Find the distance of the point P (1,2,3) from its image in the plane x + 2y + 4z = 38.

Answer

To find the distance of the point \( P(1, 2, 3) \) from its image in the plane given by the equation \( x + 2y + 4z = 38 \), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as \( x + 2y + 4z = 38 \). The normal vector \( \vec{n} \) to the plane can be derived from the coefficients of \( x \), \( y \), and \( z \):
\[ \vec{n} = (1, 2, 4) \]

### Step 2: Find the distance from point \( P \) to the plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) can be calculated using the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Rearranging the plane equation to the standard form gives:
\[ 1x + 2y + 4z - 38 = 0 \quad \text{(where \( D = -38 \))} \]
Substituting \( P(1, 2, 3) \) into the formula:
\[ d = \frac{|1(1) + 2(2) + 4(3) - 38|}{\sqrt{1^2 + 2^2 + 4^2}} = \frac{|1 + 4 + 12 - 38|}{\sqrt{1 + 4 + 16}} = \frac{|17 - 38|}{\sqrt{21}} = \frac{21}{\sqrt{21}} = \sqrt{21} \]

### Step 3: Find the coordinates of the image point \( P' \)
The image point \( P' \) can be found by moving along the normal vector from point \( P \) by a distance of \( 2d \) (since the distance from \( P \) to the plane is \( d \), the distance from \( P \) to \( P' \) is \( 2d \)):
\[ P' = P + 2d \cdot \frac{\vec{n}}{|\vec{n}|} \]
First, we calculate the unit normal vector:
\[ |\vec{n}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21} \]
Thus, the unit normal vector is:
\[ \frac{\vec{n}}{|\vec{n}|} = \left(\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}\right) \]
Now, we can find \( P' \):
\[ P' = (1, 2, 3) + 2\sqrt{21} \left(\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{4}{\sqrt{21}}\right) = (1, 2, 3) + (2, 4, 8) = (3, 6, 11) \]

### Step 4: Calculate the distance between points \( P \) and \( P' \)
Now we can find the distance \( PP' \):
\[ PP' = \sqrt{(3 - 1)^2 + (6 - 2)^2 + (11 - 3)^2} = \sqrt{2^2 + 4^2 + 8^2} = \sqrt{4 + 16 + 64} = \sqrt{84} = 2\sqrt{21} \]

### Final Answer
The distance of the point \( P(1, 2, 3) \) from its image \( P' \) in the plane \( x + 2y + 4z = 38 \) is:
\[ \boxed{2\sqrt{21}} \]