Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)/Solution 2

Translate this problem into a problem involving a system of linear equations. To do this, note that $L_1$ and $L_2$ intersect exactly when there are numbers $s$ and $t$ that satisfy

$[0,2,1]+s[-1,2,2]=[-1,0,3]+t[-2,1,1]$

rearranging this gives

$s[-1,2,2]-t[-2,1,1]=[-1,0,3]-[0,2,1]$

and

$s[-1,2,2]+t[2,-1,-1]=[-1-0,0-2,3-1]=[-1,-2,2]$

Writing as a system of linear equations gives,

$[-s+2t,2s-t,2s-t]=[-1,-2,2]$

In matrix form,

$\left(\begin{array}{cccc}-1& 2& |& -1\\ 2& -1& |& -2\\ 2& -1& |& 2\end{array}\right)$

In other words, $L_1$ and $L_2$ intersect exactly when the above system of linear equations has at least one solution. Using Gaussian Elimination, we obtain

$\left(\begin{array}{cccc}-1& 2& |& -1\\ 2& -1& |& -2\\ 2& -1& |& 2\end{array}\right)\to \left(\begin{array}{cccc}-2& 4& |& -2\\ 2& -1& |& -2\\ 2& -1& |& 2\end{array}\right)\to \left(\begin{array}{cccc}-2& 4& |& -2\\ 0& 3& |& -4\\ 0& 3& |& 0\end{array}\right)\to \left(\begin{array}{cccc}-2& 4& |& -2\\ 0& 0& |& -4\\ 0& 3& |& 0\end{array}\right)$

The second row of the last matrix reads

$0s+0t=-4$

which is impossible. Hence, the system of linear equations has no solution and the lines $L_1$ and $L_2$ do not intersect.