Question

If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then $14(\alpha + \beta + \gamma)$ is:

• A. 102
• B. 138
• C. 108
• D. 132

Answer 1

The Correct Option is C

Given:

The point \( P(3, 4, 9) \) lies on the line with parametric equations: \[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1} = \lambda \]

Step 1: Equation of any point on the line:

The general form for any point \( Q \) on the line is given by: \[ Q(3\lambda + 1, 2\lambda - 1, \lambda + 2) \]

Step 2: Equation of \( \overrightarrow{PQ} \):

The direction vector \( \overrightarrow{PQ} \) is given by: \[ \langle 3\lambda - 2, 2\lambda - 5, \lambda - 7 \rangle \quad \text{and} \quad \langle 3, 2, 1 \rangle \] Solving for the determinant: \[ 9\lambda - 6 + 4\lambda - 10 + \lambda - 7 = 0 \] Simplifying: \[ 14\lambda - 23 = 0 \] Solving for \( \lambda \): \[ \lambda = \frac{23}{14} \]

Step 3: Finding coordinates of point \( Q \):

Substituting \( \lambda = \frac{23}{14} \) into the coordinates of \( Q \): \[ Q\left( \frac{83}{14}, \frac{32}{14}, \frac{51}{14} \right) \]

Step 4: Parametric coordinates of point \( P \):

For point \( P(3, 4, 9) \), the parametric coordinates are: \[ 3 + \alpha_1 = \frac{83}{14} \quad \Rightarrow \quad x_1 = \frac{62}{7} \] \[ 4 + \beta_1 = \frac{32}{14} \quad \Rightarrow \quad y_1 = \frac{4}{7} \] \[ 9 + \gamma_1 = \frac{51}{14} \quad \Rightarrow \quad z_1 = \frac{-12}{7} \]

Step 5: Final Calculation:

Now, we calculate: \[ 14(\alpha + \beta + \gamma) = 14 \times \left( \frac{62}{7} + \frac{4}{7} - \frac{12}{7} \right) = 108 \]

Answer 2

Given Information:

The point \( P(3, 4, 9) \) lies on the line with the following parametric equations: \[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1} = \lambda \]

Step 1: Parametric Equation for Any Point on the Line:

The coordinates of any point \( Q \) on the line are expressed as: \[ Q\left( 3\lambda + 1, 2\lambda - 1, \lambda + 2 \right) \]

Step 2: Direction Vector of \( \overrightarrow{PQ} \):

The direction vector from \( P \) to \( Q \) is given by: \[ \langle 3\lambda - 2, 2\lambda - 5, \lambda - 7 \rangle \quad \text{and} \quad \langle 3, 2, 1 \rangle \] Simplifying: \[ 9\lambda - 6 + 4\lambda - 10 + \lambda - 7 = 0 \] Solving for \( \lambda \): \[ 14\lambda - 23 = 0 \quad \Rightarrow \quad \lambda = \frac{23}{14} \]

Step 3: Coordinates of Point \( Q \):

Substituting \( \lambda = \frac{23}{14} \) into the parametric equation of the line: \[ Q\left( \frac{83}{14}, \frac{32}{14}, \frac{51}{14} \right) \]

Step 4: Parametric Coordinates of Point \( P \):

For point \( P(3, 4, 9) \), the parametric coordinates are derived as: \[ 3 + \alpha_1 = \frac{83}{14} \quad \Rightarrow \quad x_1 = \frac{62}{7} \] \[ 4 + \beta_1 = \frac{32}{14} \quad \Rightarrow \quad y_1 = \frac{4}{7} \] \[ 9 + \gamma_1 = \frac{51}{14} \quad \Rightarrow \quad z_1 = \frac{-12}{7} \]

Step 5: Final Calculation:

Now, summing the values for \( \alpha \), \( \beta \), and \( \gamma \), we get: \[ 14(\alpha + \beta + \gamma) = 14 \times \left( \frac{62}{7} + \frac{4}{7} - \frac{12}{7} \right) = 108 \]