Question

If 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, then $k$ can have :

• A. any value
• B. exactly one value
• C. exactly two values
• D. exactly three values

Answer 1

Answer: (C)

Since given lines are coplanar
$\therefore \begin{vmatrix} 2-1 &3-4 & 4-5 \\[0.3em] 1 & 1 & -k \\[0.3em] k &2 & 1 \end{vmatrix} = 0 \, i.e., \begin{vmatrix} 1 & -1 & -1 \\[0.3em] 1 & 1 & -k \\[0.3em] k &2 & 1 \end{vmatrix} = 0$
$\Rightarrow $ $(1 + 2k) + (1 + k^2 ) - ( 2 - k) $ = 0
$\Rightarrow $ $k^2 + 3k = 0 \, \Rightarrow \, k = 0, -3$