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Manipal MET2014MathematicsThree Dimensional Geometry

The equation of the plane through the intersection of the planes 3 x-y+2 z-4=0 and x+y+z-2=0 and the point (2,2,1) is

Options

  1. A7 x+5 y+4 z+8=0
  2. B7 x+5 y+4 z-8=0
  3. C7 x-5 y+4 z-8=0
  4. DNone of these

Correct answer

C. 7 x-5 y+4 z-8=0

Step-by-step solution

The equation of any plane through the intersection of the planes, 3 x-y+2 z-4=0 and x+y+z-2=0 , is (3 x-y+2 z-4)+ (x+y+z-2)=0 ...(i) The plane passes through the point (2,2,1) . Therefore, this point will satisfy Eq. (i). array cc & (3 2-2+2 1-4)+ (2+2+1-2)=0 & (6-4)+3 =0 & 2+3 =0 & = -2 3 array On substituting this value of in Eq. (i), we get the required plane as array rr (3 x-y+2 z-4)- 2 3 (x+y+z-2)=0 9 x-3 y+6 z-12-2 x-2 y-2 z+4=0 7 x-5 y+4 z-8=0 array This is the required equation of the plane.

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