Position of three planes in term of real parameters

Question

Given three planes: $$\alpha: 2x+3y-z=6$$$$\beta:ax-3y+2x=5$$$$\gamma:4x-3y+3z=b$$$$a,b\in\mathbb R$$

Discuss the position of planes in terms of parameters $a,b$.

Using Kronecker-Capelli's theorem:

$$ \begin{bmatrix} 2 & 3 & -1 & | & 6 \\ a & -3 & 2 & |& 5 \\ 4 & -3 & 3 &|& b \\ \end{bmatrix}$$

Reducing the matrix to row echelon form:

1. Choosing element $a_{00}$ as a pivot, multiplying the first row with $-a/2$ and adding it to the second, and than multiplying the first row with $-2$ and adding it to the third gives:

$$ \begin{bmatrix} 2 & 3 & -1 & | & 6 \\ 0 & \frac{-3(2+a)}{2} & \frac{4+a}{2} & |& 5-3a \\ 0 & -9 & 5 &|& b-12 \\ \end{bmatrix}$$

2. Switching the second and the third row (otherwise it would give a restriction $a\neq -2$), choosing element $a_{11}=-9$ as a pivot, multiplying the second row with $(-2-a)/6$ and adding it to the third gives:

$$ \begin{bmatrix} 2 & 3 & -1 & | & 6 \\ 0 & -9 & 5 & |& 5-3a \\ 0 & 0 & \frac{1-a}{3} &|& \frac{-2b+54-ab-6a}{6} \\ \end{bmatrix}$$

If $a=1\Rightarrow rank A=2$. In that case, if $b=16\Rightarrow rank [A|x]=2$. $$rank A=rank[A|x]=r=2$$ $$n=3$$ Since $r

This means that planes intersection is a line.

If $a=1$ and $b\neq 16$, then $rank A=2,rank[A|x]=3$.

Since $rank A\neq rank [A|x]$, there are no intersections between planes.

How to check case $a\neq 1$?

$$$$

Answer 1

Important: You forgot to switch $5-3a$ with $\frac {-2b+54-ab-6a}b$ when you switch rows in passage $2$.

By the way, if $a\neq -1$ then $r(A) = 3$ so $r(A) = 3 \le r(A|x) \le 3 = n \implies r(A|x) = 3$, which means you have a unique solution, i.e. the $3$ plans intersect in a single point.

Answer 2

If the plane has a non-zero normal vector $n$, one can read the position from the normal form: $$ \beta = n_x x + n_y y + n_z z = n \cdot (x,y,z)^T \iff \\ \frac{1}{\lVert n \rVert} n \cdot (x,y,z)^T = \frac{1}{\lVert n \rVert} \beta =: d $$ If we insert a radial vector of length $r$ $$ r e_r $$ which starts at the origin and ends in the plane, we get: $$ d = \frac{1}{\lVert n \rVert} n \cdot (x,y,z)^T = \frac{1}{\lVert n \rVert} n \cdot r e_r = r \frac{1}{\lVert n \rVert} n \cdot e_r = r$$ so that $d$ is the distance of the plane to the origin (probably signed, depending on the direction of the normal vector $n$).

1. $$ \alpha: 6=2x+3y-z=(2,3,-1)^T \cdot (x,y,z)^T $$ Normal vector $(2,3-1)^T$, distance $6/\sqrt{14}$ from the origin, independent of $a$ and $b$.

2. $$ \beta: 5= ax-3y+2x=(a+2,-3,0)^T \cdot (x,y,z)^T $$

Normal vector $(a+2,-3,0)^T$, distance $5/\sqrt{(a+2)^2+9}\in (0,5/3]$ from the origin, normal vector lying in the $x$,$y$-plane. Normal vector and distance depending on $a$.

3. $$ \gamma: b= 4x-3y+3z=(4,-3,3)^T \cdot (x,y,z)^T $$ Normal vector $(4,-3,3)^T$, signed distance $b/\sqrt{34}\in \mathbb{R}$ from the origin. Only distance depending on $b$.

Here is the scene, the first plane in cyan, the second in magenta, and the third plane in green colour. The red sphere has radius $6/\sqrt{14}$.

You can fiddle with the scene here.