벡터와 공간 전체 문제

For each fixed $c\ge 0$, the isobar $p(x,y)=c$ is the curve $x^2-2cx+y^2=3c^2$, or equivalently, $(x-c)^2+y^2=4c^2$. This is a circle with centre $(c,0)$ and radius $2c$. Here is a sketch of the isobars $p(x,y)=c$ with $c=0,1,2,3$.

The parallelogram determined by the vectors $\mathbf{a}$ and $\mathbf{b}$ has vertices $\mathbf{0},\ \mathbf{a},\ \mathbf{b}$ and $\mathbf{a}+\mathbf{b}$. As $t$ varies from $0$ to $1$, $t(\mathbf{a}+\mathbf{b})$ traverses the diagonal from $\mathbf{0}$ to $\mathbf{a}+\mathbf{b}$. As $s$ varies from $0$ to $1$, $\mathbf{a}+s(\mathbf{b}-\mathbf{a})$ traverses the diagonal from $\mathbf{a}$ to $\mathbf{b}$. These two straight lines meet when $s$ and $t$ are such that

$$t(\mathbf{a}+\mathbf{b})=\mathbf{a}+s(\mathbf{b}-\mathbf{a})$$

or

$$(t+s-1)\mathbf{a}=(s-t)\mathbf{b}$$

Assuming that $\mathbf{a}$ and $\mathbf{b}$ are not parallel (i.e. the parallelogram has not degenerated to a line segment), this is the case only when $t+s-1=0$ and $s-t=0$. That is, $s=t=\tfrac{1}{2}$. So the two lines meet at their midpoints.

We may choose our coordinate axes so that $A=(0,0,0)$, $B=(s,0,0)$, $C=(s,s,0)$, $D=(0,s,0)$ and $A'=(0,0,s)$, $B'=(s,0,s)$, $C'=(s,s,s)$, $D'=(0,s,s)$.

(a) Then

$$\begin{aligned} |A'C'|&=\big|\left< s,s,s\right> -\left< 0,0,s\right> \big| &&=\big|\left< s,s,0\right> \big|&=\sqrt{2}\,s\\ |A'B|&=\big|\left< s,0,0\right> -\left< 0,0,s\right> \big| &&=\big|\left< s,0,-s\right> \big|&=\sqrt{2}\,s\\ |A'D|&=\big|\left< 0,s,0\right> -\left< 0,0,s\right> \big| &&=\big|\left< 0,s,-s\right> \big|&=\sqrt{2}\,s\\ |C'B|&=\big|\left< s,0,0\right> -\left< s,s,s\right> \big| &&=\big|\left< 0,-s,-s\right> \big|&=\sqrt{2}\,s\\ |C'D|&=\big|\left< 0,s,0\right> -\left< s,s,s\right> \big| &&=\big|\left< -s,0,-s\right> \big|&=\sqrt{2}\,s\\ |BD|&=\big|\left< 0,s,0\right> -\left< s,0,0\right> \big| &&=\big|\left< -s,s,0\right> \big|&=\sqrt{2}\,s \end{aligned}$$

(b) $E=\tfrac{1}{2}(s,s,s)$ so that $EA=\left< 0,0,0\right> -\tfrac{1}{2}\left< s,s,s\right> =-\tfrac{1}{2}\left< s,s,s\right>$ and $EC=\left< s,s,0\right> -\tfrac{1}{2}\left< s,s,s\right> =\tfrac{1}{2}\left< s,s,-s\right>$. $$\cos\theta =\frac{-\left< s,s,s\right> \cdot\left< s,s,-s\right>} {|\left< s,s,s\right> |\,|\left< s,s,-s\right> |} =\frac{-s^2}{3s^2} =-\frac{1}{3} \qquad \implies\quad \theta=109.5^\circ$$

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

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

Denote by $\big(x(t),y(t)\big)$ the position of the skier at time $t$. As long as the skier remains on the surface of the hill

$$\begin{aligned} y(t)&=h\big(x(t)\big) \\ \implies y'(t)&=h'\big(x(t)\big)\,x'(t) \\ \implies y''(t)&=h''\big(x(t)\big)\, x'(t)^2+h'\big(x(t)\big)\,x''(t) \end{aligned}$$

So the velocity and acceleration vectors of the skier are

$$\begin{aligned} \mathbf{v}(t)&=\left< 1,h'\big(x(t)\big)\right> x'(t) \\ \mathbf{a}(t)&=\left< 1,h'\big(x(t)\big)\right> x''(t) +\left< 0,h''\big(x(t)\big)\right> x'(t)^2 \end{aligned}$$

The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill and has a magnitude such that the skier remains on the surface of the hill. From the velocity vector of the skier (which remain tangential to the hill as long as the skier remains of the surface of the hill),we see that one vector normal to the hill at $\big(x(t),y(t)\big)$ is $$\mathbf{n}(t)=\left<-h'\big(x(t)\big),1\right>$$ This vector is not a unit vector, but that's ok. By Newton's law of motion

$$m\mathbf{a}=-mg\,\hat{\boldsymbol{\jmath}}+p(t)\,\mathbf{n}(t)$$

for some function $p(t)$. Dot both sides of this equation with $\mathbf{n}(t)$.

$$m\mathbf{a}(t)\cdot\mathbf{n}(t)=-mg\hat{\boldsymbol{\jmath}}\cdot\mathbf{n}(t)+p(t)|\mathbf{n}(t)|^2$$

Substituting in

$$\begin{aligned} mh''\big(x(t)\big)\,x'(t)^2&=-mg+p(t)\left[1+h'\big(x(t)\big)^2\right] \\ \implies p(t)\left[1+h'\big(x(t)\big)^2\right] &=m\Big(g+h''\big(x(t)\big)\,x'(t)^2\Big) \end{aligned}$$

As long as $p(t)\ge 0$, the hill is pushing up in order to keep the skier on the surface. When $p(t)$ becomes negative, the hill has to pull on the skier in order to keep her on the surface. But the hill can't pull, so the skier becomes airborne instead. This happens when

$$g+h''\big(x(t)\big)x'(t)^2=0$$

That is when $x'(t)=\sqrt{-g/h''\big(x(t)\big)}$. At this time $x(t)=x_0$, $y(t)=y_0$ and the speed of the skier is

$$\sqrt{x'(t)^2+y'(t)^2} =\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)}$$

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.