직선과 평면의 방정식

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 $\mathbf{b}-\mathbf{a}$.

(a) One side of the triangle is $\overrightarrow{AB}=\left< 1,0,1\right> - \left< 0,1,1\right> = \left< 1,-1,0\right>$. A second side of the triangle is $\overrightarrow{AC}=\left< 1,3,0\right> - \left< 0,1,1\right> = \left< 1,2,-1\right>$. If the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $\theta$ and if we take $\overrightarrow{AB}$ as the base of the triangle, then the triangle has base length $b=|\overrightarrow{AB}|$ and height $h=|\overrightarrow{AC}|\sin\theta$ and hence

$$\text{area}=\tfrac{1}{2} bh =\tfrac{1}{2} |\overrightarrow{AB}|\,|\overrightarrow{AC}|\sin\theta =\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| =\tfrac{1}{2}|\left< 1,-1,0\right> \times \left< 1,2,-1\right>|$$

As

$$\left< 1,-1,0\right>\times\left< 1,2,-1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 2 & -1 \end{matrix}\right] =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3\hat{\mathbf{k}}$$

we have

$$\text{area}=\tfrac{1}{2}|\left< 1,1,3\right>|=\tfrac{1}{2}\sqrt{11}\approx 1.658$$

(b) A unit vector perpendicular to the plane containing the triangle is

$$\hat{\mathbf{n}}=\frac{\overrightarrow{AB}\times\overrightarrow{AC}} {|\overrightarrow{AB}\times\overrightarrow{AC}|} =\frac{\left< 1,1,3\right>}{\sqrt{11}}$$

The distance from $P$ to the plane containing the triangle is the length of the projection of $\overrightarrow{AP}=\left< 5,-10,2\right>-\left< 0,1,1\right> =\left< 5,-11,1\right>$ on $\hat{\mathbf{n}}$. If $\theta$ the angle between $\overrightarrow{AP}$ and $\hat{\mathbf{n}}$, then this is

$$\text{distance}= |\overrightarrow{AP}|\,|\cos\theta| =\big|\overrightarrow{AP}\cdot\hat{\mathbf{n}}\big| =\left|\left< 5,-11,1\right>\cdot\frac{\left< 1,1,3\right>}{\sqrt{11}}\right| =\frac{3}{\sqrt{11}}\approx 0.9045$$

Switch to a new coordinate system with $$X=x-1\qquad Y=y-2\qquad Z=z+1$$ In this new coordinate system, the sphere has equation $X^2+Y^2+Z^2=2$. So the sphere is centred at $(X,Y,Z)=(0,0,0)$ and has radius $\sqrt{2}$. In the new coordinate system, the initial point $(x,y,z)=(2,2,0)$ has $(X,Y,Z)=(1,0,1)$ and our final point $(x,y,z)=(2,1,-1)$ has $(X,Y,Z)=(1,-1,0)$. Call the initial point $P$ and the final point $Q$. The shortest path will follow the great circle from $P$ to $Q$.

A great circle on a sphere is the intersection of the sphere with a plane that contains the centre of the sphere. Our strategy for finding the initial direction will be based on two observations.

• The shortest path lies on the plane $\Pi$ that contains the origin and the points $P$ and $Q$. Since the shortest path lies on $\Pi$, our direction vector must also lie on $\Pi$ and hence must be perpendicular to the normal vector to $\Pi$.
• The shortest path also remains on the sphere, so our initial direction must also be perpendicular to the normal vector to the sphere at our initial point $P$.

As our initial direction is perpendicular to the two normal vectors, it is parallel to their cross product.

So our main job is to find normal vectors to the plane $\Pi$ and to the sphere at $P$.

• One way to find a normal vector to $\Pi$ is to guess an equation for $\Pi$. As $(0,0,0)$ is on $\Pi$, $(0,0,0)$ must obey $\Pi$'s equation. So $\Pi$'s equation must be of the form $aX+bY+cZ=0$. That $(X,Y,Z)=(1,0,1)$ is on $\Pi$ forces $a+c=0$. That $(X,Y,Z)=(1,-1,0)$ is on $\Pi$ forces $a-b=0$. So we may take $a=1$, $b=1$ and $c=-1$. That is, $\Pi$ is $X+Y-Z=0$. (Check that all three points $(0,0,0)$, $(1,0,1)$ and $(1,-1,0)$ do indeed obey $X+Y-Z=0$.) A normal vector to $\Pi$ is $\left< 1,1,-1\right>$.

• A second way to find a normal vector to $\Pi$ is to observe that both

• the vector from $(0,0,0)$ to $(1,0,1)$, that is $\left< 1,0,1\right>$, lies completely inside $\Pi$ and

• the vector from $(0,0,0)$ to $(1,-1,0)$, that is $\left< 1,-1,0\right>$, lies completely inside $\Pi$.

So the vector

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

is perpendicular to $\Pi$.

• The vector from the centre of the sphere to the point $P$ on the sphere is perpendicular to the sphere at $P$. So a normal vector to the sphere at our initial point $(X,Y,Z)=(1,0,1)$ is $\left< 1,0,1\right>$.

Since our initial direction (Note that the change of coordinates $X=x-1$, $Y=y-2$, $Z=z+1$ has absolutely no effect on any velocity or direction vector. If our position at time $t$ is $(x(t),y(t),z(t))$ in the original coordinate system, then it is $(X(t),Y(t),Z(t)) =(x(t)-1,y(t)-2,z(t)+1)$ in the new coordinate system. The velocity vectors in the two coordinate systems $\left< x'(t),y'(t),z'(t)\right> =\left< X'(t),Y'(t),Z'(t)\right>$ are identical.) must be perpendicular to both $\left< 1,1,-1\right>$ and $\left< 1,0,1\right>$, it must be one of $\pm\left< 1,1,-1\right>\times\left< 1,0,1\right>$. To get from $(1,0,1)$ to $(1,-1,0)$ by the shortest path, our $Z$ coordinate should decrease from $1$ to $0$. So the $Z$ coordinate of our initial direction should be negative. This is the case for

$$\left< 1,1,-1\right>\times\left< 1,0,1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & 1 & -1 \\ 1 & 0 & 1 \end{matrix}\right] =\hat{\boldsymbol{\imath}} - 2\,\hat{\boldsymbol{\jmath}} -\hat{\mathbf{k}}$$