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

To solve the problem, we are given:

- A point \( A(2, 1, 2) \)
- A line \( l: \mathbf{r} = \langle 4, 2, 2 \rangle + \lambda \langle 1, -1, -1 \rangle \)
We are to find:

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

1. Direction Vector and Point on Line:
- Point \( P_0 = (4, 2, 2) \)
- Direction vector \( \mathbf{d} = \langle 1, -1, -1 \rangle \)

2. Find Foot of Perpendicular from \( A \) to Line \( l \):
Let the foot of the perpendicular be point \( F \) on the line \( l \).
Then \( F = (4 + \lambda, 2 - \lambda, 2 - \lambda) \)

Let vector \( \vec{AF} = F - A = \langle 4 + \lambda - 2,\ 2 - \lambda - 1,\ 2 - \lambda - 2 \rangle = \langle \lambda + 2,\ 1 - \lambda,\ -\lambda \rangle \)

Since \( \vec{AF} \perp \mathbf{d} = \langle 1, -1, -1 \rangle \), their dot product is 0:

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

3. Find Coordinates of Foot \( F \):
\[ F = (4 + \lambda,\ 2 - \lambda,\ 2 - \lambda) = \left(4 - \frac{1}{3},\ 2 + \frac{1}{3},\ 2 + \frac{1}{3} \right) = \left( \frac{11}{3},\ \frac{7}{3},\ \frac{7}{3} \right) \]

4. Find Image Point \( A' \):
Image is the reflection of \( A \) over foot \( F \):
Use midpoint formula: \[ F = \frac{A + A'}{2} \Rightarrow A' = 2F - A \] \[ A' = 2 \cdot \left( \frac{11}{3}, \frac{7}{3}, \frac{7}{3} \right) - (2, 1, 2) = \left( \frac{22}{3} - 2, \frac{14}{3} - 1, \frac{14}{3} - 2 \right) = \left( \frac{16}{3}, \frac{11}{3}, \frac{8}{3} \right) \]

5. Equation of Line Joining \( A \) and \( A' \):
Direction vector \( \vec{AA'} = A' - A = \left( \frac{16}{3} - 2, \frac{11}{3} - 1, \frac{8}{3} - 2 \right) = \left( \frac{10}{3}, \frac{8}{3}, \frac{2}{3} \right) \)

So parametric form of line through \( A(2,1,2) \) is:

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

Final Answers:
- 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 \( AA' \): \[ x = 2 + \frac{10}{3}t,\quad y = 1 + \frac{8}{3}t,\quad z = 2 + \frac{2}{3}t \]