Question

Find the image \( A' \) of the point \( A(2,1,2) \) in the line \[ l: \mathbf{r} = 4\hat{i} + 2\hat{j} + 2\hat{k} + \lambda (\hat{i} - \hat{j} - \hat{k}). \] Also, find the equation of the line joining \( A A' \). Find the foot of the perpendicular from point \( A \) on the line \( l \).

Hint:

To find the image of a point in a line, find the foot of the perpendicular first, then use symmetry.

Answer 1

Given:

- Point \( A(2, 1, 2) \)
- Line \( l: \mathbf{r} = \langle 4, 2, 2 \rangle + \lambda \langle 1, -1, -1 \rangle \)

To find:

• Image \( A' \) of point \( A \) in line \( l \)
• Equation of the line passing through \( A \) and \( A' \)
• Foot of the perpendicular from \( A \) to line \( l \)

1. Line Parameters:
- A point on the line: \( P_0 = (4, 2, 2) \)
- Direction vector of the line: \( \mathbf{d} = \langle 1, -1, -1 \rangle \)

2. Foot of Perpendicular Calculation:
Let the foot of the perpendicular from \( A \) to line \( l \) be \( F \). Any point on line \( l \) can be represented as \( F = (4 + \lambda, 2 - \lambda, 2 - \lambda) \).

The vector \( \vec{AF} \) is given by \( F - A = \langle (4 + \lambda) - 2, (2 - \lambda) - 1, (2 - \lambda) - 2 \rangle = \langle \lambda + 2, 1 - \lambda, -\lambda \rangle \).

Since \( \vec{AF} \) is perpendicular to the direction vector \( \mathbf{d} \), their dot product is zero:

\[ (\lambda + 2)(1) + (1 - \lambda)(-1) + (-\lambda)(-1) = 0 \]

Solving for \( \lambda \):

\[ \lambda + 2 - 1 + \lambda + \lambda = 0 \Rightarrow 3\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{3} \]

3. Coordinates of the Foot \( F \):
Substitute \( \lambda = -\frac{1}{3} \) into the parametric equation of \( F \):

\[ F = \left(4 - \frac{1}{3},\ 2 - \left(-\frac{1}{3}\right),\ 2 - \left(-\frac{1}{3}\right) \right) = \left(4 - \frac{1}{3},\ 2 + \frac{1}{3},\ 2 + \frac{1}{3} \right) = \left( \frac{11}{3},\ \frac{7}{3},\ \frac{7}{3} \right) \]

4. Image Point \( A' \) Calculation:
The foot \( F \) is the midpoint of the segment \( AA' \). Therefore, \( A' = 2F - A \).

\[ A' = 2 \cdot \left( \frac{11}{3}, \frac{7}{3}, \frac{7}{3} \right) - (2, 1, 2) = \left( \frac{22}{3} - 2, \frac{14}{3} - 1, \frac{14}{3} - 2 \right) = \left( \frac{16}{3}, \frac{11}{3}, \frac{8}{3} \right) \]

5. Equation of Line \( AA' \):
The direction vector of the line \( AA' \) is \( \vec{AA'} = A' - A \):

\[ \vec{AA'} = \left( \frac{16}{3} - 2, \frac{11}{3} - 1, \frac{8}{3} - 2 \right) = \left( \frac{10}{3}, \frac{8}{3}, \frac{2}{3} \right) \]

The parametric equation of the line passing through \( A(2,1,2) \) with direction vector \( \vec{AA'} \) is:

\[ x = 2 + \frac{10}{3}t,\quad y = 1 + \frac{8}{3}t,\quad z = 2 + \frac{2}{3}t \]

Summary of Results:
- Foot of perpendicular \( F \): \( \left( \frac{11}{3}, \frac{7}{3}, \frac{7}{3} \right) \)
- Image point \( A' \): \( \left( \frac{16}{3}, \frac{11}{3}, \frac{8}{3} \right) \)
- Equation of line \( AA' \): \[ x = 2 + \frac{10}{3}t,\quad y = 1 + \frac{8}{3}t,\quad z = 2 + \frac{2}{3}t \]