Question

Show that the lines $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\mathrm{and}\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ are coplanar. Also, find the equation of the plane containing them.

Answer 1

$$\begin{gathered} \text{We know that the lines} \\ \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\text{and}\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\text{are coplanar if} \\ \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0 \\ \text{and the equation of the plane containing these lines is} \\ \left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0 \\ \text{Here,} \\ {x}_1=-1;y_1=3;z_1=-2;x_2=0;y_2=7;z_2=-7;l_1=-3;m_1=2;n_1=1;l_2=1;m_2=-3;n_2=2 \\ \text{Now,} \\ \left|\begin{array}{ccc}0+1& 7-3& -7+2\\ -3& 2& 1\\ 1& -3& 2\end{array}\right| \\ =\left|\begin{array}{ccc}1& 4& -5\\ -3& 2& 1\\ 1& -3& 2\end{array}\right| \\ =1\left(7\right)-4\left(-7\right)-5\left(7\right) \\ =7+28-35 \\ =0 \\ \text{So, the given lines are coplanar.} \\ \text{The equation of the plane containing the given lines is} \\ \left|\begin{array}{ccc}x+1& y-3& z+2\\ -3& 2& 1\\ 1& -3& 2\end{array}\right|=0 \\ \Rightarrow \left(x+1\right)\left(7\right)-\left(y-3\right)\left(-7\right)+\left(z+2\right)\left(7\right)=0 \\ \Rightarrow 7x+7y+7z=0 \\ \Rightarrow x+y+z=0 \end{gathered}$$