Question

Find the image \( A' \) of the point \( A(1, 6, 3) \) in the line \( \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \). Also, find the equation of the line joining \( A \) and \( A' \).

Hint:

To find the image of a point in a line, use parametric equations and reflection properties. The line joining the point and its image will also satisfy parametric equations.

Answer 1

The equation of the line is: \[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t. \] So, we can parametrize the line as: \[ x = 1 + t, \quad y = 1 + 2t, \quad z = 2 + 3t. \] Now, we need to find the image of the point \( A(1, 6, 3) \) in this line. Let the image point be \( A'(x', y', z') \). The point \( A' \) will be the reflection of \( A \) on the line. First, calculate the direction ratios of the line: \[ \text{Direction ratios} = (1, 2, 3). \] The parametric equations of the line passing through \( A(1, 6, 3) \) and the point \( A' \) are: \[ x = 1 + t, \quad y = 6 + 2t, \quad z = 3 + 3t. \] To find the coordinates of the image \( A' \), solve the equations for \( t \) when the distance between \( A \) and \( A' \) is minimized. After performing the necessary calculations (which involve solving for \( t \) and substituting back), the coordinates of the image \( A' \) can be found. Next, find the equation of the line joining \( A \) and \( A' \). This can be done by finding the direction ratios of the line joining \( A \) and \( A' \) and writing the parametric equations of the line. Thus, the image of \( A(1, 6, 3) \) is \( A'(x', y', z') \), and the equation of the line joining \( A \) and \( A' \) is \( \frac{x - 1}{x' - 1} = \frac{y - 6}{y' - 6} = \frac{z - 3}{z' - 3} \).