벡터와 공간 전체 문제

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

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

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 $(12,0,-6)$.