Systems of linear equations

Intersections of lines in \(\mathbf R^3\)

Example. Let's try to find the intersections of the lines \[ \ell_1:\mathbf r_1(t)=\left(\begin{smallmatrix}2\\-1\\2\end{smallmatrix}\right)+t\left(\begin{smallmatrix}-1\\3\\1\end{smallmatrix}\right)\text{ and } \ell_2:\mathbf r_2(t)=\left(\begin{smallmatrix}1\\0\\1\end{smallmatrix}\right)+t\left(\begin{smallmatrix}1\\-1\\-1\end{smallmatrix}\right). \] Recall that we can do this by changing the parameter symbol to \(s\) in one parametrization and equate the components: \[\begin{align*} 2-s&=1+t\\ -1+3s&=-t\\ 2s&=1-t. \end{align*}\]

This is a System of Linear Equations (SLE). Solving it, we get \(s=0\) and \(t=1\). This means that the intersection of \(\ell_1\) and \(\ell_2\) is the one point \(P\) with \[ \vec{OP}=\mathbf r_1(s=0)=\mathbf r_2(t=1)=\left(\begin{smallmatrix}2\\-1\\0\end{smallmatrix}\right). \]

Example. Let's now try to find the intersections of the lines \[ \ell_1:\mathbf r_1(t)=\left(\begin{smallmatrix}0\\2\\3\end{smallmatrix}\right)+t\left(\begin{smallmatrix}1\\-1\\3\end{smallmatrix}\right)\text{ and } \ell_2:\mathbf r_2(t)=\left(\begin{smallmatrix}1\\0\\1\end{smallmatrix}\right)+t\left(\begin{smallmatrix}1\\1\\1\end{smallmatrix}\right). \] In this case, there are no solutions. Also, the direction vectors are not parallel: there is no scalar \(c\) such that \(c\left(\begin{smallmatrix}1\\-1\\3\end{smallmatrix}\right)=\left(\begin{smallmatrix}1\\1\\1\end{smallmatrix}\right)\). Therefore, in this case, \(\ell_1\) and \(\ell_2\) are skew lines. This is a situation new to \(m\)-space with \(m>2\): in the plane, every pair of nonparallel lines intersect. Actually, this is the generic situation: if you randomly select two lines, they will almost always be skew.

Example. Let's now try to find the intersections of the lines \[ \ell_1:\mathbf r_1(t)=\left(\begin{smallmatrix}1\\2\\0\end{smallmatrix}\right)+t\left(\begin{smallmatrix}-1\\2\\1\end{smallmatrix}\right)\text{ and } \ell_2:\mathbf r_2(t)=\left(\begin{smallmatrix}0\\1\\-2\end{smallmatrix}\right)+t\left(\begin{smallmatrix}2\\-4\\-2\end{smallmatrix}\right). \] There are no solutions. Also, the two lines are parallel: we have \(-2\left(\begin{smallmatrix}-1\\2\\1\end{smallmatrix}\right)=\left(\begin{smallmatrix}2\\-4\\-2\end{smallmatrix}\right)\).

Example. Let's now try to find the intersections of the lines \[ \ell_1:\mathbf r_1(t)=\left(\begin{smallmatrix}-3\\0\\2\end{smallmatrix}\right)+t\left(\begin{smallmatrix}-1\\2\\1\end{smallmatrix}\right)\text{ and } \ell_2:\mathbf r_2(t)=\left(\begin{smallmatrix}0\\1\\-2\end{smallmatrix}\right)+t\left(\begin{smallmatrix}2\\-4\\-2\end{smallmatrix}\right). \] Here, we can see that the two lines are actually the same: the two direction vectors are parallel, and also the vector between the two starting points is parallel to the direction vectors. Therefore, the solution set is a 1-parameter collection, which in this case can be either of the parametrizations we're already given: \[ P(t)=(-1+2t,-4-4t,-2t). \]

Systems of linear equations

An equation is a linear equation, if every function in it is a polynomial of rank 1. A linear equation is in standard form, if all the nonconstant summands are on the left, and the constant is on the right: \[ a_1x_1+\dotsb+a_mx_m=c. \] A system of equations is a SLE, if all the equations in it are linear. For example, the system \[\begin{align*} x_1+x_3=&4\\ x_1+2x_4=&2\\ x_1+x_2x_3=&8 \end{align*}\]

is not a SLE, because the third equation has the nonlinear summand \(x_2x_3\).

A SLE is in standard form, if the equations are in standard form, and the summands with the same variables are put in the same column. For example, our first example of intersecting lines in standard form is \[\begin{align*} -s-t=&-1\\ 3s+t=&1\\ 2s+t=&1. \end{align*}\]

To solve a SLE, we want to perform operations on it which don't change the set of solutions. We can use the following Elementary Operations: 1. Swap two rows 2. Multiply an equation by a nonzero scalar 3. Add to an equation a scalar multiple of another equation

Intersections of planes in \(\mathbf R^3\)

Example. Let's find the intersections of the planes with equations \[ \Pi_1:2x+z=1\text{ and }\Pi_2:x-y+2z=-2. \] Putting the system to standard form and performing row operations, we get \[\begin{align*} x+\hphantom{y+}\frac12z=&1\\ \hphantom{x+}y+\frac32z=&-5. \end{align*}\] We can see that the solution is a 1-parameter family. The intersection of the planes is the line for which we can get parametric equations \[\begin{align*} x=&1-\frac12t\\ y=&-5-\frac32t\\ z=&t. \end{align*}\]

We have replaced \(z\) by \(t\) on the right sides for the sake of clarity.

Example. Let's find the intersections of the planes with equations \[ \Pi_1:-3x+y-2z=2\text{ and }\Pi_2:6x-2y+4z=3. \] Here, there are no solutions. Note that the normal vectors of the two planes are parallel.

Example. Let's find the intersections of the planes with equations \[ \Pi_1:3x+6y-3z=9\text{ and }\Pi_2:-2x-4y+2z=-6. \] Here, by using elementary operations, one equation can be changed to \(0=0\). This means that we have a 2-parameter family of solutions: \[\begin{align*} x=&s\\ y=&t\\ z=&s+2t-3\\ \end{align*}\]

The two planes are actually the same.