Question

Find the equation of a line \( l_2 \) which is the mirror image of the line \( l_1 \) with respect to line \( l : \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \), given that line \( l_1 \) passes through the point \( P(1, 6, 3) \) and is parallel to line \( l \).

Hint:

To find the mirror image of a line with respect to another, calculate the foot of the perpendicular and use reflection formulas for the points.

Answer 1

Step 1: Parametric form of the given line \( l \). The given line \( l \) is: \[ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \] Its parametric equations are: \[ x = t, \quad y = 1 + 2t, \quad z = 2 + 3t, \] where \( t \) is the parameter. Step 2: Direction ratios of line \( l \). The direction ratios (DRs) of line \( l \) are \( (1, 2, 3) \). Step 3: Equation of line \( l_1 \). Line \( l_1 \) passes through the point \( P(1, 6, 3) \) and is parallel to \( l \), so its DRs are also \( (1, 2, 3) \). The equation of \( l_1 \) is: \[ \frac{x - 1}{1} = \frac{y - 6}{2} = \frac{z - 3}{3} \] Step 4: Find the foot of the perpendicular from \( P(1, 6, 3) \) to \( l \). Let the foot of the perpendicular be \( Q(t, 1 + 2t, 2 + 3t) \) on line \( l \). The DRs of \( PQ \) are \( (t - 1, 2t - 5, 3t - 1) \). Using the perpendicularity condition: \[ (t - 1)(1) + (2t - 5)(2) + (3t - 1)(3) = 0 \] Simplify: \[ t - 1 + 4t - 10 + 9t - 3 = 0 \] \[ 14t - 14 = 0 \quad \Rightarrow \quad t = 1 \] Substituting \( t = 1 \): \[ Q(1, 3, 5) \] Step 5: Reflection of \( P(1, 6, 3) \) about \( Q(1, 3, 5) \). The reflection point \( P'(x', y', z') \) of \( P(x_1, y_1, z_1) \) across \( Q(x_2, y_2, z_2) \) is: \[ x' = 2x_2 - x_1, \quad y' = 2y_2 - y_1, \quad z' = 2z_2 - z_1 \] Substituting values: \[ x' = 2(1) - 1 = 1, \quad y' = 2(3) - 6 = 0, \quad z' = 2(5) - 3 = 7 \] Thus, the reflected point is \( P'(1, 0, 7) \). Step 6: Equation of line \( l_2 \). Line \( l_2 \) passes through \( P'(1, 0, 7) \) and is parallel to \( l \), so its DRs are \( (1, 2, 3) \). The equation of \( l_2 \) is: \[ \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 7}{3} \] Final Answer: \[ \boxed{\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 7}{3}} \]