Q.A plane is parallel to $yz$-plane so it is perpendicular to :
(A) $x$-axis
(B) $y$-axis
(C) $z$-axis
(D) none of these

Concept: Coordinate Planes and Perpendicular Axes

A plane parallel to the $yz$-plane has the same orientation as the $yz$-plane itself. The $yz$-plane is defined by the equation $x = 0$ and contains all points where the $x$-coordinate can vary while $y$ and $z$ take all possible values.

The normal vector to any plane of the form $x = k$ (which includes all planes parallel to the $yz$-plane) points in the direction of the $x$-axis. Specifically, the normal vector is $\vec{n} = (1, 0, 0)$, which is precisely the direction vector of the $x$-axis.

Since a plane is perpendicular to its normal vector, any plane parallel to the $yz$-plane must be perpendicular to the $x$-axis. …

Understanding Plane Orientation in 3D Space

When we say a plane is parallel to the $yz$-plane, we're describing its orientation in three-dimensional space. The key insight is that a plane's orientation is completely determined by its normal vector — the direction perpendicular to every line lying in the plane.

The $yz$-plane itself is the set of all points where $x = 0$. Every point in this plane has coordinates of the form $(0, y, z)$. Notice that as you move around in the $yz$-plane, the $x$-coordinate never changes; only $y$ and $z$ vary.

Now, what direction is perpendicular to the $yz$-plane? If you imagine standing on the $yz$-plane, the only way to "leave" it is to move in the $x$-direction — either forward or backward along the $x$-axis. This means the normal vector to the $yz$-plane points along the $x$-axis.

Finding the Perpendicular Direction

Let's work through this systematically:

1. Identify the normal to the $yz$-plane

   The $yz$-plane has equation $x = 0$, which we can write as $1 \cdot x + 0 \cdot y + 0 \cdot z = 0$. The coefficients $(1, 0, 0)$ give us the normal vector: $\vec{n} = \hat{i}$, which points along the $x$-axis.

2. Apply the parallel plane condition

   If our plane is parallel to the $yz$-plane, it must have the same normal direction. Parallel planes share the same normal vector (or scalar multiples of it). So our plane also has normal vector along $\vec{n} = (1, 0, 0)$ or any multiple like $(k, 0, 0)$. …