Question

If the lines \( \frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k} \) and \( \frac{x-1}{2k} = \frac{y-3}{2} = \frac{z-5}{1} \) are coplanar, then \( k \) can have

• A. exactly one value, \(k=\frac{1}{2}\)
• B. exactly one value, \(k=\frac{1}{4}\)
• C. exactly two values, \(k=\frac{1}{2}, -\frac{3}{2}\)
• D. any value

Hint:

Coplanar lines $\longrightarrow$ scalar triple product = 0.

Answer 1

The Correct Option is C

Concept: Two lines are coplanar if: \[ (\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0 \]

Step 1: Direction vectors.
\[ \vec{d_1} = (1,1,-k), \quad \vec{d_2} = (2k,2,1) \]

Step 2: Points.
\[ A(2,3,4), \quad B(1,3,5) \] \[ \vec{AB} = (-1,0,1) \]

Step 3: Apply condition.
\[ \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 0 \Rightarrow k = \frac{1}{2}, -\frac{3}{2} \]