Find the co-ordinates of the foot of the perpendicular and the perpendicular distance of the point (1,3,4) from the plane 2x -y + z + 3 = 0. Find also, the image of the point in the plane.

Answer

To solve the problem of finding the coordinates of the foot of the perpendicular from the point \( (1, 3, 4) \) to the plane \( 2x - y + z + 3 = 0 \), as well as the perpendicular distance and the image of the point in the plane, we can follow these steps:

### Step 1: Identify the plane and the point
The equation of the plane is given by:
\[ 2x - y + z + 3 = 0 \]
The point from which we need to find the perpendicular is:
\[ P(1, 3, 4) \]

### Step 2: Find the normal vector of the plane
The normal vector \( \vec{n} \) of the plane can be derived from the coefficients of \( x, y, z \) in the plane equation:
\[ \vec{n} = (2, -1, 1) \]

### Step 3: Find the foot of the perpendicular
Let the foot of the perpendicular from point \( P \) to the plane be \( B(a, b, c) \). The line passing through point \( P \) in the direction of the normal vector can be expressed parametrically as:
\[ x = 1 + 2t, \quad y = 3 - t, \quad z = 4 + t \]
We substitute these expressions into the plane equation to find \( t \):
\[ 2(1 + 2t) - (3 - t) + (4 + t) + 3 = 0 \]
Simplifying this:
\[ 2 + 4t - 3 + t + 4 + t + 3 = 0 \]
\[ 4t + 4 = 0 \implies 4t = -4 \implies t = -1 \]

### Step 4: Substitute \( t \) back to find coordinates of \( B \)
Now substitute \( t = -1 \) back into the parametric equations:
\[ x = 1 + 2(-1) = 1 - 2 = -1 \]
\[ y = 3 - (-1) = 3 + 1 = 4 \]
\[ z = 4 + (-1) = 4 - 1 = 3 \]
Thus, the coordinates of the foot of the perpendicular \( B \) are:
\[ B(-1, 4, 3) \]

### Step 5: Calculate the perpendicular distance
The perpendicular distance \( d \) from point \( P(1, 3, 4) \) to the plane can be calculated using the formula:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Where \( A = 2, B = -1, C = 1, D = 3 \) and \( (x_1, y_1, z_1) = (1, 3, 4) \):
\[ d = \frac{|2(1) - 1(3) + 1(4) + 3|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{|2 - 3 + 4 + 3|}{\sqrt{4 + 1 + 1}} = \frac{|6|}{\sqrt{6}} = \sqrt{6} \]

### Step 6: Find the image of the point in the plane
The image \( A' \) of point \( P \) in the plane can be found using the midpoint formula. The foot of the perpendicular \( B \) is the midpoint of \( P \) and its image \( A' \):
\[ B = \left( \frac{x + x'}{2}, \frac{y + y'}{2}, \frac{z + z'}{2} \right) \]
Where \( A' \) is the image point. We know \( B(-1, 4, 3) \) and \( P(1, 3, 4) \):
\[ -1 = \frac{1 + x'}{2} \implies x' = -2 - 1 = -3 \]
\[ 4 = \frac{3 + y'}{2} \implies y' = 8 - 3 = 5 \]
\[ 3 = \frac{4 + z'}{2} \implies z' = 6 - 4 = 2 \]
Thus, the image of the point \( P \) in the plane is:
\[ A'(-3, 5, 2) \]

### Summary of Results
- Foot of the perpendicular \( B \): \( (-1, 4, 3) \)
- Perpendicular distance \( d \): \( \sqrt{6} \)
- Image of the point \( P \): \( A'(-3, 5, 2) \)