Question

Prove that the lines $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$ and $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ are coplanar. Also, find the equation of the plane containing these two lines.

Answer 1

If two line are coplanar then

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ 1_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$

Here

$(x_1,y_1,z_1)=(2,4,6)$

$(x_2,y_2,z_2)=(-1,-3,-5)$

$(l_1,m_1,n_1)=(1,4,7)$

$(l_2,m_2,n_2)=(3,5,7)$

Substituting the values

$=\left|\begin{array}{ccc}-3& -7& -11\\ 1& 4& 7\\ 3& 5& 7\end{array}\right|$

$=-3(28-35)+7(7-21)-11(5-12)$

$=21-98+77$

$=98-98$

$=0$

Hence the lines are coplanar.

The equation of the plane containing these lines is

$\left|\begin{array}{ccc}x-2& y-4& z-6\\ 1& 4& 7\\ 3& 5& 7\end{array}\right|=0$

$\Rightarrow (x-2)(28-35)-(y-4)(7-21)+(z-6)(5-12)=0$

$\Rightarrow (x-2)(-7)-(y-4)(-14)+(z-6)(-7)=0$

$\Rightarrow -7x+14+14y-56-7z+42=0$

$\Rightarrow -7x+14y-7z=0$

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

Hence this is the equation of the plane.