Question

Find the values of $\lambda$ for which the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^2}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^2}=\frac{z-1}{2}$ are coplanar.

Answer 1

The lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$.

The given lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^2}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^2}=\frac{z-1}{2}$ are coplanar.

$\therefore \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$
$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}3-1& 2-2& 1-\left(-3\right)\\ 1& 2& {\lambda }^2\\ 1& {\lambda }^2& 2\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}2& 0& 4\\ 1& 2& {\lambda }^2\\ 1& {\lambda }^2& 2\end{array}\right|=0 \\ \Rightarrow 2\left(4-{\lambda }^4\right)-0+4\left({\lambda }^2-2\right)=0 \\ \Rightarrow -2{\lambda }^4+4{\lambda }^2=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow {\lambda }^2\left({\lambda }^2-2\right)=0 \\ \Rightarrow {\lambda }^2=0\mathrm{or}{\lambda }^2-2=0 \\ \Rightarrow \lambda =0\mathrm{or}\lambda =\pm \sqrt{2} \end{gathered}$$
Thus, the values of $\lambda$ are 0, $-\sqrt{2}$ and $\sqrt{2}$.