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

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