The shortest distance between the lines x+1 3 = y-2 2 = z+1 2 and x-2 1 = y-2 2 = z+3 3 is

Step-by-step solution

The shortest distance between skew lines is given by d = |( a₂ - a₁ ) ( b₁ b₂ )| | b₁ b₂ | . From x+1 3 = y-2 2 = z+1 2 , we have point a₁ = (-1, 2, -1) and direction b₁ = (3, 2, 2) . From x-2 1 = y-2 2 = z+3 3 , we have point a₂ = (2, 2, -3) and direction b₂ = (1, 2, 3) . The connecting vector is a₂ - a₁ = (3, 0, -2) . The cross product evaluates to: b₁ b₂ = vmatrix i & j & k 3 & 2 & 2 1 & 2 & 3 vmatrix = 2 i - 7 j + 4 k = (2, -7, 4) . Its magnitude is | b₁ b₂ | = 4 + 49 + 16 = 69 . The scalar triple product yield