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

Answer 2

Given:
Line \( L \) through points \( P(1, 2, 1) \) and \( Q(2, 1, -1) \)
Point \( A(2, 2, 2) \)
Find mirror image \( A'(\alpha, \beta, \gamma) \) of \( A \) about line \( L \) and calculate \( \alpha + \beta + 6\gamma \). 
Step 1: Direction vector of line \( L \)
\[ \vec{d} = Q - P = (2-1, 1-2, -1-1) = (1, -1, -2) \] Step 2: Vector \( \vec{PA} \)
\[ \vec{PA} = A - P = (2-1, 2-2, 2-1) = (1, 0, 1) \] Step 3: Find projection of \( \vec{PA} \) on \( \vec{d} \)
\[ \text{proj}_{\vec{d}} (\vec{PA}) = \frac{\vec{PA} \cdot \vec{d}}{\vec{d} \cdot \vec{d}} \vec{d} \] Calculate dot products: \[ \vec{PA} \cdot \vec{d} = 1 \times 1 + 0 \times (-1) + 1 \times (-2) = 1 + 0 -2 = -1 \] \[ \vec{d} \cdot \vec{d} = 1^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6 \] So, \[ \text{proj}_{\vec{d}} (\vec{PA}) = \frac{-1}{6} (1, -1, -2) = \left(-\frac{1}{6}, \frac{1}{6}, \frac{1}{3}\right) \] Step 4: Find point \( R \) on line closest to \( A \)
\[ R = P + \text{proj}_{\vec{d}} (\vec{PA}) = \left(1 - \frac{1}{6}, 2 + \frac{1}{6}, 1 + \frac{1}{3}\right) = \left(\frac{5}{6}, \frac{13}{6}, \frac{4}{3}\right) \] Step 5: Find vector \( \vec{AR} \)
\[ \vec{AR} = R - A = \left(\frac{5}{6} - 2, \frac{13}{6} - 2, \frac{4}{3} - 2\right) = \left(-\frac{7}{6}, \frac{1}{6}, -\frac{2}{3}\right) \] Step 6: Find mirror image \( A' \)
Mirror image \( A' = A + 2 \vec{AR} \): \[ A' = \left(2 - \frac{14}{6}, 2 + \frac{2}{6}, 2 - \frac{4}{3}\right) = \left(2 - \frac{7}{3}, 2 + \frac{1}{3}, 2 - \frac{4}{3}\right) = \left(\frac{-1}{3}, \frac{7}{3}, \frac{2}{3}\right) \] Step 7: Calculate \( \alpha + \beta + 6\gamma \)
\[ \alpha + \beta + 6 \gamma = -\frac{1}{3} + \frac{7}{3} + 6 \times \frac{2}{3} = \frac{6}{3} + 4 = 2 + 4 = 6 \] Final answer: 6