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MHT CET20239 May 2023Morning ShiftMathematicsThree Dimensional GeometryActual

If the Cartesian equation of a line is 6 x-2=3 y+1=2 z-2 , then the vector equation of the line is

Options

  1. Ar = ( 1 3 i - 1 3 j + k )+ ( i +2 j +3 k )
  2. Br =( i + j + k )+ ( i +2 j +3 k )
  3. Cr = ( -1 3 i + 1 3 j + k )+ ( i -2 j +3 k )
  4. Dr = ( 1 3 i - 1 3 j - k )+ ( i - j + k )

Correct answer

A. r = ( 1 3 i - 1 3 j + k )+ ( i +2 j +3 k )

Step-by-step solution

Given Cartesian equation of the line is aligned & 6 x-2=3 y+1=2 z-2 & 6 (x- 1 3 )=3 (y+ 1 3 )=2(z-1) aligned aligned & x- 1 3 1 6 = y+ 1 3 1 3 = z-1 1 2 & x- 1 3 1 = y+ 1 3 2 = z-1 3 aligned The given line passes through ( 1 3 , -1 3 , 1 ) and has direction ratios proportional to 1,2,3 . Vector equation is r = ( 1 3 i - 1 3 j + k )+ ( i +2 j +3 k )

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