how do you prove that two lines in 3D are coplanar? Apply that to Prove the following question
Question : Prove that the lines (x-1)/2=(y-2)/3=(z-3)/4 and (x-2)/3=(y-3)/4=(z-4)/5 are coplanar.
Solution:
We know that lines \[\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\ and\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\] are coplanar if \[\left|\begin{matrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\\\end{matrix}\right|=0\] Therefore for the lines \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\ and \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\] we have \[x_1=1,\ y_1=2,\ z_1=3,\ a_1=2.\ b_1=3,\ c_1=4\] and \[x_2=2,\ y_2=3,\ z_3=4,\ a_2=3,\ b_2=4,\ c_2=5\] Substituting into the condition \[\left|\begin{matrix}2-1&3-2&4-3\\2&3&4\\3&4&5\\\end{matrix}\right|\] \[=\left|\begin{matrix}1&1&1\\2&3&4\\3&4&5\\\end{matrix}\right|\] \[=1\left(15-16\right)-1\left(10-12\right)+1\left(8-9\right)\] =-1+2-1=2-2=0 Therefore the given two lines are coplanar.