Question

Let \( L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \) and \( L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \), where \( \alpha \in \mathbb{R} \), be two lines which intersect at the point \( B \). If \( P \) is the foot of the perpendicular from the point \( A(1, 1, -1) \) on \( L_2 \), then the value of \( 26 \alpha (PB)^2 \) is:

Hint:

For perpendicular distance problems, use the parametric equations of the lines and the formula for the distance from a point to a line to find the required distances.

Answer 1

Correct Answer: 216

Geometry Problem Solution

Given two lines \( L_1 \) and \( L_2 \) with equations:

Line \( L_1 \): \[ \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \]

Line \( L_2 \): \[ \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \]

The intersection point \( B \) of \( L_1 \) and \( L_2 \) is determined, and the foot of the perpendicular \( P \) from point \( A(1, 1, -1) \) to line \( L_2 \) is also found.

Step 1: Parametric Forms of the Lines

For \( L_1 \), the parametric equations are:

\[ x = 1 + 3t, \quad y = 1 - t, \quad z = -1 \] where \( t \) is the parameter.

For \( L_2 \), the parametric equations are:

\[ x = 2 + 2s, \quad y = 0, \quad z = -4 + \alpha s \] where \( s \) is the parameter.

Step 2: Determining the Intersection Point \( B \)

Equating the corresponding coordinates of the lines to find the intersection point \( B \):

1. \( 1 + 3t = 2 + 2s \)

2. \( 1 - t = 0 \)

3. \( -1 = -4 + \alpha s \)

Solving the system:

- From equation 2: \( t = 1 \)

- Substitute \( t = 1 \) into equation 1: \( 1 + 3(1) = 2 + 2s \implies 4 = 2 + 2s \implies s = 1 \)

- Substitute \( s = 1 \) into equation 3: \( -1 = -4 + \alpha(1) \implies \alpha = 3 \)

The intersection point is \( B(5, 0, -1) \).

Step 3: Finding the Foot of the Perpendicular \( P \)

The coordinates of \( P \), the foot of the perpendicular from \( A(1, 1, -1) \) to \( L_2 \), are found using the perpendicularity condition. The vector \( \overrightarrow{AP} \) is orthogonal to the direction vector of \( L_2 \), which is \( (2, 0, 3) \). Solving for the parameter \( s \) yields \( s = \frac{7}{13} \). Substituting \( s = \frac{7}{13} \) into \( L_2 \)'s parametric equations gives:

\[ x_P = \frac{40}{13}, \quad z_P = -\frac{31}{13} \]

The coordinates of \( P \) are \( P\left( \frac{40}{13}, 0, -\frac{31}{13} \right) \).

Step 4: Calculating the Distance \( PB \)

The distance between \( P \) and \( B(5, 0, -1) \) is calculated as:

\[ PB = \sqrt{\left( 5 - \frac{40}{13} \right)^2 + \left( 0 - 0 \right)^2 + \left( -1 + \frac{31}{13} \right)^2} \]

Simplifying the terms:

\[ 5 - \frac{40}{13} = \frac{25}{13}, \quad -1 + \frac{31}{13} = \frac{18}{13} \]

Thus,

\[ PB = \sqrt{\left( \frac{25}{13} \right)^2 + \left( \frac{18}{13} \right)^2} = \sqrt{\frac{625}{169} + \frac{324}{169}} = \sqrt{\frac{949}{169}} = \frac{\sqrt{949}}{13} \]

Step 5: Evaluating \( 26 \alpha (PB)^2 \)

The expression \( 26 \alpha (PB)^2 \) is evaluated with \( \alpha = 3 \):

\[ 26 \alpha (PB)^2 = 26 \times 3 \times \left( \frac{\sqrt{949}}{13} \right)^2 = 26 \times 3 \times \frac{949}{169} \]

Simplifying the expression:

\[ 26 \times 3 \times \frac{949}{169} = \frac{78 \times 949}{169} = \frac{73962}{169} = 216 \]

Final Answer:

The value of \( 26 \alpha (PB)^2 \) is:

\[ \boxed{216} \]