The shortest distance between the lines aligned & r =(4 i - j )+ ( i +2 j -3 k ) and & r =( i - j +2 k )+ (2 i +4 j -5 k ) is aligned

Step-by-step solution

The lines are given by r = (4 i - j ) + ( i + 2 j - 3 k ) and r = ( i - j + 2 k ) + (2 i + 4 j - 5 k ) . The shortest distance between skew lines is found using d = |( a₂ - a₁ ) ( b₁ b₂ )| | b₁ b₂ | . Computing a₂ - a₁ = ( i - j + 2 k ) - (4 i - j ) = -3 i + 2 k . The cross product yields b₁ b₂ = 2 i - j with magnitude 5 . The scalar triple product is -6 , so the distance becomes d = 6 5 . The shortest distance is 6 5 units.