Construct the cross section of a cube by a plane passing through three given points

Question

I want to find/construct the cross section of a cube that includes the three points shown below. In other words, if a plane went through the cube such that it would slice where the points are, what would the whole cross section look like in general?

Since the back two points lie on the same plane, I thought about connecting the back two points with a line. However, what is throwing me off is the fact that the line is tilted so when looking at the right side of the top face, if I extended that line, I am pretty sure it would not intersect the first line I constructed.

_{[Original picture was lost; the picture above is a reduced copy of an image from the answer by John Hughes]}

Answer 1

It looks to me as if in this case, the intersection will be a hexagon. The plane will, of course, intersect the cube in OTHER points than just these three. But you can get a pretty good sense of things by drawing the triangle that contains the three points; the plane is the unique plane containing that triangle.

Let's label the top four edges A, B, C, D, with A being the one with the dot, and the others being read clockwise, so that B starts at the right-hand end of A. Then the next four edges --- the vertical ones --- call them P, Q, R, and S, again reading clockwise from the one with the dot. And then the bottom four edges: call those W, X, Y, Z, with W being immediately below A, so that the dot is on edge C.

Then to fill in the picture, I'd place a dot midway along edge B (or perhaps a little nearer the back), another a quarter of the way up edge 3, another 2/5 of the way up edge 4. Having done this, I'd connect any pair of dots that are on the same face with a straight line, resulting in a hexagonal intersection.

Here's some more detail (an initial attempt, followed by a complete solution)

(1) Draw the diagonal line on the back face.

(2) Draw a parallel diagonal on the front face, to determine where the plane intersects that front face. Mark a vertex along the vertical edge.

(3) Now we get to the part where I can't give a precise recipe: draw an intersection dot (I've used purple) on the right-top edge, and another on the left bottom. The edges joining these to the two nearby intersections (I've drawn orange) have to be parallel.

(4) And now when you add the last two edges (I've used aqua), those too must be parallel:

On the other hand, they also have to close the loop, which mine do not. If you move one purple point back a bit, the other has to move forward a bit as well (to keep the orange lines parallel) and the aqua line at the bottom won't close the hexagon at the point you're hoping for. So you're forced to adjust that purple point back and forth until you hit a sweet spot. Here's an example with shifted purple points that just about works:

I've drawn another example (where the initial point on the front is further left) which better demonstrates what goes wrong when your purple point location guess is bad:

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I believe that the perfect location for the purple point can probably be determined by something like Pascal's theorem, but I don't see, offhand, how to make that work. I hope that these musings are of some use to you, however.

Here's a complete solution:

1. Draw the line through the two points on the back face; extend this until it meets the extension of the right-hand vertical side of the back face, and put a green point there.
2. Make a parallel copy of this line, passing through the front-face point; mark the intersection with the right vertical edge with a red dot; then extend the line until it meets the extension of the left vertical edge of the front face. Put a blue point there.
3. The green point and the rightmost red point both lie on the slicing plane AND the plane of the right face; therefore the edge between them does as well; draw that in orange. Do a corresponding thing on the right face.
4. The intersections between these orange segments and the front-to-back segments, marked with purple dots, are the last two verts of the hexagon.
5. Draw the hexagon.

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Thanks for asking this question -- it made me think hard for a bit, and learn something new!

Answer 2

Here's my construction. You may refer to this image to follow along.

From top to bottom, call the input points $A$, $B$, $C$.

$D$, the fourth point, is easy: draw a line through $C$ parallel to $AB$, and the intersection of that line with the front right edge is $D$. ...it occurs to me that this doesn't always happen correctly, but in this case it does.

Now it gets harder; our goal is to figure out how far along the top right edge our plane intersects. Fortunately, we have tools that can answer this question.

Draw a line parallel to the front-back lines, through $A$, and find its intersection with the top front line, call it $G$. $G$ is as far from the top-right-front corner as $A$ is from the top-right-back corner.

Draw a circle around the top-front-right corner through $G$; find the upper intersection of this circle and the extended right-front line, $H$. $H$ now is as far up from $D$ as $A$ is, measuring around the cube and only up and left (not backward-forward).

A horizontal line through $H$ and meeting the extended right-rear line makes $I$; this line will match distance around the cube with distance forward and backward along the cube.

Draw $DI$, and find $J$ as the intersection of $DI$ and the extended top-front. This point has the right ratio of Finally, a vertical line through $J$ intersects top-right at $E$, which is (finally) the point on the hexagon.

Then, finally, $F$ can be gotten thus: A line through $B$, parallel to $DE$, intersects the bottom-left edge at $F$.