1,-1) · (4,1,-3) = -4 + 1 + 3 = 0. Shortest distance = |0|/√(4²+1²+(-3)²) = 0. The lines intersect, so the shortest distance is 0. Asked in long-answer form with two vector equations or Cartesian equations. The question may require identifying whether lines are skew, intersecting, or parallel before selecting the distance formula. Using |a₂ - a₁| as the shortest distance without checking perpendicularity to both line directions.