Question

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

• A. \(-1\)
• B. \( \frac{2}{9} \)
• C. \( \frac{9}{2} \)
• D. \(0\)

Hint:

For intersecting lines → equate coordinates and solve parameters systematically.

Answer 1

The Correct Option is C

Concept: For intersection → coordinates must satisfy both lines.

Step 1: Parametric form of first line \[ x = 1 + 2\lambda,\quad y = -1 + 3\lambda,\quad z = 1 + 4\lambda \]

Step 2: Parametric form of second line \[ x = 3 + \mu,\quad y = k + 2\mu,\quad z = \mu \]

Step 3: Equate coordinates From \(z\): \[ 1 + 4\lambda = \mu \] From \(x\): \[ 1 + 2\lambda = 3 + \mu \] Substitute \(\mu\): \[ 1 + 2\lambda = 3 + (1 + 4\lambda) \] \[ 1 + 2\lambda = 4 + 4\lambda \Rightarrow -3 = 2\lambda \Rightarrow \lambda = -\frac{3}{2} \] \[ \mu = 1 + 4\left(-\frac{3}{2}\right) = 1 - 6 = -5 \]

Step 4: Use \(y\)-coordinate \[ -1 + 3\lambda = k + 2\mu \] \[ -1 + 3\left(-\frac{3}{2}\right) = k + 2(-5) \] \[ -1 - \frac{9}{2} = k - 10 \] \[ -\frac{11}{2} = k - 10 \Rightarrow k = 10 - \frac{11}{2} = \frac{20 - 11}{2} = \frac{9}{2} \] Conclusion \[ {k = \frac{9}{2}} \]