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MHT CET202613 April 2026Evening ShiftMathematicsThree Dimensional GeometryActual

The equation of a line passing through a point (4, -2, 3) and perpendicular to the XZ-plane is....

Options

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

Correct answer

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

Step-by-step solution

The position vector of the given point is a = 4 i - 2 j + 3 k . Since the line is perpendicular to the XZ-plane, it is parallel to the Y-axis. Therefore, the direction vector of the line is b = j . The vector equation of a line passing through a point with position vector a and parallel to a vector b is given by r = a + b . Substituting the values, we get r = (4 i - 2 j + 3 k ) + j . Answer: r = (4 i - 2 j + 3 k ) + ( j )

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