# Cube standing on a corner

## Cube standing on a corner

This question has arisen from this post and the picture and the insights have been taken from the answers mentioned there.

$$\quad\qquad\qquad\qquad\qquad\qquad\qquad$$

In the above picture, a cube is standing on one of its corners. Let the cone lie on a $$xz \text{ plane}$$ and assume $$y$$-axis (the red line) passes through the centre of the cone and through one of the body diagonals of the cube. Now for the cube, imagine a plane passing though the points $$QS$$ and parallel to the $$xz \text{ plane}$$.

What I thought:

I thought, after looking at a dice for some time, that the $$3$$ vertices $$QST$$ would indeed form a triangle but the $$3^{rd}$$ vertex will be at a different height (not in the same plane), like the triangle $$QST'$$ below. So when a $$xz \text{ plane}$$ will cut the cube, what we'll get will be some polygon which with my level of understanding and imagination wasn't possible to fathom then.

$$\qquad$$
Image on the Left: Front View; $$\qquad\qquad\qquad$$Image on the Right: Top View

What it turned out to be:

It turned out that all the $$3$$ vertices, $$QST$$ indeed lie on single $$xz \text{ plane}$$, like $$QST$$ above, and thus form a triangle. Similarly it's true for the $$3$$ vertices above these (the vertices $$U,V,W$$). And add to that all the $$4$$ $$xz \text{ planes}$$ upon which all these vertices lie are at intervals of equal height!

I want to know that:

1. How can we know whether a triangle forms and not some other polygon for the vertices $$Q,S,T$$ on the $$xz \text{ plane}$$ parallel to cone's base and passing through $$1$$ or more of the $$3$$ vertices (which may be an unfamiliar figure, maybe a pentagon or some quadrilateral, as below)?
2. Whether all the 3 vertices of the cube will lie on a plane parallel to $$xz \text{ plane}$$ and not on 2 planes parallel to $$xz \text{ plane}$$
3. How to determine the height of these $$xz \text{ planes}$$ from one another?
4. How will one calculate the centre and the radius for the obtained polygon in point 1?
[Note: Point 4 has a bit to do with the linked post (in case you don't get what I'm saying)]
• 1. For the 3 vertices $Q, S, T$? Maybe I am misunderstanding, but three vertices (that are not collinear) would form a $3$-gon, not some other polygons. Aug 10, 2022 at 21:51
• @peterwhy thanks a lot! made the correction in my question. Its xz plane instead of xy. Aug 11, 2022 at 4:25
• @peterwhy 3 vertices will form a triangle as in front view (and top view) but when the xz plane cuts the cube, we'll have something else on the xy plane if all 3 vertices are not on the same height. Aug 11, 2022 at 4:27
• Kindly don't "close vote" for "needs more focus" as these seemingly multiple questions are actually a single question itself, regarding cube standing on a corner, but so that the answerer may know the confusion I'm having, I have broken it down. Example: one may solve the question with addressing 1. or 2. or 3. or 4.. Aug 11, 2022 at 12:01
• @hardmath great observation! Thank you! Made the edit. Luckily saw a way to correct the question and so that peterwhy don't need to change anything in their answer. Aug 12, 2022 at 18:25

To find lengths, one may work with another cube rotated to a more convenient orientation.

Consider a cube with side length $$a$$, with two neighbouring vertices at $$O(0,0,0)$$ and $$E(a, 0, 0)$$, and a body diagonal from $$O$$ to $$D(a,a,a)$$.

## Question 3

In your diagram, you are looking for the height of point $$Q$$ (or $$S$$ or $$T$$) along the $$y$$-direction, relative to point $$P$$ in your diagram.

In my rotated cube, this would be to find the projected length of $$OE$$ along the $$OD$$ direction. One may perform this projection by dot product:

• The dot product of $$\overrightarrow{OD}$$ and $$\overrightarrow{OE}$$ is

$$\overrightarrow{OD}\cdot \overrightarrow{OE} = (a,a,a)\cdot (a,0,0) = a^2$$

• Dividing this by the length of the body diagonal $$OD$$, and calling the result $$s$$:

\begin{align*} \overrightarrow{OD}\cdot \overrightarrow{OE} &= \left\|\overrightarrow{OD}\right\| \left\|\overrightarrow{OE}\right\| \cos\theta = \left\|\overrightarrow{OD}\right\|s\\ s &= \frac{\overrightarrow{OD}\cdot \overrightarrow{OE}} {\left\|\overrightarrow{OD}\right\|} = \frac{a^2}{\sqrt{3a^2}} = \frac{a}{\sqrt 3} \end{align*}

where $$\theta$$ is the angle between $$\overrightarrow{OD}$$ and $$\overrightarrow{OE}$$, as in the linked Wikipedia page.

$$s$$ is the projected length of $$OE$$ along the $$OD$$ direction, which in your orientation is the height of $$Q$$ along the $$y$$-direction relative to $$P$$. Repeat the same process for other vertices of my cube, e.g. $$(0,a,0)$$, $$(a,a,0)$$, etc.

## Question 4

From Q3 above we obtained $$s$$, the projected length of $$\overrightarrow{OE}$$ along the $$\overrightarrow{OD}$$ direction. If $$r$$ is the perpendicular distance from $$E$$ to line $$OD$$, then

\begin{align*} OE^2 &= s^2+r^2\\ r^2 &= OE^2 - s^2\\ &= a^2- \left(\frac{a}{\sqrt3}\right)^2\\ r&= a\sqrt{\frac23} \end{align*}

In your orientation, $$r = a\sqrt{\frac23}$$ would be the radial component of $$Q$$ away from the $$y$$-axis (or from $$P$$).

(If we find the exact "projected" vector with signed length $$s$$ and direction $$\overrightarrow{OD}$$, this will be the vector projection of $$\overrightarrow{OE}$$ onto $$\overrightarrow{OD}$$. Then subtract this vector from $$\overrightarrow{OE}$$, this will return a vector with length $$r$$ and perpendicular from $$OD$$. This is also the first steps of the Gram-Schmidt process.)

## Question 1

From the same calculation as in Q3 above, points $$Q, S,T$$ all have the same $$y$$-coordinates, i.e. the same "height" from $$P$$. So they lie on the same plane parallel to the $$xz$$-plane.

The face diagonals $$QS$$, $$ST$$ and $$TQ$$ are all on this plane, and all have length $$a\sqrt 2$$. So $$\triangle QST$$ is an equilateral triangle on this plane.

## Question 2

The cube has 3-fold rotational symmetry centred at a body diagonal.

For cross sections of a cube on a plane perpendicular to the body diagonal, the cross section shape maintains the same symmetry and may be either

• an equilateral triangle, or
• a convex hexagon.
• (or just a single point at the two ends $$P$$ and $$V$$)

One interactive demo of cube cross sections from Google search is https://www.geogebra.org/m/X6GYjh2U (not by me).

• Wow! Amazing! +1. Too clear and understandable explanation. Thanks a lot! Also, the links were a great help. Aug 11, 2022 at 15:57