The vector equation of two lines are aligned & r =(1-t) +(t-2) +(3-2 t) k & r =(s+1) +(2 s-1) -(2 s+1) k aligned Then the shortest distance between them is

Step-by-step solution

The first line is r = ( i - 2 j + 3 k ) + t(- i + j - 2 k ) . Let a ₁ = i - 2 j + 3 k and b ₁ = - i + j - 2 k . The second line is r = ( i - j - k ) + s( i + 2 j - 2 k ) . Let a ₂ = i - j - k and b ₂ = i + 2 j - 2 k . The shortest distance d between two lines r = a ₁ + t b ₁ and r = a ₂ + s b ₂ is given by d = |( a ₂ - a ₁) ( b ₁ b ₂)| | b ₁ b ₂| . Calculate a ₂ - a ₁ = ( i - j - k ) - ( i - 2 j + 3 k ) = 0 i + j - 4 k . Calculate b ₁ b ₂ = vmatrix i & j & k -1 & 1 & -2 1 & 2 & -2 vmatrix = i (-2 + 4) - j (2 + 2) +