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MHT CET202012 Oct 2020Evening ShiftMathematicsThree Dimensional GeometryActual

The vector equation of the line x+3 2 = 2 y-3 5 ; z=-1 is

Options

  1. Ar = (3 - 3 2 - k )+ (4 +5 )
  2. Br = (-3 + 3 2 - k )+ (4 +5 )
  3. Cr = (-3 + 3 2 + k )+ (4 +5 )
  4. Dr = (3 + 3 2 - k )+ (4 + 5 2 )

Correct answer

B. r = (-3 + 3 2 - k )+ (4 +5 )

Step-by-step solution

Equation of line is x+3 2 = 2 y-3 5 ; z=-1 x+3 2 = 2 (y- 3 2 ) 5 ; z=-1 x+3 2 = y- 3 2 ( 5 2 ) ; z=-1 This line passes through point (-3, 3 2 ,-1 ) and d.r.s. are 2, 5 2 , 0 i.e. 4,5,0 Hence vector equation of given line is r = (-3 i + 3 2 j - k )+ (4 i +5 j )

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