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MHT CET202616 April 2026Morning ShiftMathematicsThree Dimensional GeometryActual

The vector equation of the line whose cartesian equations are x = 2, 2y - 3z + 7 = 0

Options

  1. Ar = (2 i + 2 j - 3 k ) + (2 j - 3 k )
  2. Br = (2 i + 2 j - 3 k ) + (2 i - 3 j )
  3. Cr = (2 i + 7 3 k ) + (3 j + 2 k )
  4. Dr = (-2 i + 7 3 k ) + (3 j + 2 k )

Correct answer

C. r = (2 i + 7 3 k ) + (3 j + 2 k )

Step-by-step solution

The given cartesian equations of the line are x = 2 and 2y - 3z + 7 = 0 . We can rewrite these equations in symmetric form to find a point on the line and its direction ratios. x = 2 x - 2 = 0 x - 2 0 2y - 3z + 7 = 0 2y = 3z - 7 y - 0 3 = z - 7 3 2 Equating them, we get the symmetric form of the line: x - 2 0 = y - 0 3 = z - 7 3 2 This shows that the line passes through the point with position vector a = 2 i + 0 j + 7 3 k and is parallel to the vector b = 0 i + 3 j + 2 k . The vector equation of the line is given b

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