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MHT CET202124 Sep 2021Evening ShiftMathematicsThree Dimensional GeometryActual

The equation of the plane which passes through (2,-3,1) and is normal to the line joining the points (3,4,-1) and (2,-1,5) is given by

Options

  1. Ax+5 y-6 z+19=0
  2. Bx-5 y+6 z-23=0
  3. Cx+5 y+6 z+7=0
  4. Dx-5 y-6 z-11=0

Correct answer

A. x+5 y-6 z+19=0

Step-by-step solution

The required plane passes through (2,-3,1) . It is normal to the line having d.r.s. (1,5,-6) . x +5 y -6 z = k 2+5(-3)-6(1)=1 i.e. k =-19 Hence equation of plane is x+5 y-6 z+19=0

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