Question

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

• A. \( x = -2 ;\ y = 3 + \lambda ;\ z = 4 \)
• B. \( \dfrac{x - 2}{1} = \dfrac{z + 4}{1} ;\ y = 3 \)
• C. \( x = -2 ;\ y = -3 + \lambda ;\ z = 4 \)
• D. \( x = 2 ;\ y = 3 + \lambda ;\ z = -4 \)

Hint:

A line perpendicular to a plane is always parallel to the normal vector of that plane.

Answer 1

The Correct Option is D

Step 1: Understand the XOZ plane.
The XOZ plane is defined by the equation \[ y = 0 \] The normal vector to this plane is along the \( y \)-axis, i.e., \[ \vec{n} = \hat{j} \]

Step 2: Direction of the required line.
A line perpendicular to the XOZ plane must be parallel to the normal vector \( \hat{j} \). Hence, the direction ratios of the line are \( (0, 1, 0) \).

Step 3: Use the point–direction form of a line.
The given point is \( (2, 3, -4) \). Therefore, the parametric equations of the line are \[ x = 2 \] \[ y = 3 + \lambda \] \[ z = -4 \]

Step 4: Final conclusion.
The equation of the required line is \[ \boxed{x = 2 ;\ y = 3 + \lambda ;\ z = -4} \]