Question

The lines \(\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}\) and \(\frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1}\) are coplanar if

• A. \(k = 0,-3\)
• B. \(k = -1,3\)
• C. \(k = 1,2\)
• D. \(k = 2,4\)

Hint:

Use the determinant condition for coplanar lines.

Answer 1

The Correct Option is A

Step 1: Test option A directly:
For $k = 0$: directions $(1,1,0)$ and $(0,2,1)$, connecting vector $(-1,1,1)$. Determinant: $-1(1) - 1(1) + 1(2) = 0$.

Step 2: Test k = -3:
Directions $(1,1,3)$ and $(-3,2,1)$. Determinant: $-1(1 - 6) - 1(1 + 9) + 1(2 + 3) = 5 - 10 + 5 = 0$.

Step 3: Other options:
For $k=1$ the determinant is $-4$, and for $k=-1$ it is $2$, not zero, so (B), (C), (D) fail. Option (A).

Final Answer:
The lines are coplanar for k = 0 or -3. \[ \boxed{\text{(A) }k=0,\,-3} \]