벡터와 공간 전체 문제

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

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

The marble is subject to two forces. The first, gravity, is $-mg\,\hat{\mathbf{k}}$ with $m$ being the mass of the marble. The second is the normal force imposed by the plane. This forces acts in a direction perpendicular to the plane. One vector normal to the plane is $a\,\hat{\boldsymbol{\imath}}+b\,\hat{\boldsymbol{\jmath}}+c\,\hat{\mathbf{k}}$. So the force due to the plane is $T\left< a,b,c\right>$ with $T$ determined by the property that the net force perpendicular to the plane must be exactly zero, so that the marble remains on the plane, neither digging into nor flying off of it. The projection of the gravitational force onto the normal vector $\left< a,b,c\right>$ is

$$\frac{-mg\left< 0,0,1\right>\cdot\left< a,b,c\right>}{|\left< a,b,c\right>|^2}\left< a,b,c\right> =\frac{-mgc}{a^2+b^2+c^2}\left< a,b,c\right>$$

The condition that determines $T$ is thus

$$T\left< a,b,c\right>+\frac{-mgc}{a^2+b^2+c^2}\left< a,b,c\right> =0 \implies T=\frac{mgc}{a^2+b^2+c^2}$$

The total force on the marble is then (ignoring friction – which will have no effect on the direction of motion)

$$\begin{aligned} T\left< a,b,c\right>-mg\left< 0,0,1\right>&=\frac{mgc}{a^2+b^2+c^2}\left< a,b,c\right>-mg\left< 0,0,1\right>\\ &=mg\frac{c\left< a,b,c\right>-\left< 0,0,a^2+b^2+c^2\right>}{a^2+b^2+c^2}\\ &=mg\frac{\left< ac,bc,-a^2-b^2\right>}{a^2+b^2+c^2}\\ \end{aligned}$$

The direction of motion $\left< ac,bc,-a^2-b^2\right>$. If you want to turn this into a unit vector, just divide by $\sqrt{(a^2+b^2)(a^2+b^2+c^2)}$. Note that the direction vector in perpendicular $\left< a,b,c\right>$ and hence is parallel to the plane. If $c=0$, the plane is vertical. In this case, the marble doesn't roll – it falls straight down. If $a=b=0$, the plane is horizontal. In this case, the marble doesn't roll — it remains stationary.

The two planes $x+y+z=3$ and $x+y+z=9$ are parallel. The centre must be on the plane $x+y+z=6$ half way between them. So, the center is on $x+y+z=6$, $2x-y=0$ and $3x-z=0$. Solving these three equations, or equvalently,

$$y=2x,\ z=3x,\ x+y+z=6x=6$$

gives $(1,2,3)$ as the centre. $(1,1,1)$ is a point on $x+y+z=3$. $(3,3,3)$ is a point on $x+y+z=9$. So $\left< 2,2,2\right>$ is a vector with tail on $x+y+z=3$ and head on $x+y+z=9$. Furthermore $\left< 2,2,2\right>$ is perpendicular to the two planes. So the distance between the planes is $|\left< 2,2,2\right>|=2\sqrt{3}$ and the radius of the sphere is $\sqrt{3}$. The sphere is

$$(x-1)^2+(y-2)^2+(z-3)^2=3$$

Set $y=0$ and then solve $2x+3y-z=0,\ x-4y+2z=-5$, i.e. $2x-z=0$, $x+2z=-5$, or equvalently

$$z=2x,\ x+2z=5x=-5$$

The result, $(-1,0,-2)$, is one point on the plane. Set $y=5$ and then solve $2x+3y-z=0,\ x-4y+2z=-5$, i.e. $2x+15-z=0$, $x-20+2z=-5$, or equivalently

$$z=2x+15,\ x-20+4x+30=-5$$

The result, $(-3,5,9)$, is another point on the plane. So three points on the plane are $(-2,0,1), \ (-1,0,-2)$ and $(-3,5,9)$. $\left< -2+1\,,\,0-0\,,\, 1+2 \right>=\left< -1,0,3\right>$ and $\left< -2+3\,,\,0-5\,,\, 1-9 \right> = \left< 1,-5,-8\right>$ are two vectors having both head and tail in the plane.

$$\begin{aligned} \left< -1,0,3\right> \times \left< 1,-5,-8\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -1 & 0 & 3 \\ 1 & -5 & -8 \end{matrix}\right] =\left< 15 , -5 ,5 \right> \end{aligned}$$

is a vector perpendicular to the plane. $\frac{1}{5}\left< 15 , -5 ,5 \right>=\left< 3,-1,1\right>$ is also a vector perpendicular to the plane. The plane is

$$3(x+1) -(y-0) + (z+2)=0\quad\text{or}\quad 3x-y+z=-5$$

The vector $\mathbf{n}$ is perpendicular to the plane $\mathbf{n}\cdot\mathbf{x}= c$. So the line

$$\mathbf{x}(t)=\mathbf{p}+t\mathbf{n}$$

passes through $\mathbf{p}$ and is perpendicular to the plane.

It crosses the plane at the value of $t$ which obeys

$$\mathbf{n}\cdot\mathbf{x}(t)= c \text{or} \mathbf{n}\cdot[\mathbf{p}+t\mathbf{n}]= c$$

namely

$$t=[c-\mathbf{n}\cdot\mathbf{p}]/|\mathbf{n}|^2$$

The vector

$$\mathbf{x}(t)-\mathbf{p}=t\,\mathbf{n}=\mathbf{n}\,[c-\mathbf{n}\cdot\mathbf{p}]/|\mathbf{n}|^2$$

has head on the plane $\mathbf{n}\cdot\mathbf{x}= c$, tail at $\mathbf{p}$, and is perpendicular to the plane. So the distance is the length of that vector, which is

$$|c-\mathbf{n}\cdot\mathbf{p}|/|\mathbf{n}|$$

The distance from the point $(x,y,z)$ to $(1,2,3)$ is $\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}$ and the distance from $(x,y,z)$to $(5,2,7)$ is $\sqrt{(x-5)^2+(y-2)^2+(z-7)^2}$. Hence $(x,y,z)$ is equidistant from $(1,2,3)$ and $(5,2,7)$ if and only if

$$\begin{aligned} & & (x-1)^2+(y-2)^2+(z-3)^2&=(x-5)^2+(y-2)^2+(z-7)^2\\ &\iff & x^2-2x+1+z^2-6z+9&=x^2-10x+25+z^2-14z+49\\ &\iff & 8x+8z&=64\\ &\iff & x+z&=8 \end{aligned}$$

This is the plane through $(3,2,5)=\tfrac{1}{2}(1,2,3)+\tfrac{1}{2}(5,2,7)$ with normal vector $\left< 1,0,1\right>=\frac{1}{4}\big(\left< 5,2,7\right>-\left< 1,2,3\right> \big)$.

The distance from the point $\mathbf{x}$ to $\mathbf{a}$ is $\sqrt{(\mathbf{x}-\mathbf{a})\cdot(\mathbf{x}-\mathbf{a})}$ and the distance from $\mathbf{x}$ to $\mathbf{b}$ is $\sqrt{(\mathbf{x}-\mathbf{b})\cdot(\mathbf{x}-\mathbf{b})}$. Hence $\mathbf{x}$ is equidistant from $\mathbf{a}$ and $\mathbf{b}$ if and only if

$$\begin{aligned} & & (\mathbf{x}-\mathbf{a})\cdot(\mathbf{x}-\mathbf{a})&=(\mathbf{x}-\mathbf{b})\cdot(\mathbf{x}-\mathbf{b}) \\ &\iff\qquad & |\mathbf{x}|^2-2\mathbf{a}\cdot\mathbf{x}+|\mathbf{a}|^2&=|\mathbf{x}|^2-2\mathbf{b}\cdot\mathbf{x}+|\mathbf{b}|^2 \\ &\iff & 2(\mathbf{b}-\mathbf{a})\cdot\mathbf{x}&=|\mathbf{b}|^2-|\mathbf{a}|^2 \end{aligned}$$

This is the plane through $\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{b}$ with normal vector $\mathbf{b}-\mathbf{a}$.