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

Answer 1

The correct option is B exactly two values

We have, $\left|\begin{array}{ccc}1& -1& -1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$

$\Rightarrow 1(1+2k)+1(1+k^2)-1(2-k)=0$

$\Rightarrow k^2+1+2k+1-2+k=0$

$\Rightarrow k^2+3k=0$

$\Rightarrow \left(k\right)(k+3)=0$

Thus $k$ has $2$ values.