Question

Consider a line \( L \) passing through the points \( P(1, 2, 1) \) and \( Q(2, 1, -1) \). If the mirror image of the point \( A(2, 2, 2) \) in the line \( L \) is \( (\alpha, \beta, \gamma) \), then \( \alpha + \beta + 6\gamma \) is equal to .

Answer 1

Correct Answer: 6

Step 1: Direction Ratios of Line \( L \):

The direction ratios of line \( L \) passing through \( P(1,2,1) \) and \( Q(2,1,-1) \) are: \[ \text{DR's of } L = (2 - 1, 1 - 2, -1 - 1) = 1 : -1 : -2. \]

Step 2: Direction Ratios of \( AB \):

Let \( B(\alpha, \beta, \gamma) \) be the mirror image of \( A(2,2,2) \) in line \( L \). Then, the direction ratios of \( AB \) are: \[ (\alpha - 2, \beta - 2, \gamma - 2). \]

Step 3: Perpendicular Condition:

Since \( AB \) is perpendicular to line \( L \), we have: \[ 1(\alpha - 2) - 1(\beta - 2) - 2(\gamma - 2) = 0. \] Simplifying, we get: \[ \alpha - \beta - 2\gamma = -4. \tag{1} \]

Step 4: Finding the Midpoint \( C \) of \( AB \):

The midpoint \( C \) of \( AB \) lies on the line \( L \), so: \[ C = \left(\frac{\alpha + 2}{2}, \frac{\beta + 2}{2}, \frac{\gamma + 2}{2}\right). \]

Step 5: Direction Ratios of \( PC \) and Parallel Condition:

The direction ratios of \( PC \) are: \[ \left(\frac{\alpha - 2}{2}, \frac{\beta - 2}{2}, \frac{\gamma - 2}{2}\right). \] Since line \( L \) is parallel to \( PC \), we have: \[ \frac{\alpha - 2}{2} : \frac{\beta - 2}{2} : \frac{\gamma - 2}{2} = 1 : -1 : -2. \] This gives: \[ \alpha - 2 = -2K, \quad \beta - 2 = 2K, \quad \gamma - 2 = 4K. \] Solving for \( \alpha \), \( \beta \), and \( \gamma \), we get: \[ \alpha = -2K + 2, \quad \beta = 2K + 2, \quad \gamma = 4K + 2. \]

Step 6: Substitute into Equation (1):

Substitute these values into equation (1): \[ -2K + 2 - (2K + 2) - 2(4K + 2) = -4. \] Solving for \( K \), we find: \[ K = \frac{1}{6}. \]

Step 7: Calculating \( \alpha + \beta + 6\gamma \):

Substitute \( K = \frac{1}{6} \) into the expressions for \( \alpha \), \( \beta \), and \( \gamma \): \[ \alpha = -2 \times \frac{1}{6} + 2 = \frac{10}{6} = \frac{5}{3}, \] \[ \beta = 2 \times \frac{1}{6} + 2 = \frac{14}{6} = \frac{7}{3}, \] \[ \gamma = 4 \times \frac{1}{6} + 2 = \frac{16}{6} = \frac{8}{3}. \] Therefore, \[ \alpha + \beta + 6\gamma = \frac{5}{3} + \frac{7}{3} + 6 \times \frac{8}{3} = \frac{24}{3} = 6. \]

Answer: 6