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)

Condition for two lines are coplanar.
$\begin{vmatrix} x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\l_{1} & m_{1} & n_{1} \\l_{2} & m_{2} & n_{2}\end{vmatrix}=0$
where, $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ are the points lie on lines (i) and (ii) respectively and $ < l_{1}, m_{1}, n_{1}>$ and $ < l_{2}, m_{2}, n_{2}>$ are the direction cosines of the line (i) and line (ii), respectively.
$\therefore \begin{vmatrix} 2-1 & 3-4 & 4-5 \\1 & 1 & -k \\k & 2 & 1\end{vmatrix}=0$
$\Rightarrow \begin{vmatrix} 1 & -1 & -1 \\1 & 1 & -k \\k & 2 & 1\end{vmatrix} =0$
$\Rightarrow 1(1+2 k)+\left(1+k^{2}\right)-(2-k) =0$
$\Rightarrow k^{2}+2 k+k =0 $
$\Rightarrow k^{2}+3 k =0$
$\Rightarrow k =0,-3$
If $0$ appears in the denominator, then the correct way of representing the equation of straigh.t line is
$\frac{x-2}{1}=\frac{y-3}{1} ; z=4$