벡터와 공간 전체 문제

$\left< d_x,d_y\right>$ 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 $7x_0+24=9y_0$.

(a) The vector parametric equation is $\left< x,y\right>=\left< 1,2\right>+t\left< 3,2\right>$. The scalar parametric equations are $x=1+3t,\ y=2+2t$. The symmetric equation is $\frac{x-1}{3}=\frac{y-2}{2}$.

(b) The vector parametric equation is $\left< x,y\right>=\left< 5,4\right>+t\left< 2,-1\right>$. The scalar parametric equations are $x=5+2t,\ y=4-t$. The symmetric equation is $\frac{x-5}{2}=\frac{y-4}{-1}$.

(c) The vector parametric equation is $\left< x,y\right>=\left< -1,3\right>+t\left< -1,2\right>$. The scalar parametric equations are $x=-1-t,\ y=3+2t$. The symmetric equation is $\frac{x+1}{-1}=\frac{y-3}{2}$.

(a) The midpoint of the side opposite $\mathbf{a}$ is $\tfrac{1}{2}(\mathbf{b}+\mathbf{c})$. The vector joining $\mathbf{a}$ to that midpoint is $\tfrac{1}{2}\mathbf{b}+\tfrac{1}{2}\mathbf{c}-\mathbf{a}$. The vector parametric equation of the line through $\mathbf{a}$ and $\tfrac{1}{2}(\mathbf{b}+\mathbf{c})$ is

$$\mathbf{x}(t)=\mathbf{a}+t\big(\tfrac{1}{2}\mathbf{b}+\tfrac{1}{2}\mathbf{c}-\mathbf{a}\big)$$

Similarly, for the other two medians (but using $s$ and $u$ as parameters, rather than $t$)

$$\begin{aligned} \mathbf{x}(s)&=\mathbf{b}+s\big(\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{c}-\mathbf{b}\big)\\ \mathbf{x}(u)&=\mathbf{c}+u\big(\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{b}-\mathbf{c}\big) \end{aligned}$$

(b) The three medians meet at a common point if there are values of $s,t$ and $u$ such that

$$\begin{aligned} \mathbf{a}+t\big(\tfrac{1}{2}\mathbf{b}+\tfrac{1}{2}\mathbf{c}-\mathbf{a}\big) &\ =\ \mathbf{b}+s\big(\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{c}-\mathbf{b}\big) &&\ =\ \mathbf{c}+u\big(\tfrac{1}{2}\mathbf{a}+\tfrac{1}{2}\mathbf{b}-\mathbf{c}\big)\\ (1-t)\mathbf{a}+\frac{t}{2}\mathbf{b}+\frac{t}{2}\mathbf{c} &\ =\ \frac{s}{2}\mathbf{a}+(1-s)\mathbf{b}+\frac{s}{2}\mathbf{c} &&\ =\ \frac{u}{2}\mathbf{a}+\frac{u}{2}\mathbf{b}+(1-u)\mathbf{c} \end{aligned}$$

Assuming that the triangle has not degenerated to a line segment, this is the case if and only if the coefficients of $\mathbf{a},\ \mathbf{b}$ and $\mathbf{c}$ match

$$\begin{aligned} 1-t&=\frac{s}{2}&&=\frac{u}{2}\\ \frac{t}{2}&=1-s&&=\frac{u}{2}\\ \frac{t}{2}&=\frac{s}{2}&&=1-u \end{aligned}$$

or

$$s=t=u,\ 1-t=\frac{t}{2} \implies s=t=u=\frac{2}{3}$$

The medians meet at $\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$.

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) Each point on the $y$-axis is of the form $(0,y,0)$. Such a point is on the plane $P$ if

$$3(0)+\frac{1}{2}y+0=4 \iff y=8$$

So the intersection of $P$ with the $y$-axis is the single point $(0,8,0)$. (b) Each point on the $z$-axis is of the form $(0,0,z)$. Such a point is on the plane $P$ if

$$3(0)+\frac{1}{2}(0)+z=4 \iff z=4$$

So the intersection of $P$ with the $z$-axis is the single point $(0,0,4)$. (c) The intersection of the plane $P$ with the $yz$-plane is a line. We have shown in parts (a) and (b) that the points $(0,8,0)$ and $(0,0,4)$ are on that line. Here is a sketch of the part of that line that is in the first octant.

(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.

(a) The plane must be parallel to $\left< 2,4,6\right> -\left< 1,0,1\right> =\left< 1,4,5\right>$ and to $\left< 1,2,-1\right> -\left< 1,0,1\right> =\left< 0,2,-2\right>$. So its normal vector must be perpendicular to both $\left< 1,4,5\right>$ and $\left< 0,2,-2\right>$ and hence parallel to

$$\left< 1,4,5\right> \times\left< 0,2,-2\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & 4 & 5 \\ 0 & 2 & -2 \end{matrix}\right] =\left< -18,2,2\right>$$