Finding a cross section of a cube [closed]

$\begingroup$

I am given the following cube where points $A,B,C$ lie on the bottom, left and back face respectively.

I know how to find cross sections when the points lie on the edges. But, in this case, I dont know how to start. Is there a useful trick to handle this type of problems? Thanks.

$\endgroup$ 3

2 Answers

$\begingroup$

Let $\pi$ be the plane of the triangle $ABC$. Drop perpendiculars from $B$ and $C$ to the bottom plane $\beta$, resulting in the auxiliary points $B'$, $C'\in\beta$. The lines $B\vee C\subset \pi$ and $B'\vee C'\subset\beta$ intersect in the point $F\in(\pi\wedge\beta)$. It follows that the line $F\vee A$ is in fact $=\pi\wedge\beta$. In the situation of the following figure we now have enough material to construct the cross section we are after by connecting known points and intersecting with the proper edges. Depending on the given disposition the cross section could, e.g., be a hexagon. In such a case a second auxiliary step will be necessary.

$\endgroup$ 2 $\begingroup$

I'm assuming you want the cross-section given by the plane through $A$, $B$, $C$.

If so, first note that the plane normal is in the direction $(B-A)\times(C-A)$, so its equation can be written in the form $$ (X - A) \cdot \big[ (B-A)\times(C-A)\big] = 0 $$ Intersect this plane with each face of the cube.

$\endgroup$ 1