Practicing Success
The distance between the lines r ̅ = ( ̂i + 2 ̂j -4 ̂k) + λ (2 ̂i + 3 ̂j +6 ̂k) and r ̅ = (3 ̂i + 3 ̂j -5 ̂k) + μ (2 ̂i + 3 ̂j +6 ̂k) is- |
d= (√296 /7 ) d= (√299 /7 ) d= (√291 /7 ) d= (√293 /7 ) |
d= (√293 /7 ) |
The given lines are r ̅ = ( ̂i + 2 ̂j -4 ̂k) + λ (2 ̂i + 3 ̂j +6 ̂k) and r ̅ = (3 ̂i + 3 ̂j -5 ̂k) + μ (2 ̂i + 3 ̂j +6 ̂k) The given lines are parallel a1 ̅ = ( ̂i + 2 ̂j -4 ̂k), a2 ̅ = (3 ̂i + 3 ̂j -5 ̂k) and b ̅ = (2 ̂i + 3 ̂j +6 ̂k) Therefore the distance between the two lines is given by d= mod{ b ̅ x (a2 ̅ - a1 ̅ )/ mod(b) } ⇒ d= mod{ (2 ̂i + 3 ̂j +6 ̂k) x ( 2 ̂i + ̂j - ̂k)/ √4+9+36 } ⇒ d= mod{ ( -9 ̂i + 14 ̂j -4 ̂k)/ √4+9+36 } ⇒ d= (√293/ √49 ) ⇒ d= (√293 /7 )
|