To solve the problem, we need to find the foot of the perpendicular from the point \( P(1, 2, -3) \) to the line given by the equation
\[
\frac{x + 1}{2} = \frac{y - 3}{-2} = \frac{z}{-1}
\]
### Step 1: Parametrize the Line
The line can be expressed in parametric form. Letting \( t \) be the parameter, we can write:
\[
x = 2t - 1, \quad y = -2t + 3, \quad z = -t
\]
Thus, any point \( Q \) on the line can be represented as:
\[
Q(t) = (2t - 1, -2t + 3, -t)
\]
### Step 2: Find the Vector \( \overrightarrow{PQ} \)
The vector \( \overrightarrow{PQ} \) from point \( P(1, 2, -3) \) to point \( Q(t) \) is given by:
\[
\overrightarrow{PQ} = Q(t) - P = (2t - 1 - 1, -2t + 3 - 2, -t + 3)
\]
This simplifies to:
\[
\overrightarrow{PQ} = (2t - 2, -2t + 1, -t + 3)
\]
### Step 3: Direction Vector of the Line
The direction vector of the line can be derived from the parametric equations. It is:
\[
\mathbf{d} = (2, -2, -1)
\]
### Step 4: Use the Perpendicular Condition
For \( \overrightarrow{PQ} \) to be perpendicular to the direction vector \( \mathbf{d} \), their dot product must equal zero:
\[
\overrightarrow{PQ} \cdot \mathbf{d} = 0
\]
Calculating the dot product:
\[
(2t - 2) \cdot 2 + (-2t + 1) \cdot (-2) + (-t + 3) \cdot (-1) = 0
\]
Expanding this gives:
\[
4t - 4 + 4t - 2 + t - 3 = 0
\]
Combining like terms:
\[
9t - 9 = 0
\]
Thus, we find:
\[
t = 1
\]
### Step 5: Find the Foot of the Perpendicular
Substituting \( t = 1 \) back into the parametric equations for the line:
\[
Q(1) = (2(1) - 1, -2(1) + 3, -1) = (1, 1, -1)
\]
So, the foot of the perpendicular is:
\[
Q(1) = (1, 1, -1)
\]
### Step 6: Calculate the Length of the Perpendicular
The length of the perpendicular \( PQ \) can be calculated using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
Substituting \( P(1, 2, -3) \) and \( Q(1, 1, -1) \):
\[
d = \sqrt{(1 - 1)^2 + (1 - 2)^2 + (-1 + 3)^2} = \sqrt{0 + 1 + 4} = \sqrt{5}
\]
### Step 7: Find the Image of Point \( P \)
To find the image of point \( P \) across the line, we can use the midpoint formula. Let \( P' \) be the image of point \( P \):
\[
\frac{P + P'}{2} = Q
\]
This leads to the equations:
\[
\frac{1 + x'}{2} = 1, \quad \frac{2 + y'}{2} = 1, \quad \frac{-3 + z'}{2} = -1
\]
Solving these:
1. \( 1 + x' = 2 \) gives \( x' = 1 \)
2. \( 2 + y' = 2 \) gives \( y' = 0 \)
3. \( -3 + z' = -2 \) gives \( z' = 1 \)
Thus, the image \( P' \) is:
\[
P' = (1, 0, 1)
\]
### Summary of Results
1. Foot of the perpendicular: \( (1, 1, -1) \)
2. Length of the perpendicular: \( \sqrt{5} \)
3. Image of the point \( P \): \( (1, 0, 1) \)
