직선과 평면의 방정식

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

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

Note $12\left< \frac13,-\frac12,\frac16 \right>=\left< 4,-6,2 \right>$. So, we actually only know one normal direction to the line we're supposed to be describing. That means there are actually infinitely many lines satisfying the given conditions.

There are infinitely many correct answers. One is

$$\begin{aligned} L_1:\ \left< x\,,\,y\,,\,z-1\right> &= t\left< 1,0,0\right> & L_2:\ \left< x\,,\,y\,,\,z-2\right> &= t\left< 0,1,0\right> \\ L_3:\ \left< x\,,\,y\,,\,z-3\right> &= t\left< 1,1,0\right> & L_4:\ \left< x\,,\,y\,,\,z-4\right> &= t\left< 1,-1,0\right> \end{aligned}$$

No two of the lines are parallel because

• $L_1$ and $L_2$ are not parallel because $\left< 1,0,0\right>\times \left< 0,1,0\right>=\left< 0,0,1\right>\ne\mathbf{0}$.
• $L_1$ and $L_3$ are not parallel because $\left< 1,0,0\right>\times \left< 1,1,0\right>=\left< 0,0,1\right>\ne\mathbf{0}$.
• $L_1$ and $L_4$ are not parallel because $\left< 1,0,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_2$ and $L_3$ are not parallel because $\left< 0,1,0\right>\times \left< 1,1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_2$ and $L_4$ are not parallel because $\left< 0,1,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_3$ and $L_4$ are not parallel because $\left< 1,1,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-2\right>\ne\mathbf{0}$.

No two of the lines intersect because

• every point on $L_1$ has $z=1$ and
• every point on $L_2$ has $z=2$ and
• every point on $L_3$ has $z=3$ and
• every point on $L_4$ has $z=4$.

(a) Note that the value of the parameter $t$ in the equation $\left< x,y,z\right> = \left< -3,2,4\right>+t\left< -4,2,1\right>$ need not have the same value as the parameter $t$ in the equation $\left< x,y,z\right> = \left< 2,1,2\right>+t\left< 1,1,-1\right>$. So it is much safer to change the name of the parameter in the first equation from $t$ to $s$. In order for a point $(x,y,z)$ to lie on both lines we need

$$\left< -3,2,4\right>+s\left< -4,2,1\right> = \left< 2,1,2\right>+t\left< 1,1,-1\right>$$

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

$$\begin{aligned} -4s -t &= 5\\ 2s -t &= -1\\ s +t &= -2 \end{aligned}$$

Adding the last two equations together gives $3s=-3$ or $s=-1$. Substituting this into the last equation gives $t=-1$. Note that $s=t=-1$ does indeed satisfy all three equations so that $\left< x,y,z\right>=\left< -3,2,4\right>-\left< -4,2,1\right>=\left< 1,0,3\right>$ lies on both lines. Any plane that contains the two lines must be parallel to both direction vectors $\left< -4,2,1\right>$ and $\left< 1,1,-1\right>$. So its normal vector must be perpendicular to them, i.e. must be parallel to $\left< -4,2,1\right>\times\left< 1,1,-1\right>=\left< -3,-3,-6\right>=-3\left< 1,1,2\right>$. The plane must contain $(1,0,3)$ and be perpendicular to $\left< 1,1,2\right>$. Its equation is $\left< 1,1,2\right>\cdot\left< x-1,y,z-3\right>=0$ or $x+y+2z=7$. This can be checked by verifying that $\left< -3,2,4\right>+s\left< -4,2,1\right>$ and $\left< 2,1,2\right>+t\left< 1,1,-1\right>$ obey $x+y+2z=7$ for all $s$ and $t$ respectively.

(b) In order for a point $(x,y,z)$ to lie on both lines we need

$$\left< -3,2,4\right>+s\left< -4,2,1\right> = \left< 2,1,-1\right>+t\left< 1,1,-1\right>$$

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

$$\begin{aligned} -4s -t &= 5\\ 2s -t &= -1\\ s +t &= -5 \end{aligned}$$

Adding the last two equations together gives $3s=-6$ or $s=-2$. Substituting this into the last equation gives $t=-3$. However, substituting $s=-2,\ t=-3$ into the first equation gives $11=5$, which is impossible. The two lines do not intersect. In order for two lines to lie in a common plane and not intersect, they must be parallel. So, in this case no plane contains the two lines.

(c) In order for a point $(x,y,z)$ to lie on both lines we need

$$\left< -3,2,4\right>+s\left< -2,-2,2\right> = \left< 2,1,-1\right>+t\left< 1,1,-1\right>$$

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

$$\begin{aligned} -2s -t &= 5\\ -2s -t &= -1\\ 2s +t &= -5 \end{aligned}$$

The first two equations are obviously contradictory. The two lines do not intersect. Any plane containing the two lines must be parallel to $\left< 1,1,-1\right>$ (and hence automatically parallel to $\left< -2,-2,2\right>=-2\left< 1,1,-1\right>$) and must also be parallel to the vector from the point $(-3,2,4)$, which lies on the first line, to the point $(2,1,-1)$, which lies on the second. The vector is $\left< 5,-1,-5\right>$. Hence the normal to the plane is $\left< 5,-1,-5\right>\times\left< 1,1,-1\right>=\left< 6,0,6\right>=6\left< 1,0,1\right>$. The plane perpendicular to $\left< 1,0,1\right>$ containing $(2,1,-1)$ is $\left< 1,0,1\right>\cdot\left< x-2,y-1,z+1\right>=0$ or $x+z=1$.

(d) Again the two lines are parallel, since $\left< -2,-2,2\right>=-2\left< 1,1,-1\right>$. Furthermore the point $\left< 3,2,-2\right>=\left< 3,2,-2\right>+0\left< -2,-2,2\right> =\left< 2,1,-1\right>+1\left< 1,1,-1\right>$ lies on both lines. So the two lines not only intersect but are identical. Any plane that contains the point $(3,2,-2)$ and is parallel to $\left< 1,1,-1\right>$ contains both lines. In general, the plane $ax+by+cz=d$ contains $(3,2,-2)$ if and only if $d=3a+2b-2c$ and is parallel to $\left< 1,1,-1\right>$ if and only if $\left< a,b,c\right>\cdot\left< 1,1,-1\right>=a+b-c=0$. So, for arbitrary $a$ and $b$ (not both zero) $ax+by+(a+b)z=a$ works.

First observe that

• $\left< 1,1,0\right>$ is perpendicular to $x+y=0$ and hence to the line, and
• $\left< 1,-1,2\right>$ is perpendicular to $x-y+2z=0$ and hence to the line.

Consequently

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

is perpendicular to both $\left< 1,1,0\right>$ and $\left< 1,-1,2\right>$. So $\frac{1}{2}\left< 2 \,,\, -2 \,,\, -2 \right>=\left< 1,-1,-1\right>\>$ is also perpendicular to both $\left< 1,1,0\right>$ and $\left< 1,-1,2\right>$ and hence is parallel to the line. As the point $(2,-1,-1)$ is on the line, the vector equation of the line is

$$\left< x-2,y+1,z+1\right>= t\left< 1,-1,-1\right>$$

The scalar parametric equations for the line are

$$\begin{aligned} x-2=t,\ y+1=-t,\ z+1=-t \text{or} x=2+t,\ y=-1-t,\ z=-1-t \end{aligned}$$