If the lines x-21=y-31=z-4-k and x-1k=y-42=z-51 are coplanar, then k can have

Question

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

Accepted Answer

Given lines will be coplanar

If $$\,\,\,\,\left| {\begin{pmatrix} { - 1} & 1 & 1  \
   1 & 1 & { - k}  \
   k & 2 & 1  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow  - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$

$$ \Rightarrow k = 0, - 3$$

Solutions · 1