벡터의 연산

Suppose that the cube has height, length and width $s$. We may choose our coordinate axes so that the vertices of the cube are at $(0,0,0)$, $(s,0,0)$, $(0,s,0)$, $(0,0,s)$, $(s,s,0)$, $(0,s,s)$, $(s,0,s)$ and $(s,s,s)$.

We'll start with a couple of examples. The diagonal from $(0,0,0)$ to $(s,s,s)$ is $\left< s,s,s\right>$. One face of the cube has vertices $(0,0,0)$, $(s,0,0)$, $(0,s,0)$ and $(s,s,0)$. One diagonal of this face runs from $(0,0,0)$ to $(s,s,0)$ and hence is $\left< s,s,0\right>$. The angle between $\left< s,s,s\right>$ and $\left< s,s,0\right>$ is

$$\begin{aligned} \cos^{-1}\left(\frac{\left< s,s,s\right>\cdot\left< s,s,0\right>} {|\left< s,s,s\right>|\,|\left< s,s,0\right>|}\right) =\cos^{-1}\left(\frac{2s^2}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)\approx 35.26^\circ \end{aligned}$$

A second diagonal for the face with vertices $(0,0,0)$, $(s,0,0)$, $(0,s,0)$ and $(s,s,0)$ is that running from $(s,0,0)$ to $(0,s,0)$. This diagonal is $\left< -s,s,0\right>$. The angle between $\left< s,s,s\right>$ and $\left< -s,s,0\right>$ is

$$\begin{aligned} \cos^{-1}\left(\frac{\left< s,s,s\right>\cdot\left< -s,s,0\right>} {|\left< s,s,s\right>|\,|\left< -s,s,0\right>|}\right) =\cos^{-1}\left(\frac{0}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}(0)=90^\circ \end{aligned}$$

Now we'll consider the general case. Note that every component of every vertex of the cube is either $0$ or $s$. In general, two vertices of the cube are at opposite ends of a diagonal of the cube if all three components of the two vertices are different. For example, if one end of the diagonal is $(s,0,s)$, the other end is $(0,s,0)$. The diagonals of the cube are all of the form $\left< \pm s,\pm s,\pm s\right>$. All of these diagonals are of length $\sqrt{3}s$. Two vertices are on the same face of the cube if one of their components agree. They are on opposite ends of a diagonal for the face if their other two components differ. For example $(0,s,s)$ and $(s,0,s)$ are both on the face with $z=s$. Because the $x$ components $0,\ s$ are different and the $y$ components $s,\ 0$ are different, $(0,s,s)$ and $(s,0,s)$ are the ends of a diagonal of the face with $z=s$. The diagonals of the faces with $z=0$ or $z=s$ are $\left< \pm s,\pm s,0\right>$. The diagonals of the faces with $y=0$ or $y=s$ are $\left< \pm s,0, \pm s\right>$. The diagonals of the faces with $x=0$ or $x=s$ are $\left< 0,\pm s,\pm s\right>$. All of these diagonals have length $\sqrt{2}s$. The dot product of one the cube diagonals $\left< \pm s,\pm s,\pm s\right>$ with one of the face diagonals $\left< \pm s,\pm s,0\right>$, $\left< \pm s,0, \pm s\right>$, $\left< 0,\pm s,\pm s\right>$ is of the form $\pm s^2\pm s^2+0$ and hence must be either $2s^2$ or $0$ or $-2s^2$. In general, the angle between a cube diagonal and a face diagonal is

$$\begin{aligned} \cos^{-1}\left(\frac{\text{$2s^2$ or 0 or $-2s^2$}}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}\left(\frac{\text{$2$ or $0$ or $-2$}}{\sqrt{6}}\right)\approx \text{$35.26^\circ$ or $90^\circ$ or $144.74^\circ$}. \end{aligned}$$