MHT CET202019 Oct 2020Evening ShiftMathematicsThree Dimensional GeometryActual
The equation of the line passing through (1,2,3) and perpendicular to the lines x-1= y+2 2 = z+4 4 and x-1 2 = y-2 2 =z+3 is
Options
- Ax-1 6 = 2-y 7 = z-3 2
- Bx-1 6 = y-2 7 = z-3 2
- Cx-1 4 = 2-y 5 = z-3 2
- Dx-1= y-2 2 = z-3 4
Correct answer
A. x-1 6 = 2-y 7 = z-3 2
Step-by-step solution
(D) The vector perpendicular to both the given lines is given by | array lll i & j & k 1 & 2 & 4 2 & 2 & 1 array |= i (-6)- j (-7)+ k (-2)=-6 i +7 j -2 k Hence d.r.s. of required line are 6,-7,2 . Thus eq. of required line is x-1 6 = y-2 -7 = 2-3 2 i.e. x-1 6 = 2-y 7 = z-3 2