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.
$\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$
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.
$\therefore \left|\begin{array}{ccc}2-1& 3-4& 4-5\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& -1& -1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0$
$\Rightarrow 1(1+2k)+\left(1+k^2\right)-(2-k)=0$
$\Rightarrow k^2+2k+k=0$
$\Rightarrow k^2+3k=0$
$\Rightarrow k=0,-3$
If $0$ appears in the denominator, then the correct way of representing the equation of straigh.t line is
$\frac{x-2}{1}=\frac{y-3}{1};z=4$