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MHT CET202122 Sep 2021Evening ShiftMathematicsVector AlgebraActual

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

Options

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

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

Step-by-step solution

We have 4 x-3 z+5=0 and y=2 aligned & 4 x=3 z-5 4 x=3 (z- 5 3 ) & 4 x 12 = 3 (z- 5 3 ) 12 x 3 = 3 (z- 5 3 ) 4 aligned Thus line passes through point (0,2, 5 3 ) , and has direction ratios 3,0,4 . Hence required equation of line is (2 j + 5 3 k )+ (3 i +4 k )

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