Browse Questions

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

Solution :
If two lines are Coplanar then
$\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_@-z_1\\ l_1 & m_1 & n_1 \\l_2 & m_2 & n_2\end{vmatrix}$
given $(x_1,y_1,z_1)$ is $(-3,1,5)$ and
$(x_2,y_2,z_2)$ is $(-1,2,5)$
$l_1,m_1,n_1=-3,1,5$
and $l_1,m_2,n_2=-1,2,5$
Substituting the values
$\begin{vmatrix} -1+3 & 2-1 & 5-5 \\ -3 & 1 & +5 \\ -1 & 2 & 5\end{vmatrix}$
On expanding we get
$\begin{vmatrix} 2 &1 & 0 \\ -3 &1 &5 \\ -1 & 2 & 5 \end{vmatrix}$
$2(5-10-1(-15+5)=0$
Hence the lines are coplanar.
The equation of the plane containing these lines is
$\begin{vmatrix} x+3 & y-1 & z-5 \\ -3 & 1 &5 \\ -1 & 2 &5 \end{vmatrix}$
(ie) $(x+3)(5-10)-(y-1)(-15+5)+(2-5)(-6+1)$
=> $-5(x+3)+10(y-1)-5(z-5)$
=> $-5x+10y-5z-30=0$
or $5x-10y+5z+30=0$
Hence this is the equation of the plane.
edited Apr 28 by meena.p