Question

Show that the lines: ${\displaystyle \frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1}}$ and ${\displaystyle \frac{x-4}{3}=\frac{2y-2}{-4}=z-1}$ are coplanar.

Answer 1

Given lines are:  ${\displaystyle \frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1}}$ and ${\displaystyle \frac{x-4}{3}=\frac{2y-2}{-4}=z-1}$ 

or ${\displaystyle \frac{x-1}{-2}=\frac{y-3}{4}=\frac{z-0}{1}}$ and ${\displaystyle \frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1}}$

There lines will be coplanar if ${\displaystyle \left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}$ = 0

${\displaystyle \left|\begin{array}{ccc}4-1& 1-3& 1-0\\ -2& 4& -1\\ 3& -2& 1\end{array}\right|=\left|\begin{array}{ccc}3& -2& 1\\ -2& 4& -1\\ 3& -2& 1\end{array}\right|}$ = 0  ...[Since, R₁ = R₃]

Thus, given lines are coplanar.