I am trying to solve the following problem:
"Find out if the plane $hx+3y-2hz+4=0$ contains a line for every value of $h\in\mathbb{R}$ and if so, write parametric and cartesian equations of said line."
I have solved the equation $hx+3y-2hz+4=0$ and I have found that:
if $h\neq 0$: $\pmatrix{x\\ y\\ z}=y\pmatrix{-\frac{3}{h}\\ 1\\0}+z\pmatrix{2\\ 0\\ 1}+\pmatrix{-\frac{4}{h}\\0\\0};\ y,z\in\mathbb{R}$
if $h=0$: $\pmatrix{x\\ y\\ z}=x\pmatrix{1\\ 0\\0}+z\pmatrix{0\\ 0\\ 1}+\pmatrix{0\\-\frac{4}{3}\\0};\ x,z\in\mathbb{R}$
but it doesn't look obvious to me that there is a line common to both planes.
I have also tried to find such a line $\begin{cases}a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\end{cases}$ by studying the system $\begin{cases}a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\\hx+3y-2hz+4=0 \end{cases}$ and imposing the condition that it has rank $2$ but it become very messy and while the method seems reasonable the results don't look promising.
Is there a better way to solve this problem? Thanks.
From the equation of the plane $hx + 3 y - 2 h z + 4 = 0$, we can see that for any value of $h$ the point $(0, -4/3, 0)$ lies on the plane. Now if we vary $h$, say to $h'$ then the intersection of these two planes will be along the cross product of the normals of the two planes:
$(h, 3, -2 h) \times (h', 3, -2h') = (6 (h - h'), 0, 3 (h - h') ) $
Since $h' \ne h $ , this vector (the direction vector of the intersection line) is along $(2, 0, 1)$
Therefore the line common to all such planes is given parametrically by
$ (0, -4/3, 0) + t (2, 0, 1) $