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 coplaner, then $k$ can have :

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

Answer 1

The correct option is C exactly two values

$\left|\begin{array}{ccc}2-1& 3-4& 4-5\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}1& -1& -1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$
$\Rightarrow 1(1+2k)+1(1+k^2)-1(2-k)=0$
$\Rightarrow k^2+3k=0$
$\Rightarrow k=0,k=-3$
Hence, $k$ has exactly two values.