I have the following here…
Let $\mathcal{P}_1:a_1x+b_1y+c_1z=d_1$ and $\mathcal{P}_2:a_2x+b_2y+c_2z=d_2$ be planes in $\mathbb{R}^3$. Use Linear algebra to prove that the intersection of these two planes is either empty, a line, or a plane but not a singular point. Note that since $\mathcal{P}_1$ is a plane, one of $a_1,b_1,c_1$ must be nonzero. The same is true for $\mathcal{P}_2$.
The student I am helping has only done a few weeks of linear algebra and does not know the concept of rank or dimension. All they really know is RREF, linear independence, span and finding euations of lines and planes in different forms (Vector, parametric, general etc…)
To show the intersection of two planes could be a line:
$c_1 z = d_1-a_1 x – b_1 y$ which implies $z=\frac{d_1-a_1x-b_1y}{c1}$
and $c_2 z = d_2-a_2 x – b_2 y$ which implies $z=\frac{d_2-a_2x-b_2y}{c2}$
By equating them, I get $z=\frac{d_1-a_1x-b_1y}{c1}=\frac{d_2-a_2x-b_2y}{c2}$
So if I solve for $y$, I get:
$y=\frac{(a_1c_2-a_2c_1)}{(b_2c_1-b_1c_2)}x+\frac{(c_1d_2-c_2d_1}{(b_2c_1-b_1c_2)}$
which is a line.
To show the intersection of two planes could be a Plane:
For a plane I reasoned the following. If the intersection of two planes is a plane, then $d_1 = d_2$ meaning that $a_1x+b_1y+c_1z=a_2x+b_2y+c_2z$ so that means $(a_1-a_2)x+(b_1-b_2)y+(c_1-c_2)z=0$ which is a plane.
I'm not sure if this is correct though…
What do they mean by "empty point" though?
Best Answer
Note that the number of non-zero rows in RREF($X$) equals the rank of matrix $X$, you may find it easy to look at this problem using augmented matrix $C=[A:B]$; where $A$ is the $2\times 3$ coefficient matrix for the given system and $B$ is $2\times 1$ column vector i.e.
$C=\begin{pmatrix}a_1&b_1&c_1:d_1\\a_2&b_2&c_2:d_2\end{pmatrix}$
so that the following cases may arise:
(i) RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&0:*\end{pmatrix}$ which implies rank($A$)<rank($C$) i.e. NO SOLUTION or empty set.
(ii) RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&*:*\end{pmatrix}$ which implies rank($A$)=rank($C$)<n (n0. of variables) i.e. INFINITELY MANY SOLUTIONS; out of which exactly $n-rank(A)+1=2$ solutions are linearly independent. These $2$ LI solutions i.e. $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ can form a line in $\mathbb R^3$.
(iii)RREF($C$)=$\begin{pmatrix}a_1&b_1&c_1:d_1\\0&0&0:0\end{pmatrix}$ which implies rank($A$)=rank($C$)<n (n0. of variables) i.e. INFINITELY MANY SOLUTIONS; out of which exactly $n-rank(A)+1=3$ solutions are linearly independent. These $3$ LI solutions i.e. $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_2,y_2,z_2)$ can form a plane in $\mathbb R^3$.