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 given line has the equation: \[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t. \] This implies the parametric form of the line is: \[ x = 1 + t, \quad y = 1 + 2t, \quad z = 2 + 3t. \] We aim to find the image of point \( A(1, 6, 3) \) reflected across this line. Let the image point be \( A'(x', y', z') \). The direction ratios of the line are \( (1, 2, 3) \). The parametric equations for a line passing through \( A(1, 6, 3) \) with the same direction ratios are: \[ x = 1 + t, \quad y = 6 + 2t, \quad z = 3 + 3t. \] To find \( A' \), we determine the value of \( t \) for which the distance between \( A \) and the point on the line is minimized. This involves solving for \( t \) and substituting back into the parametric equations. Subsequently, we determine the equation of the line segment connecting \( A \) and \( A' \). This is achieved by finding the direction ratios of the line segment \( AA' \) and expressing its parametric equations. Consequently, the image of \( A(1, 6, 3) \) is \( A'(x', y', z') \), and the line segment connecting \( A \) and \( A' \) is represented by the equation \( \frac{x - 1}{x' - 1} = \frac{y - 6}{y' - 6} = \frac{z - 3}{z' - 3} \).