벡터와 공간 전체 문제

(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 $L$.

So $\mathbf{d}$ must be perpendicular to both $\left< 0,1,-1\right>$ and $\left< -3,1,2 \right>$ and so must be a nonzero constant times

$$\begin{aligned} \left< 0,1,-1\right> \times \left< -3,1,2 \right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 1 & -1 \\ -3 & 1 & 2 \end{matrix}\right] =\left< 3 \,,\, 3 \,,\, 3 \right> \end{aligned}$$

We choose $\mathbf{d}=\frac{1}{3}\left< 3 \,,\, 3 \,,\, 3 \right>=\left< 1\,,\,1\,,\,1 \right>$. So

$$\left< x,y,z\right> = \left< 10,11,13\right> + t \left< 1\,,\,1\,,\,1 \right>$$

is a vector parametric equation for the line. We can also write this as $x=10+t$, $y=11+t$, $z=13+t$.

Let's use $z$ as the parameter and call it $t$. Then $z=t$ and

$$\begin{aligned} x+y&=11-t \\ x-y&=13+t \end{aligned}$$

Adding the two equations gives $2x=24$ and subtracting the second equation from the first gives $2y=-2-2t$. So

$$(x,y,z) = \big(12\,,\,-1-t\,,\,t\big)$$

(a) The point $(0,y,0)$, on the $y$–axis, is equidistant from $(2, 5, -3)$ and $(-3, 6, 1)$ if and only if

$$\begin{aligned} &\big|\left< 2, 5, -3\right>-\left< 0,y,0\right>\big| =\big|\left< -3, 6, 1\right>-\left< 0,y,0\right>\big| \\ &\iff 2^2+(5-y)^2+(-3)^2=(-3)^2+(6-y)^2+1^2 \\ &\iff 2y=8 \\ &\iff y=4 \end{aligned}$$

(b) The points $(1,3,1)$ and $\mathbf{r}(0)=(0,0,2)$ are both on the plane. Hence the vector $\left< 1,3,1\right>-\left< 0,0,2\right>=\left< 1,3,-1\right>$ joining them, and the direction vector of the line, namely $\left< 1,1,1\right>$ are both parallel to the plane. So

$$\left< 1,3,-1\right>\times \left< 1,1,1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ 1&3&-1 \\ 1&1&1 \end{matrix}\right] = \left< 4,-2,-2\right>$$

is perpendicular to the plane. As the point $(0,0,2)$ is on the plane and the vector $\left< 4,-2,-2\right>$ is perpendicular to the plane, the equation of the plane is

$$4(x-0)-2(y-0)-2(z-2)=0\text{ or } 2x-y-z=-2$$

We have

$$\mathbf{v}(t) = \mathbf{r}'(t) = 5 \sqrt{2}\,\hat{\boldsymbol{\imath}} + 5e^{5t}\,\hat{\boldsymbol{\jmath}} +5 e^{-5t}\,\hat{\mathbf{k}}$$

and hence

$$|\mathbf{v}(t)| = |\mathbf{r}'(t)| = 5 \big|\sqrt{2}\,\hat{\boldsymbol{\imath}} + e^{5t}\,\hat{\boldsymbol{\jmath}} + e^{-5t}\,\hat{\mathbf{k}}\big| = 5\sqrt{2+ e^{10t}+ e^{-10t}}$$

Since $2+ e^{10t}+ e^{-10t} = \big(e^{5t}+e^{-5t}\big)^2$, that's (d).

(a) By definition,

$$\begin{aligned} \mathbf{r}(t)&= a \cos t\,\hat{\boldsymbol{\imath}}+a\sin t\,\hat{\boldsymbol{\jmath}}+ct\,\hat{\mathbf{k}}\\ \mathbf{v}(t)=\mathbf{r}'(t)&= -a \sin t\,\hat{\boldsymbol{\imath}}+a\cos t\,\hat{\boldsymbol{\jmath}}+c\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}t}(t)=|\mathbf{v}(t)|&= \sqrt{a^2+c^2}\\ \mathbf{a}(t)=\mathbf{r}''(t)&= -a \cos t\,\hat{\boldsymbol{\imath}}-a\sin t\,\hat{\boldsymbol{\jmath}} \end{aligned}$$

The $(x,y)= a(\cos t,\sin t)$ coordinates go around a circle of radius $a$ and centre $(0,0)$ counterclockwise. One circle is completed for each increase of $t$ by $2\pi$. At the same time, the $z$ coordinate increases at a constant rate. Each time the $(x,y)$ coordinates complete one circle, the $z$ coordinate increases by $2 \pi c$. The path is a helix with radius $a$ and with each turn having height $2\pi c$.

(b) By definition,

$$\begin{aligned} \mathbf{r}(t)&= a \cos t\sin t\,\hat{\boldsymbol{\imath}} +a\sin^2 t\,\hat{\boldsymbol{\jmath}}+a\cos t\,\hat{\mathbf{k}}\\ &= \frac{a}{2} \sin 2t\,\hat{\boldsymbol{\imath}}+a\frac{1-\cos 2t}{2}\,\hat{\boldsymbol{\jmath}}+a\cos t\,\hat{\mathbf{k}}\\ \mathbf{v}(t)=\mathbf{r}'(t)&= a \cos 2t\,\hat{\boldsymbol{\imath}} +a\sin 2t\,\hat{\boldsymbol{\jmath}}-a\sin t\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}t}(t)=|\mathbf{v}(t)|&= a\sqrt{1+\sin^2t}\\ \mathbf{a}(t)=\mathbf{r}''(t)&= -2a \sin 2t\,\hat{\boldsymbol{\imath}} +2a\cos 2t\,\hat{\boldsymbol{\jmath}}-a\cos t\,\hat{\mathbf{k}} \end{aligned}$$

Write

$$\begin{aligned} x(t)=\tfrac{a}{2} \sin 2t \qquad y(t)=a\tfrac{1-\cos 2t}{2} \qquad z(t)=a\cos t \end{aligned}$$

Then

$$\begin{aligned} x(t)^2+\big[y(t)-\tfrac{a}{2}\big]^2&=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} \cos^2 2t =\tfrac{a^2}{4} \end{aligned}$$

Thus the $(x,y)$ coordinates go around a circle of radius $\frac{a}{2}$ and centre $\big(0,\frac{a}{2}\big)$ counterclockwise. At the same time the $z$ coordinate oscillates over the interval between $1$ and $-1$ half as fast. In fact, one can say more about the curve.

$$\begin{aligned} x(t)^2+y(t)^2&=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4}\big(1-\cos 2t\big)^2 \\ &=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} - \tfrac{a^2}{2}\cos 2t + \tfrac{a^2}{4} \cos^2 2t \\ &= \tfrac{a^2}{2} - \tfrac{a^2}{2} (1- 2\sin^2 t) \\ &= a^2\sin^2 t \\ x(t)^2+y(t)^2+z(t)^2 &= a^2\sin^2 t + a^2\cos^2 t =a^2 \end{aligned}$$

So our curve lies on the intersection of the cylinder $x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4}$ and the sphere $x^2+y^2+z^2=a^2$. (Don't worry if you didn't think of trying to evaluate $x(t)^2+y(t)^2+z(t)^2$.)