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MHT CET20244 May 2024Evening ShiftMathematicsThree Dimensional GeometryActual

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

Options

  1. Ar = ( i - j 3 + k 3 )+ (3 i +2 j + k )
  2. Br = (- i + j 3 - k 3 )+ ( 1 2 i + 1 3 j + 1 6 k )
  3. Cr =(3 i - j - k )+ (3 i +2 j + k )
  4. Dr =( i - j + k )+ ( 1 2 i + 1 3 j + 1 6 k )

Correct answer

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

Step-by-step solution

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

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