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This is plane and simple

When dealing with planes in a 3D space, one must not only know how to solve 3x3 linear systems (most of which depend on a given variable) but also how to interpret the results that were found. Each type of system (consistent, inconsistent or having infinitely many solutions) has a geometrical meaning. We'll see a couple of those cases in the exercise below.

Question 🤔

[Maximum marks: 16]

Consider the system below where each equation represents a plane in a 3D space:

\begin{cases} x+2y-z=0 \\ x-my-3z=0 \\ x+3y+mz=m \end{cases}

a) Show that the reduced system (in row echelon form) is

\begin{cases} x+2y-z=0 \\ \quad (2+m)y+2z=0 \\ \qquad m(m+3)z=m(2+m) \end{cases}

[4 marks]

b) For the given system:

i) Find the values of $m$ for which the system has infinitely many solutions and find the line of intersection between the planes [5 marks]

ii) Find the values of $m$ for which the system has no solutions. Interpret this result geometrically. [2 marks]

iii) Find the unique solution in terms of $m$ . [2 marks]

c) For $m=1$ find the cartesian equation of the line which crosses the intersection between the three planes perpendicularly to the plane $x+3y+mz=m$ . [3 marks]

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Solution 🤓

a) By reducing the system to row echelon form, we’ll be able to better analise the conditions that would make it have one solution, infinite solutions or no solution (all three possibilities will show up in this example). It is a long process, but the sooner we start, the sooner we finish. Let’s proceed to row operations:

\left( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 1 & -m & -3 & 0 \\ 1 & 3 & m & m \\ \end{array} \right)

R_1 - R_2 \rarr R_2\\R_1-R_3 \rarr R_3

\left( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 2+m & 2 & 0 \\ 0 & -1 & -1-m & -m \\ \end{array} \right)

(2+m) R_3 +R_2 \rarr R_3

\left( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 2+m & 2 & 0 \\ 0 & 0 & -m^2-3m & -m(2+m) \\ \end{array} \right)

(-1) R_3 \rarr R_3

\left( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 2+m & 2 & 0 \\ 0 & 0 & m(m+3) & m(2+m) \\ \end{array} \right)

Therefore the initial system is equivalent to:

\begin{cases} x+2y-z=0 \\ (2+m)y+2z=0 \\ m(m+3)z=m(2+m) \end{cases}

b) In this item we’ll simulate all cases using Geogebra (the link to the simulation can be found at the end of this post)

b) i) We must zero the last line, having for that case three variables for only two equations. That is done by setting $m = 0$ . The equivalent system is

\begin{cases} x+2y-z=0 \\ 2y+2z=0 \\ 0 = 0 \end{cases}

By setting $z = t$ and substituting it into the second row, we get $y = -t$ . By substituting both on the first row, we get $x = 3t$ . The solution is a line of intersection between the three planes (none of which is parallel). The parametric equations of the line found are:

L_1: x = 3t \quad y = -t \quad z = t \quad, \quad t \in \mathbb{R}

b) ii) In this case, to find a system that has no solution, we’ll set $m = -3$ . The equivalent system is the following (in the right).

We have two parallel planes and an intersecting one, where the intersection of the first plane and any of the two forms a line parallel to the third. The last line implies that $0 = 3$ , which is absurd.

Therefore we just proved mathematically that the system has no solution and therefore, in a geometric interpretation, the intersection of the three planes is neither a point (which would imply in having a solution) or a line (infinitely many solutions).

\begin{cases} x+2y-z=0 \\ -y+2z=0 \\ 0z=3 \end{cases}

c) We can find a unique solution when $m$ is not equal to $-3$ or to $0$ . Using row 3, we can find

z = \frac{m(m+2)}{m(m+3)} = \frac{m+2}{m+3}

Substituting z on the second equation and isolating $y$ gives us

y = -2 \cdot \frac{m+2}{m+3} \cdot \frac{1}{m+2} = - \frac{2}{m+3}

Lastly, using the first row and isolating $x$ gives us

x = \frac{m+2}{m+3} - 2\cdot \left( - \frac{2}{m+3} \right) = \frac{m+6}{m+3}

Therefore, the unique solution (as a function of $m$ , is a point given by the coordinates

\left( \frac{m+6}{m+3} , - \frac{2}{m+3}, \frac{m+2}{m+3} \right)

d) For $m = 1$ the unique solution will be

\left(\frac{m+6}{m+3}, - \frac{2}{m+3}, \frac{m+2}{m+3} \right) = \left( \frac{7}{4} , - \frac{1}{2}, \frac{3}{4} \right)

(notice that we do not need to solve the system from scratch)

That particular solution is illustrated in the image on the right (point A being the solution). Given the line is perpendicular to the plane $x+3y+z=1$ (after substituting m) we already have its direction vector, which is the same as the normal vector to the plane given. Therefore, the cartesian equation of the line is

\frac{x-\frac{7}{4}}{1}=\frac{y+\frac{1}{2}}{3}=\frac{z-\frac{3}{4}}{1}

Find in the link below the animation built in Geogebra.

This is plane and simple

Question 🤔

Solution 🤓

www.geogebra.org