Question

Find the equation of a plane containing the lines $\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and

$\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$.

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,

$\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}$ ------ ( 1 )

$\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$ ------ ( 2 )

We get,

$\Rightarrow$ $x_1=5,y_1=7,z_1=-3$ and $a_1=4,b_1=4,c_1=-5$

$\Rightarrow$ $x_2=8,y_2=4,z_2=5$ and $a_2=7,b_2=1,c_2=3$

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$ $=\left|\begin{array}{ccc}3& -3& 8\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|$

$=3(12+5)+3(12+35)+8(4-28)$

$=51+141-192$

$=0$

$\therefore$ The lines are co-planar.

The equation of the plane containing the given lines is

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

$\Rightarrow$ $(x-5)(12+5)-(y-7)(12+35)-(z+3)(4-28)=0$

$\Rightarrow$ $(x-5)\left(17\right)-(y-7)\left(47\right)-(z+3)(-24)=0$

$\Rightarrow$ $17x-85-47y+329+24z+72=0$

$\Rightarrow$ $17x-47y+24z+316=0$