직선과 평면의 방정식

Since $t$ can be any real number, these equation describe the same line. They're both valid. For example, the point given by the first parametric equation with $t=7$, namely $\mathbf c + 7\mathbf d$, is exactly the same as the point given by the second parametric equation with $t=-7$, namely $\mathbf c -(-7)\mathbf d$.

$\left< d_x,d_y\right>$ is the direction of the line, so it can be any non-zero scalar multiple of $\left< 9,7\right>$.

$\left< x_0,y_0\right>$ can be any point on the line. Describing these is the same as describing the line itself. We're trying to find all doubles $\left< x_0,y_0\right>$ that obey

$$\begin{aligned} \begin{cases} x_0-3&=9t\\ y_0-5&=7t \end{cases} \end{aligned}$$

for some real number $t$. That is,

$$\begin{aligned} t=\frac{x_0-3}{9}&=\frac{y_0-5}{7}\\ 7(x_0-3)&=9(y_0-5)\\ 7x_0+24&=9y_0 \end{aligned}$$

Any of these steps could specify the possible values of $\left< x_0,y_0\right>$. Say, they can be any pair satisfying $7x_0+24=9y_0$.

We are looking for a vector that is perpendicular to $z=0$ and hence is parallel to $\hat{\mathbf{k}}$. To be parallel of $\hat{\mathbf{k}}$, the vector has to be of the form $c\,\hat{\mathbf{k}}$ for some real number $c$. For the vector to be nonzero, we need $c\ne 0$ and for the vector to be different from $\hat{\mathbf{k}}$, we need $c\ne 1$. So three possible choices are $-\hat{\mathbf{k}}$, $2\,\hat{\mathbf{k}}$, $7.12345\,\hat{\mathbf{k}}$.

(a) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,0)$, namely $\left< x,y,z\right>$, lie in the plane. So the vector $\left< x,y,z\right>$ must be perpendicular to $\left< 1,2,3\right>$ and

$$0=\left< x,y,z\right>\cdot \left< 1,2,3\right> =x+2y+3z$$

(b) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,1)$, namely $\left< x,y,z-1\right>$, lie in the plane. So the vector $\left< x,y,z-1\right>$ must be perpendicular to $\left< 1,1,3\right>$ and

$$0=\left< x,y,z-1\right>\cdot \left< 1,1,3\right> =x+y+3(z-1) \iff x+y+3z=3$$

(c) If both $(1,2,3)$ and $(1,0,0)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(1,0,0)$, namely $\left< 0,2,3\right>$, lie in the plane. So the vector $\left< 0,2,3\right>$ must be perpendicular to $\left< 4,5,6\right>$. As

$$\left< 0,2,3\right> \cdot \left< 4,5,6\right> = 28\ne 0$$

the vector $\left< 0,2,3\right>$ is not perpendicular to $\left< 4,5,6\right>$. So there is no plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\left< 4,5,6\right>$.

(d) If both $(1,2,3)$ and $(0,3,4)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(0,3,4)$, namely $\left< 1,-1,-1\right>$, lie in the plane. So the vector $\left< 1,-1,-1\right>$ must be perpendicular to $\left< 2,1,1\right>$. As

$$\left< 1,-1,-1\right> \cdot \left< 2,1,1\right> = 0$$

the vector $\left< 1,-1,-1\right>$ is indeed perpendicular to $\left< 2,1,1\right>$. So there is a plane that passes through both $(1,2,3)$ and $(0,3,4)$ and has normal vector $\left< 2,1,1\right>$. We now just have to build its equation.

If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(1,2,3)$, namely $\left< x-1,y-2,z-3\right>$, lie in the plane. So the vector $\left< x-1,y-2,z-3\right>$ must be perpendicular to $\left< 2,1,1\right>$ and

$$0=\left< x-1,y-2,z-3\right>\cdot \left< 2,1,1\right> =2(x-1)+(y-2)+(z-3) \iff 2x+y+z=7$$

As a check, note that both $(x,y,z)=(1,2,3)$ and $(x,y,z)=(0,3,4)$ obey the equation $2x+y+z=7$.

Solution 1: That's too easy. We just guess. The plane $x+y+z=1$ contains all three given points.

Solution 2: The plane does not pass through the origin. (You can see this by just making a quick sketch.) So the plane has an equation of the form $ax+by+cz=1$.

• For $(1,0,0)$ to be on the plane we need that

$$a(1) +b(0) +c(0) =1 \implies a=1$$

• For $(0,1,0)$ to be on the plane we need that

$$a(0) +b(1) +c(0) =1 \implies b=1$$

• For $(0,0,1)$ to be on the plane we need that

$$a(0) +b(0) +c(1) =1 \implies c=1$$

So the plane is $x+y+z=1$.

Solution 3: Both the head and the tail of the vector from $(1,0,0)$ to $(0,1,0)$, namely $\left< -1,1,0\right>$, lie in the plane. Similarly, both the head and the tail of the vector from $(1,0,0)$ to $(0,0,1)$, namely $\left< -1,0,1\right>$, lie in the plane. So the vector

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

is a normal vector for the plane. As $(1,0,0)$ is a point in the plane,

$$\begin{aligned} \left< 1 , 1 ,1 \right> \cdot \left< x-1\,,\,y-0\,,\, z-0 \right> = 0\qquad \text{or}\quad x+y+z=1 \end{aligned}$$

is an equation for the plane.

The vector from $(1,2,3)$ to $(2,3,4)$, namely $\left< 1,1,1\right>$ is parallel to the vector from $(1,2,3)$ to $(3,4,5)$, namely $\left< 2,2,2\right>$. So the three given points are collinear. Precisely, all three points $(1,2,3),\ (2,3,4)$ and $(3,4,5)$ are on the line $\left< x-1,y-2,z-3\right>=t\left< 1,1,1\right>$. There are many planes through that line.

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 $z=4$.