[Math] Finding the intersection points of a line with a cube

analytic geometrygeometry

The following is an old high school exercise:

Let $A = (5, 4, 6)$ and $B = (1,0,4)$ be two adjacent vertices of a cube in $\mathbb{R}^3$. The vertex $C$ lies in the $xy$-plane.

a) Compute the coordinates of the other vertices of the cube such that all $x$– and $z$-coordinates are positive.

b) Let $$g: \vec{r} = \begin{pmatrix} 10\\1\\5 \end{pmatrix} + \lambda \begin{pmatrix} 1\\1\\-1 \end{pmatrix}$$
be a line. Compute the coordinates of the intersection points of $g$ and the cube.

Question: What is the most efficient way using high school level maths to compute the intersection points by hand?

The most obvious approach would be to intersect the line with all six planes containing the cube's faces. This requires me to solve six $3 \times 3$ systems of linear equations which will probably take a while. Even worse, I will then have to check if the intersection points really belong to the surface of the cube. These are another six $3 \times 2$ systems of linear equations. Of course, the first six systems of linear equations are much easier to compute using the coordinate form of the planes, but still it will take a while.

A maybe more elegant approach would be to translate and rotate the cube such that one vertex coincides with the origin and the edges lie on the (positive) coordinate axes. Apply the same translation and rotation to $g$ and the intersection points are much easier to compute. It is also much easier to check whether or not the resulting points lie on the surface of the cube. However, in general, the rotation matrix will be very messy and therefore not advisable to do by hand.

Is there a different approach that avoids a lot of (messy) computation? I feel like there has to be some geometric property I am missing which allows for a completely different approach. After all, this is meant to be solved by high schoolers by hand.

Best Answer

Let $C(p,q,0)$. Since $AB^2=BC^2$ and $2AB^2=AC^2$, solving the system $$6^2=(1-p)^2+(0-q)^2+(4-0)^2\quad\text{and}\quad 2\times 6^2=(5-p)^2+(4-q)^2+(6-0)^2$$gives $$(p,q)=(-1,4),(5,-2)$$

Let $D(r,s,t)$. Since the midpoint of the side $AC$ is the same as the midpoint of the side $BD$, we get $$\frac{5+p}{2}=\frac{1+r}{2},\qquad \frac{4+q}{2}=\frac{0+s}{2},\qquad \frac{6+0}{2}=\frac{4+t}{2}$$ i.e. $$r=p+4,\qquad s=q+4,\qquad t=2$$

Therefore, we have either $C(-1,4,0),D(3,8,2)$ or $C(5,-2,0),D(9,2,2)$.

Case 1 : $A(5,4,6),B(1,0,4),C(-1,4,0),D(3,8,2)$ which are on the plane $2x-y-2z+6=0$.

Solving $(2t)^2+(-t)^2+(-2t)^2=6^2$ gives $t=\pm 2$.

For $t=2$, the other vertics are $A'(9,2,2),B'(5,-2,0),C'(3,2,-4),D'(7,6,-2)$.
Note here that if a point $(a,b,c)$ is either in or on the cube whose vertices are $P_i(X_i,Y_i,Z_i)\ (i=1,2,\cdots, 8)$ , then $$\min(X_1,X_2,\cdots, X_8)\le a\le \max(X_1,X_2,\cdots, X_8),$$ $$\min(Y_1,Y_2,\cdots, Y_8)\le b\le \max(Y_1,Y_2,\cdots, Y_8),$$ $$\min(Z_1,Z_2,\cdots, Z_8)\le c\le \max(Z_1,Z_2,\cdots, Z_8)$$ Solving the system of inequalities $$-1\le 10+\lambda\le 9\quad \text{and}\quad -2\le 1+\lambda\le 8\quad\text{and}\quad -4\le 5-\lambda\le 6$$gives$$\lambda=-1$$Then, however, $(9,0,6)$ is not on the cube since $A'(9,2,2)$ is the only vertex whose $x$-coordinate is $9$ which is the maximum of $x$.

For $t=-2$, the other vertices are $A'(1,6,10),B'(-3,2,8),C'(-5,6,4),D'(-1,10,6)$.
Then, however, there are no $\lambda$ satisfying the system of inequalities $$-5\le 10+\lambda\le 5\quad \text{and}\quad 0\le 1+\lambda\le 10\quad\text{and}\quad 0\le 5-\lambda\le 10$$
Case 2 : $A(5,4,6),B(1,0,4),C(5,-2,0),D(9,2,2)$ which are on the plane $x-2y+2z-9=0$.

Solving $t^2+(-2t)^2+(2t)^2=6^2$ gives $t=\pm 2$.

For $t=-2$, we get the same vertices as those in the case for $t=2$ in Case 1.
For $t=2$, the other vertices are $A'(7,0,10),B'(3,-4,8),C'(7,-6,4),D'(11,-2,6)$. Now, solving the system of inequalities $$1\le 10+\lambda\le 11\quad \text{and}\quad -6\le 1+\lambda\le 4\quad\text{and}\quad 0\le 5-\lambda\le 10$$gives$$-5\le\lambda\le 1\tag1$$The four vertices $A',B',C',D'$ are on the plane $x-2y+2z-27=0$. Solving $(10+\lambda)-2(1+\lambda)+2(5-\lambda)-27=0$ gives $\lambda=-3$ which satisfies $(1)$ to have $P(7,-2,8)$ which is on the cube since $\vec{A'P}=\frac 13\vec{A'B'}+\frac 13\vec{A'D'}$.

The four vertices $A,D,D',A'$ are on the plane $2x+2y+z-24=0$. Solving $2(10+\lambda)+2(1+\lambda)+(5-\lambda)-24=0$ gives $\lambda=-1$ which satisfies $(1)$ to have $Q(9,0,6)$ which is on the cube since $\vec{AQ}=\frac 23\vec{AA'}+\frac 23\vec{AD}$.

Since three vectors $\vec{AA'}=(2,-4,4),\vec{AB}=(-4,-4,-2)$ and $\vec{AD}=(4,-2,-4)$ are not parallel to $(1,1,-1)$, we see that the number of the intersection points of $g$ and the cube $ABCD$-$A'B'C'D'$ is at most $2$.

Therefore, the intersection points are $$\color{red}{(7,-2,8)\qquad\text{and}\qquad (9,0,6)}$$

Added : I've just noticed that somehow I had skipped the question a). The condition in a) can make the solution above simpler avoiding separating it into cases.

The key point in the solution above is the inequality $(1)$. We can say that for $\lambda$ which does not satisfy $(1)$, the point $(10+\lambda,1+\lambda,5-\lambda)$ is not on the cube. Also, the number of the intersection points is at most $2$. Using these information, we could avoid a lot of computations even if you started to check the planes other than $A'B'C'D'$ and $ADD'A'$ on which the intersection points exist as follows :

$ABCD$ whose equation is $x-2y+2z-9=0$. Then, we get $\lambda=3$ which does not satisfy $(1)$.
$BCC'B'$ whose equation is $2x+2y+z-6=0$. Then, we get $\lambda=-7$ which does not satisfy $(1)$.
$CC'D'D$ whose equation is $2x-y-2z-12=0$. Then, we get $\lambda=1$ to have $(11,2,4)$ which is not on the cube since $D'(11,-2,6)$ is the only vertex whose $x$-coordinate is $11$ which is the maximum of $x$.
$ABB'A'$ whose equation is $2x-y-2z+6=0$. Then, we get $\lambda=-5$ to have $(5,-4,10)$ which is not on the cube since $A'(7,0,10)$ is the only vertex whose $z$-coordinate is $10$ which is the maximum of $z$.

Also, after knowing all the coordinates of the eight vertices, you don't need to solve a system of linear equations in order to find the equations of planes. For example, since $\vec{AA'}=(2,-4,4)$, we see that the equation of the plane $A'B'C'D'$is of the form $x-2y+2z+k=0$. Substituting the coordinates of $A'$ gives $k=-27$.

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Here is the code for the first problem:

  x1 = ?  ;  y1 = ? ;    // First diagonal point
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  xc = (x1 + x2)/2  ;  yc = (y1 + y2)/2  ;    // Center point
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If you think about it a bit, you’ll see that the intersection points of the rotation axis with the cube are independent of its rotation—points along this axis are fixed. So, you only need to compute the intersection of the axis with the starting position of the cube, which we assume is centered on the origin and axis-aligned. The outward unit face normals are then $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$. By symmetry, you only need consider three faces that meet at a vertex, so you might as well pick the three standard unit basis vectors.

With a unit outward normal $\mathbf n$, the point-normal form of the equation of a face plane is $\mathbf n\cdot\mathbf x = s/2$, where $s$ is the cube’s side length. You have a unit direction vector $\mathbf k=(\cos\alpha\sin\phi,\sin\alpha\sin\phi,\cos\phi)$ for the rotation axis, so the distance to the intersection with the plane is given by the value of $t$ for which $\mathbf n\cdot(t\mathbf k)=s/2$, or $$t = {s\over2\mathbf n\cdot\mathbf k}.\tag{*}$$ However, the dot product of $\mathbf k$ with a standard basis vector simply picks out the corresponding coordinate of $\mathbf k$, therefore the distance to the nearest face is found by dividing half the edge length by the coordinate of $\mathbf k$ with the greatest absolute value.

If you can’t assume that the cube is ever in the convenient “home” position above, you can still use (*) but you’ll have to compute three face normals from the data at hand. You don’t need to resort to cross products to do this, though: Pick any vertex. The direction vectors of the three edges that meet at that vertex give you the required face normals. These vectors are just the differences of the coordinates of the chosen vertex and the coordinates of its three neighbors. Since the vectors are all the same length, you can avoid normalizing until you’ve selected the one that minimizes the magnitude of $t$ (multiplying unit normals by the same nonzero scalar doesn’t affect the relative magnitudes of their dot products with $\mathbf k$). Also, since you’re only interested in the magnitudes of these dot products, you can be fairly cavalier about whether you use inward or outward normals. In other words, the order of the points when you subtract coordinates doesn’t make any difference.

Best Answer

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