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MHT CET2008MathematicsThree Dimensional Geometry

The symmetric equation of lines 3 x+2 y+z-5=0 and x+y-2 z-3=0 , is

Options

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

Correct answer

C. x+1 -5 = y-4 7 = z-0 1

Step-by-step solution

Let a, b, c be the direction ratios of required line. 3 a+2 b+c=0 and a+b-2 c=0 a -4-1 = b 1+6 = c 3-2 a -5 = b 7 = c 1 In order to find a point on the required line we put z=0 in the two given equations to obtain, 3 x+2 y=5 and x+y=3 . Solving these two equations, we get x=-1, y=4 . Coordinates of point on required line are (-1,4,0) Hence, required line is x+1 -5 = y-4 7 = z-0 1

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