If the shortest distance between the lines r ₁= i +2 j +2 k + ( i -2 j +2 k ), R , >0 and r ₂=-4 i - k + (3 i -2 j -2 k ), R , is 9, then the value of is

Step-by-step solution

Given the skew lines: r ₁ = ( i + 2 j + 2 k ) + ( i - 2 j + 2 k ) r ₂ = (-4 i - k ) + (3 i - 2 j - 2 k ) The shortest distance between skew lines is given by d = |( a ₂ - a ₁) ( b ₁ b ₂)| | b ₁ b ₂| . The direction vectors are b ₁ = i - 2 j + 2 k and b ₂ = 3 i - 2 j - 2 k . The cross product evaluates to b ₁ b ₂ = 8 i + 8 j + 4 k with magnitude 12. The difference between position vectors is a ₂ - a ₁ = (-4 - ) i - 2 j - 3 k . The dot product simplifies to ( a ₂ - a ₁) ( b ₁ b ₂) = -8 - 60 . Setting the distance equ