Question

Let the line $ L $ pass through $ (1, 1, 1) $ and intersect the lines $ \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} $ and $ \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}. $ Then, which of the following points lies on the line $ L $?

• A. \( (4, 22, 7) \)
• B. \( (5, 4, 3) \)
• C. \( (10, -29, -50) \)
• D. \( (7, 15, 13) \)

Hint:

When solving problems involving parametric equations of lines, always substitute the values into the system of equations to find the intersection point, then check which point satisfies all conditions.

Answer 1

The Correct Option is D

Line \( L \) passes through \( (1, 1, 1) \). Its parametric equations are \( x = 1 + 2t, y = 1 + 3t, z = 1 + 4t \).

Consider the first line: \( \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \). Let the common ratio be \( k \). Its parametric equations are \( x = 1 + 2k, y = -1 + 3k, z = 1 + 4k \).

Consider the second line: \( \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1} \). Let the common ratio be \( m \). Its parametric equations are \( x = 3 + m, y = 4 + 2m, z = m \).

We need to find the value of \( t \) where line \( L \) intersects the other two lines.

Equating the parametric equations of line \( L \) with the first line yields a system of equations. Similarly, equating line \( L \) with the second line yields another system.

Solving these systems determines the value of \( t \) and the intersection point on line \( L \).

The intersection point \( (7, 15, 13) \) lies on line \( L \) and satisfies the conditions for intersection with both given lines.

Therefore, the correct answer is: \[ (7, 15, 13). \]