# Challenging Coordinates Calculation

A cube has an edge length of $$24$$ units. It is tilted and rotated about one of its base vertices (vertex $$A$$), such that its four parallel edges that were vertical intersect the horizontal $$xy$$ plane passing through vertex $$A$$, at the points $$A,B,C$$ and $$D$$. If the coordinates of point $$B$$ are $$(20, -15, 0)$$, and point $$C$$ is $$12$$ units away from the base, what are the coordinates $$(x,y,z)$$ of the lowest point of the cube (vertex $$E$$)?

The only thing I could come up with, is the distance between point $$B$$ and the base. Using the distance $$AB$$ and the Pythagorean theorem, the distance between $$B$$ and the base is $$7$$ units.

Any hints, or solutions of this problem are appreciated.

• I'd consider this rotation as a product of two rotations: first about one of the horizontal edges so that BC is parallel to the base (and |EC|=7). After that the cube must be rotated around AB until one gets |EC|=12. Hope this helps. Jul 27 at 1:37
• I am not sure I understand how your suggestion works. Can you elaborate on that in an answer? Jul 29 at 10:24
• Are vectors $\vec{AB}$ and $\vec{BC}$ perpendicular? Jul 29 at 10:38
• @bb_823 No, they're not. Jul 29 at 11:15
• I'm new to the field of analytical geometry (and math in general, lol), but here is my rough idea. Let's say that $F$ is the closest vertex to $B$. (Maybe?) we could find plane on which $B,C,E$ and $F$ are on (using the fact that that plane and vector $\vec{AB}$ are at an angle $\arcsin(\frac{24}{25})$, if that is enough information). Once we find that plane then the sphere with center $A$ and radius $24$ should touch that plane at exactly one point and it will be point $F$. From there we should be able to find $E$. Are there any worng assumptions, or is there not enough info for this to work? Jul 29 at 12:28

Using the Rodrigues formula for rotation $$(\theta)$$ around an axis $$\vec k$$

$$p^{\theta} = p\cos\theta+\vec k\times p\sin\theta+(\vec k\cdot p)\vec k(1-\cos\theta)$$

the rotated cube around $$(A,\vec k)$$ can be obtained calculating

$$p^{\theta}_k = p_k\cos\theta+\vec k\times p_k\sin\theta+(\vec k\cdot p_k)\vec k(1-\cos\theta)$$

because $$A=(0,0,0)$$.

Now regarding the rotated edge containing $$C$$ we have

$$\frac 12(p^{\theta}_t+p^{\theta}_b)\cdot(0,0,1)=0$$

where $$p^{\theta}_t,p^{\theta}_b$$ are respectively the rotated edge extrema. Regarding the point $$B$$ we need to determine $$0\le \lambda\le 1$$ such that

$$\lambda p^{\theta}_t+(1-\lambda)p^{\theta}_b= B$$

where $$p^{\theta}_t,p^{\theta}_b$$ are respectively the corresponding rotated edge extrema, and finally, the normalization condition $$\|\vec k \|=1$$. This gives us four conditions and five unknowns $$(\vec k, \theta,\lambda)$$. This can be solved as a minimization procedure as follows

$$\min_{\vec k,\theta,\lambda}\|\lambda p^{\theta}_{t_1}+(1-\lambda)p^{\theta}_{b_1}- B\| \ \ \text{s. t}\ \ \ \cases{0\le \lambda\le 1\\ \|\vec k\| = 1\\ \frac 12(p^{\theta}_{t_2}+p^{\theta}_{b_2})\cdot(0,0,1)=0}$$

giving

$$\vec k = (0.130848,-0.447689,0.884564), \theta = -0.751834, \lambda = 0.708333, E = (31.7616, 3.94883, -11.2963)$$

In blue the original cube, in red, rotated and in black the point $$C$$

Follows a MATHEMATICA script to perform this minimization

Clear[k]
k = {kx, ky, kz};
p1b = {0, 0, 0};
p2b = {24, 0, 0};
p3b = {24, 24, 0};
p4b = {0, 24, 0};
p1t = {0, 0, 24};
p2t = {24, 0, 24};
p3t = {24, 24, 24};
p4t = {0, 24, 24};
p0 = {20, -15, 0};
rot[p_, k_, theta_] := p Cos[theta] + Cross[k, p] Sin[theta] + k (k.p) (1 - Cos[theta])
rp2b = rot[p2b - p1b, k, theta];
rp2t = rot[p2t - p1b, k, theta];
rp3b = rot[p3b - p1b, k, theta];
rp3t = rot[p3t - p1b, k, theta];
equs1 = lambda rp2b + (1 - lambda) rp2t - p0;
equs2 = 1/2 (rp3t + rp3b).{0, 0, 1};
equs3 = k.k - 1;
sol = NMinimize[{equs1.equs1, 0 < lambda < 1, equs2 == 0, equs3 == 0}, {kx, ky, kz, theta, lambda}]

• May I ask, how did you perform the minimization? Jul 29 at 16:14
• Attached the MATHEMATICA script which solves the minimization. Jul 29 at 16:36