직선과 평면의 방정식

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

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

The symmetric equations for the line are

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

(a) The line $L$ must be perpendicular both to $\left< 2\,,\,1\,,\,-1\right>$, which is a normal vector for the plane $2x + y - z = 5$, and to $\left< -1\,,\,-2\,,\,3\right>$, which is a direction vector for the line $x = 3 - t$, $y = 1 - 2t$ and $z = 3t$. Any such vector must be a nonzero constant times

$$\begin{aligned} \left< 2\,,\,1\,,\,-1\right> \times \left< -1\,,\,-2\,,\,3\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ -1 & -2 & 3 \end{matrix}\right] =\left< 1 \,,\, -5 \,,\, -3 \right> \end{aligned}$$

(b) For the point $Q(a, b, c)$

• to be a distance $2$ from the $xy$–plane, it is necessary that $|c|=2$, and
• to be a distance $3$ from the $xz$–plane, it is necessary that $|b|=3$, and
• to be a distance $4$ from the $yz$–plane, it is necessary that $|a|=4$.

As $a < 0$, $b > 0$, $c > 0$, the point $Q$ is $(-4, 3, 2)$ and the line $L$ is

$$x = -4 + t \qquad y = 3 - 5t \qquad z = 2 - 3t$$

(a) The line $L$ intersects the $xy$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $z=0$. When $z=0$ the equations of $L$ reduce to $x + y = 6$, $x - y = 0$. So the intersection point is $(3,3,0)$.

The line $L$ intersects the $xz$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $y=0$. When $y=0$ the equations of $L$ reduce to $x + z = 6$, $x +2z = 0$. Substituting $x=-2z$ into $x+z=6$ gives $-z=6$. So the intersection point is $(12,0,-6)$.

The line $L$ intersects the $yz$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $x=0$. When $x=0$ the equations of $L$ reduce to $y + z = 6$, $-y +2z = 0$. Substituting $y=2z$ into $y+z=6$ gives $3z=6$. So the intersection point is $(0,4,2)$.

(b) Our main job is to find a direction vector $\mathbf{d}$ for the line.

• Since the line is to be parallel to $y=z$, $\mathbf{d}$ must be perpendicular to the normal vector for $y=z$, which is $\left< 0,1,-1\right>$.
• $\mathbf{d}$ must also be perpendicular to $L$. For a point $(x,y,z)$ to be on $L$ it must obey $x+y = 6-z$ and $x-y =-2z$. Adding these two equations gives $2x=6-3z$ and subtracting the second equation from the first gives $2y= 6+z$. So for a point $(x,y,z)$ to be on $L$ it must obey $x=3-\frac{3z}{2}$, $y=3+\frac{z}{2}$. The point on $L$ with $z=0$ is $(3,3,0)$ and the point on $L$ with $z=2$ is $(0,4,2)$. So $\left< 0-3,4-3,2-0 \right>=\left< -3,1,2 \right>$ is a direction vector for $L$.