The lines \(\displaystyle\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-K}\) and \(\displaystyle\frac{x-1}{K} = \frac{y-4}{2} = \frac{z-5}{1}\) are coplanar if:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(K = 0\) or \(-3\)
The first line passes through \(A(2, 3, 4)\) with direction \(\vec{b}_1 = (1, 1, -K)\); the second through \(B(1, 4, 5)\) with direction \(\vec{b}_2 = (K, 2, 1)\).
The directions are never parallel (\(\frac{1}{K} = \frac{1}{2}\) would need K = 2, and then \(\frac{-K}{1} = -2 \ne \frac{1}{2}\)). So the lines lie in one plane exactly when they meet, that is when the shortest distance between them is 0:
\( \displaystyle d = \frac{\left|\overrightarrow{AB}\cdot(\vec{b}_1\times\vec{b}_2)\right|}{|\vec{b}_1\times\vec{b}_2|} = 0 \)
\(\overrightarrow{AB} = (1 - 2,\ 4 - 3,\ 5 - 4) = (-1, 1, 1)\)
\( \displaystyle \vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & -K \\ K & 2 & 1 \end{vmatrix} \)
\( = (1 + 2K)\hat{\imath} - (1 + K^2)\hat{\jmath} + (2 - K)\hat{k}\)
So
\(\overrightarrow{AB}\cdot(\vec{b}_1\times\vec{b}_2) = -(1 + 2K) - (1 + K^2) + (2 - K)\)
\(= -K^2 - 3K\)
Setting it to 0:
\(K(K + 3) = 0\)
\(K = 0 \text{ or } K = -3\)