To find the image of a point in a given plane, we can follow these steps:
### Step-by-Step Solution:
#### (i) Finding the image of the point (2, -3, 2) in the plane \(2x + y - 3z = 10\)
1. **Identify the point and the plane equation**:
- Point \( P(2, -3, 2) \)
- Plane equation: \( 2x + y - 3z = 10 \)
2. **Find the normal vector of the plane**:
- The normal vector \( \mathbf{n} \) can be derived from the coefficients of \( x, y, z \) in the plane equation:
\[
\mathbf{n} = (2, 1, -3)
\]
3. **Find the foot of the perpendicular from point P to the plane**:
- The formula for the foot of the perpendicular from point \( P(x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
\left( x_1 - \frac{A \cdot D}{A^2 + B^2 + C^2}, y_1 - \frac{B \cdot D}{A^2 + B^2 + C^2}, z_1 - \frac{C \cdot D}{A^2 + B^2 + C^2} \right)
\]
- Rearranging the plane equation gives \( D = -10 \). Thus, we have:
\[
A = 2, B = 1, C = -3, D = -10
\]
- Calculate \( A^2 + B^2 + C^2 \):
\[
A^2 + B^2 + C^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14
\]
- Now, substituting into the foot of the perpendicular formula:
\[
\text{Foot} = \left( 2 - \frac{2 \cdot (-10)}{14}, -3 - \frac{1 \cdot (-10)}{14}, 2 - \frac{-3 \cdot (-10)}{14} \right)
\]
- Simplifying gives:
\[
\text{Foot} = \left( 2 + \frac{20}{14}, -3 + \frac{10}{14}, 2 - \frac{30}{14} \right) = \left( 2 + \frac{10}{7}, -3 + \frac{5}{7}, 2 - \frac{15}{7} \right)
\]
- Thus, the foot of the perpendicular is:
\[
\left( \frac{24}{7}, -\frac{16}{7}, -\frac{1}{7} \right)
\]
4. **Find the image point**:
- The image point \( I \) can be found using the midpoint formula:
\[
I = (2 \cdot \text{Foot} - P)
\]
- Calculating gives:
\[
I = \left( 2 \cdot \frac{24}{7} - 2, 2 \cdot -\frac{16}{7} + 3, 2 \cdot -\frac{1}{7} - 2 \right)
\]
- Simplifying:
\[
I = \left( \frac{48}{7} - \frac{14}{7}, -\frac{32}{7} + \frac{21}{7}, -\frac{2}{7} - \frac{14}{7} \right) = \left( \frac{34}{7}, -\frac{11}{7}, -\frac{16}{7} \right)
\]
#### (ii) Finding the image of the point (1, 2, 3) in the plane \(x + 2y + 4z = 38\)
1. **Identify the point and the plane equation**:
- Point \( P(1, 2, 3) \)
- Plane equation: \( x + 2y + 4z = 38 \)
2. **Find the normal vector of the plane**:
- Normal vector \( \mathbf{n} = (1, 2, 4) \)
3. **Find the foot of the perpendicular from point P to the plane**:
- Rearranging gives \( D = -38 \).
- Calculate \( A^2 + B^2 + C^2 \):
\[
A^2 + B^2 + C^2 = 1^2 + 2^2 + 4^2 = 1 + 4 + 16 = 21
\]
- Foot of the perpendicular:
\[
\text{Foot} = \left( 1 - \frac{1 \cdot (-38)}{21}, 2 - \frac{2 \cdot (-38)}{21}, 3 - \frac{4 \cdot (-38)}{21} \right)
\]
- Simplifying gives:
\[
\text{Foot} = \left( 1 + \frac{38}{21}, 2 + \frac{76}{21}, 3 + \frac{152}{21} \right) = \left( \frac{59}{21}, \frac{118}{21}, \frac{215}{21} \right)
\]
4. **Find the image point**:
- The image point \( I \):
\[
I = (2 \cdot \text{Foot} - P)
\]
- Calculating gives:
\[
I = \left( 2 \cdot \frac{59}{21} - 1, 2 \cdot \frac{118}{21} - 2, 2 \cdot \frac{215}{21} - 3 \right)
\]
- Simplifying:
\[
I = \left( \frac{118}{21} - \frac{21}{21}, \frac{236}{21} - \frac{42}{21}, \frac{430}{21} - \frac{63}{21} \right) = \left( \frac{97}{21}, \frac{194}{21}, \frac{367}{21} \right)
\]
#### (iii) Finding the image of the point (2, -1, 3) in the plane \(3x - 2y - z = 9\)
1. **Identify the point and the plane equation**:
- Point \( P(2, -1, 3) \)
- Plane equation: \( 3x - 2y - z = 9 \)
2. **Find the normal vector of the plane**:
- Normal vector \( \mathbf{n} = (3, -2, -1) \)
3. **Find the foot of the perpendicular from point P to the plane**:
- Rearranging gives \( D = -9 \).
- Calculate \( A^2 + B^2 + C^2 \):
\[
A^2 + B^2 + C^2 = 3^2 + (-2)^2 + (-1)^2 = 9 + 4 + 1 = 14
\]
- Foot of the perpendicular:
\[
\text{Foot} = \left( 2 - \frac{3 \cdot (-9)}{14}, -1 - \frac{-2 \cdot (-9)}{14}, 3 - \frac{-1 \cdot (-9)}{14} \right)
\]
- Simplifying gives:
\[
\text{Foot} = \left( 2 + \frac{27}{14}, -1 - \frac{18}{14}, 3 - \frac{9}{14} \right) = \left( \frac{55}{14}, -\frac{32}{14}, \frac{33}{14} \right)
\]
4. **Find the image point**:
- The image point \( I \):
\[
I = (2 \cdot \text{Foot} - P)
\]
- Calculating gives:
\[
I = \left( 2 \cdot \frac{55}{14} - 2, 2 \cdot -\frac{32}{14} + 1, 2 \cdot \frac{33}{14} - 3 \right)
\]
- Simplifying:
\[
I = \left( \frac{110}{14} - \frac{28}{14}, -\frac{64}{14} + \frac{14}{14}, \frac{66}{14} - \frac{42}{14} \right) = \left( \frac{82}{14}, -\frac{50}{14}, \frac{24}{14} \right)
\]
### Final Results:
- (i) The image of the point (2, -3, 2) in the plane \(2x + y - 3z = 10\) is \(\left( \frac{34}{7}, -\frac{11}{7}, -\frac{16}{7} \right)\).
- (ii) The image of the point (1, 2, 3) in the plane \(x + 2y + 4z = 38\) is \(\left( \frac{97}{21}, \frac{194}{21}, \frac{367}{21} \right)\).
- (iii) The image of the point (2, -1, 3) in the plane \(3x - 2y - z = 9\) is \(\left( \frac{82}{14}, -\frac{50}{14}, \frac{24}{14} \right)\).
