Question

If the lines \( \frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4} \) and \( \frac{x-3}{1} = \frac{y - 9/2}{2} = z/1 \) intersect, then the value of \( k \) is:

• A. \( \frac{1}{2} \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( \frac{3}{2} \)

Hint:

Verify intersection by equating parametric forms of the lines.

Answer 1

The Correct Option is C

Step 1: Parametric equations of the first line: 
The parametric equations of the first line are obtained by setting: \[ \frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4} = t. \] From this, we get: \[ x = 2t + k, \quad y = 3t - 1, \quad z = 4t + 1. \] 
Step 2: Parametric equations of the second line: 
For the second line, set: \[ \frac{x-3}{1} = \frac{y - 9/2}{2} = z = s. \] 
Step 3: Condition for intersection: 
For the two lines to intersect, their parametric forms must satisfy: \[ x_1 = x_2, \quad y_1 = y_2, \quad z_1 = z_2. \] Substitute the equations: \[ 2t + k = s + 3, \quad 3t - 1 = 2s + \frac{9}{2}, \quad 4t + 1 = s. \] 
Step 4: Solve the system of equations: 
From \( 4t + 1 = s \), we get: \[ s = 4t + 1. \] Substitute \( s = 4t + 1 \) into \( 2t + k = s + 3 \): \[ 2t + k = (4t + 1) + 3. \] Simplify: \[ 2t + k = 4t + 4. \] Rearrange to find \( k \): \[ k = 2t + 4 - 4t = 4 - 2t. \] Substitute \( s = 4t + 1 \) into \( 3t - 1 = 2s + \frac{9}{2} \): \[ 3t - 1 = 2(4t + 1) + \frac{9}{2}. \] Simplify: \[ 3t - 1 = 8t + 2 + \frac{9}{2}. \] Combine terms: \[ 3t - 1 = 8t + \frac{13}{2}. \] Multiply through by 2: \[ 16t - 2 = 6t + 13. \] Solve for \( t \): \[ 16t - 6t = 13 + 2. \] \[ 10t = 15 \implies t = \frac{3}{2}. \] 
Step 5: Find \( k \): 
Substitute \( t = \frac{3}{2} \) into \( k = 4 - 2t \): \[ k = 4 - 2\left(\frac{3}{2}\right). \] Simplify: \[ k = 4 - 3 = 1. \] 
Final Answer: \[ \boxed{1}. \] \end{enumerate}