Question

Prove that the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ are coplanar. Find also the point of intersection.

Answer 1

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}.....\left(1\right)$ and $\frac{x-2}{3}=\frac{y=3}{4}=\frac{z-4}{5}.....\left(2\right)$
Hence $x_1=1,y_1=2,z_1=3$
$x_2=2,y_2=3,z_2=4$
and $l_1=2,m_1=3,n_1=4$
$l_2=3,m_2=4,n_2=5$
if lines are coplanar
$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$
now
$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$
$=\left|\begin{array}{ccc}2-1& 3-2& 4-3\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 2& 3& 4\\ 3-2& 4-3& 5-4\end{array}\right|$ by $R_3\to R_3-R_2$
$=\left|\begin{array}{ccc}1& 1& 1\\ 2& 3& 4\\ 1& 1& 1\end{array}\right|$
$=0(\because R_1=R_3)$
$\therefore$ Given lines are coplanar again
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$
$\Rightarrow x=2\lambda +1,y=3\lambda +2,z=4\lambda +3$
On line (i) point are $(2\lambda +1,3\lambda +2,4\lambda +3)$. If line $\left(1\right)$ and $\left(2\right)$ are intersect the other point will be on line $\left(2\right)$.
$\therefore \frac{2\lambda +1-2}{3}=\frac{3\lambda +2-3}{4}$
$\Rightarrow \frac{2\lambda -1}{3}=\frac{3\lambda -1}{4}$
$\Rightarrow 8\lambda -4=9\lambda -3$
$\Rightarrow -4+3=\lambda$
$\Rightarrow \lambda =-1$
$\therefore (2\times -1+1,3\times -1+2,4\times -1+3)$
$=(-1,-1,-1)$.