Question

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, find the equations of the planes containing them.

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 $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar.

$$\begin{gathered} \therefore \left|\begin{array}{ccc}-1-1& -1-\left(-1\right)& 0-0\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}-2& 0& 0\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0 \\ \Rightarrow -2\left(k^2-4\right)-0+0=0 \\ \Rightarrow k^2-4=0 \\ \Rightarrow k=\pm 2 \end{gathered}$$
The equation of the plane containing the given lines is $\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0$.

For k = 2, $\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=$$\left|\begin{array}{ccc}x-1& y+1& z\\ 2& 2& 2\\ 2& 2& 2\end{array}\right|=0$
So, no plane exists for k = 2.

For k = −2,
$\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0$
$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}x-1& y+1& z\\ 2& -2& 2\\ 2& 2& -2\end{array}\right|=0 \\ \Rightarrow \left(x-1\right)\left(4-4\right)-\left(y+1\right)\left(-4-4\right)+z\left(4+4\right)=0 \\ \Rightarrow 8\left(y+1\right)+8z=0 \\ \Rightarrow y+z+1=0 \end{gathered}$$
Thus, the equation of the plane containing the given lines is y + z + 1 = 0.