Question

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

Answer 1

If two lines are coplanar then

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|$

Given $(x_1,y_1,z_1)$ is $(-3,1,5)$ and $(x_2,y_2,z_2)$ is $(-1,2,5)$ and $l_1,m_1,n_1=-3,1,5$

$l_2,m_2,n_2=-1,2,5$

Substituting the values

$\left|\begin{array}{ccc}-1+3& 2-1& 5-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|$

on expanding we get

$\left|\begin{array}{ccc}2& 1& 0\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|$

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

Hence the lines are coplanar.

The equation of the plane containing these lines is

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

$(x+3)(5-10)-(y-1)(-15+5)+(2-5)(-6+1)$

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

$\Rightarrow -5x+10y-5z-30=0$

$5x-10y+5z+30=0$

$5(x-2y+z+6)=0$

$x-2y+z+6=0$

or

$5x-10y+5z+30=0$

$\therefore$ The equation of the plane is $5x-10y+5z+30=0$.