JEE Main20246 Apr 2024Evening ShiftMathematicsThree Dimensional GeometryActual
If the shortest distance between the lines x- 3 = y-2 -1 = z-1 1 and x+2 -3 = y+5 2 = z-4 4 is 44 30 , then the largest possible value of | | is equal to _________
Correct answer
0
Step-by-step solution
aligned & a ₁= i +2 j + k & a ₂=-2 i -5 j +4 k & p -=3 i - j + k & q -=-3 i +2 j +4 k & ( +2) i +7 j -3 k = a ₁- a ₂ & p q -=-6 i -15 j +3 k aligned aligned & 44 30 = |-6 -12-105-9| (-6)^2+(-15)^2+3^2 & 44 30 = |6 +126| 3 30 & 132=|6 +126| & =1, =-43 & | |=43 aligned