MHT CET202312 May 2023Evening ShiftMathematicsThree Dimensional GeometryActual
The shortest distance (in units) between the lines x+1 3 = y+2 1 = z+1 2 and r =(2 i -2 j +3 k )+ ( i +2 j ) is
Options
- A8 3 5
- B1 3 5
- C7 3 5
- D2 3 5
Correct answer
A. 8 3 5
Step-by-step solution
Given lines are: x+1 3 = y+2 1 = z+1 2 and x-2 1 = y+2 2 = z-3 0 Required distance = | array lll 3 & 0 & 4 3 & 1 & 2 1 & 2 & 0 array | (6-1)^2+(0-2)^2+(0-4)^2 aligned & = | 3(0-4)+0+4(6-1) 25+4+16 | & = | 8 45 | & = 8 3 5 aligned