Question

Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5};\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar. Also find the equation of plane containing the lines.

Answer 1

The lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_1}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ 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 lines are,

$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ ------ ( 1 )

$\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ ------ ( 2 )

We get,

$\Rightarrow$ $x_1=-3,y_1=1,z_1=5$ and $a_1=-3,b_1=1,c_1=5$

$\Rightarrow$ $x_2=-1,y_2=2,z_2=5$ and $a_2=-1,b_2=2,c_2=5$

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$ $=\left|\begin{array}{ccc}2& 1& 0\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|$

$=2(5-10)-1(-15+5)+0$

$=-10+10+0$

$=0$

$\therefore$ The lines are co-planar.

The equation of the plane containing the given lines is

$\left|\begin{array}{ccc}x-(-3)& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=0$

$\Rightarrow$ $\left|\begin{array}{ccc}x+3& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=0$

$\Rightarrow$ $(x+3)(5-10)-(y-1)(-15+5)+(z-5)(-6+1)=0$

$\Rightarrow$ $-5(x+3)+10(y-1)-5(z-5)=0$

$\Rightarrow$ $x-2y+z=0$

$\therefore$ Th equation of the plane containing the given lines is $x-2y+z=0$