A plane passes through thee points `P(4, 0, 0) and Q(0, 0, 4)` and is parallel to the Y-axis. The distance of the plane from the origin is

Options

A. `2`
B. `4`
C. `sqrt(2)`
D. `2sqrt(2)`

Answer

To find the distance of the plane from the origin, we will follow these steps:

### Step 1: Identify the Equation of the Plane
Given that the plane is parallel to the Y-axis and passes through the points \( P(4, 0, 0) \) and \( Q(0, 0, 4) \), we can use the general form of the equation of a plane parallel to the Y-axis:
\[ Ax + Bz + C = 0 \]

### Step 2: Substitute the Points into the Plane Equation
Substituting the coordinates of point \( P(4, 0, 0) \):
\[ A(4) + B(0) + C = 0 \implies 4A + C = 0 \quad \text{(1)} \]

Substituting the coordinates of point \( Q(0, 0, 4) \):
\[ A(0) + B(0) + C = 0 \implies C = 0 \quad \text{(2)} \]

### Step 3: Solve for A and B
From equation (2), we have \( C = 0 \). Substituting \( C = 0 \) into equation (1):
\[ 4A + 0 = 0 \implies A = 0 \]

Since \( A = 0 \) and \( C = 0 \), we can express \( B \) in terms of \( C \). The equation of the plane simplifies to:
\[ Bz = 0 \]
This means that the plane can be described by:
\[ x + z - 4 = 0 \quad \text{(3)} \]

### Step 4: Find the Distance from the Origin to the Plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For our plane \( x + z - 4 = 0 \), we can rewrite it as:
\[ 1x + 0y + 1z - 4 = 0 \]
Here, \( A = 1 \), \( B = 0 \), \( C = 1 \), and \( D = -4 \). The coordinates of the origin are \( (0, 0, 0) \).

### Step 5: Substitute into the Distance Formula
Substituting into the distance formula:
\[ d = \frac{|1(0) + 0(0) + 1(0) - 4|}{\sqrt{1^2 + 0^2 + 1^2}} = \frac{|-4|}{\sqrt{1 + 0 + 1}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Final Answer
The distance of the plane from the origin is:
\[ \boxed{2\sqrt{2}} \]