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MHT CET20242 May 2024Morning ShiftMathematicsVector AlgebraActual

The vector equation of a line whose Cartesian equations are y=2,4 x-3 z+5=0 is

Options

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

Correct answer

C. r = (2 j + 5 3 k )+ (3 i +4 k )

Step-by-step solution

Given cartesian equation of line is aligned & 4 x-3 z+5=0, y=2 & 4 x=3 z-5, y=2 & 4 x=3 (z- 5 3 ), y=2 & x 3 = z- 5 3 4 , y=2 aligned The given line passes (0,2, 5 3 ) and the direction ratios are proportional to 3,0,4 . The vector equation is r = (2 j + 5 3 k )+ (3 i +4 k )

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