Cut a cube with a plane defined by a vector, having 2 heights

Question

I need to cut a cube by an inclined plane.

1. The cube is placed in the center of the space.
2. The cutter geometry must have $2$ heights: $H_1$ and $H_2$ and $W$ (the side length).

I would like to define the plane by a vector as $(\mathbf V_x, \mathbf V_y, \mathbf V_z)$, $d$ where $d$ is the distance from the cube center and the plane.

$\mathbf V_y$ is supposed to be $0$ and $\displaystyle d=\sqrt{\mathbf V_x^2+\mathbf V_z^2}$.

Can you help me? to define $\mathbf V_x$ and $\mathbf V_z$ if I know $H_1$ and $H_2$ and $W$?

Thanks!

Answer 1

Hint: As you've assumed the centre of cube as origin, the coordinates of the point where the green line intersects the plane be $A(\frac{W}{2},-\frac{W}{2}+h_2,\frac{W}{2})$ and $B(-\frac{W}{2},-\frac{W}{2}+h_1,\frac{W}{2})$. So, the coordinates of the point below $A$ (inside the paper) will be $C(\frac{W}{2},-\frac{W}{2}+h_2,-\frac{W}{2})$.

Now, from the three position vectors, find the normal vector of the plane. To find the constant term in the equation of plane, use the fact that $d$ is distance between plane and origin.

Answer 2

You mention you already know $h_1, h_2$ and $w$. If $(0,0,0)$ is the center of the cube with $w$ as the side of the cube, you already know $4$ points on the plane: $(\frac{w}{2}, \pm \frac{w}{2},h_2-\frac{w}{2})$ and $(-\frac{w}{2}, \pm \frac{w}{2},h_1-\frac{w}{2})$. That should help you to come up with the equation of the plane.

Say, equation of the plane is $ax+by+cz = D$.

By solving for $a, b, c$, given the above points on the plane, you get

$a = 2(h_1-h_2), b = 0, c = 2w, D = w(h_1+h_2-w)$.