Course: 5.10 线条的矢量方等量

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Air traffic control is tracking two planes in the vicinity of their airport. At a given moment, one plane is at a location 45 km east and 120 km north of the airport at an altitude of 7.5 km. The second plane is located 63 km east and 96 km south of the airport at an altitude of 6.0 km. The first plane is flying directly toward the airport while the second plane is continuing at a constant altitude with a heading defined by the vector $\begin{aligned} \vec{h_{2}} = ⟨ 3, 4, 0 ⟩ \end{aligned}$ to land eventually at another airport to the northwest of our air traffic controllers. Do the paths of these two aircraft cross?
::在某一时刻,一架飞机位于机场以东45公里和以北120公里处,高度为7.5公里;第二架飞机位于机场以东63公里和以南96公里处,高度为6.0公里;第一架飞机正直接飞往机场,而第二架飞机正以恒定高度飞行,航向由矢量h2340确定,最后降落在我国空中交通管制员西北的另一机场。

Vector Equation of a Line
::线条矢量方对数

You are probably very familiar with using y = mx + b , the slope-intercept form, as the equation of a line. While this equation works well in two-dimensional space, it is insufficient to completely define the equation of a line in higher order spaces.
::您可能非常熟悉使用 y = mx + b, 斜度界面, 作为线的方程式。虽然该方程式在二维空间中运作良好, 但不足以完全定义更高顺序空格中的线的方程式。

Lines in Space
::空间线

In a two-dimensional plane, a line can be represented by the equation y = mx + b , where m is the slope of the line and b is the y-intercept. In three-dimensional space, this equation can represent a line or a plane or the projection of a 3D line onto the x-y plane. For example, in the diagram below, the line on the left passes through points P and Q. The equation in 2D space for this line is $\begin{aligned} y = \frac{- P}{Q} x + P \end{aligned}$
::在二维平面中,一线可以用公式y = mx + b 表示,其中 m 是线的斜度, b 是 y 界面。在三维空格中,此方程式可以代表线或平面或三维线投射到x-y平面上。例如,在下图中,左过点P和Q的线。此线的 2D 空格是 yPQx+P 。

The three orange lines in the diagram at right below can also be described by this equation. All three have slope $\begin{aligned} M = \frac{- P}{Q} \end{aligned}$ and y-intercept B = P . If we want to specify one of these lines in particular, we need to identify the line as the intersection of two planes. For example, the line where $\begin{aligned} y = \frac{- P}{Q} x + P \end{aligned}$ and z = 0 for all points is the intersection of the orange and green planes in the diagram below right. (Remember that the orange and green planes are drawn with edges for convenience, but that they represent planes that extend outward toward infinity.)
::下方图表中的三条橙色线也可以用此方程式描述。这三个线都有斜坡 MPQ 和 y- intercept B = P 。如果我们特别要指定其中一条线, 我们需要将这条线指定为两平面的交叉点。例如, yPQx+P 和 z = 0 在所有点上的线是下方图表中的橙色和绿色平面的交叉点。 (记住, 橙色和绿色平面都是为方便而绘制的边缘, 但它们代表向无限延伸的平面。 )

Therefore we can specify a particular line by requiring that the equations for both planes be simultaneously true. In a previous section we developed one equation for a plane given by d = - n _x x _o - n _y y _o - n _z z _o . In this case Qy + Px - QP = 0 and z = 0 must both be satisfied for all points on the line passing through points P and Q .
::因此,我们可以通过要求两架飞机的方程同时为正方程来指定特定的直线。在前一节中,我们为 d = -nxo -nyyo -nzzo 给出的平面开发了一个方程。在这种情况下, Qy + Px - QP = 0 和 z = 0 都必须既满足通过点P 和 Q 的线上的所有点。

Finding the Vector Equation of a Line
::查找一线的矢量等量

Another way to identify points on a line can be found using the vector addition method we discussed earlier in this chapter. To find the position vector, $\begin{aligned} \vec{r} \end{aligned}$ , for any point along a line, we can add the position vector of a point on the line which we already know and add to that a vector, $\begin{aligned} \vec{v} \end{aligned}$ , that lies on the line as shown in the diagram below.
::也可以使用我们在本章前面讨论过的矢量添加方法找到线上的点。要找到线上任何一点的位置矢量,我们可以在线上添加我们已经知道的点的位置矢量,并在下面图表显示的线上添加一个矢量,即 v。

The position vector $\begin{aligned} \vec{R} \end{aligned}$ for a point between P and Q is given by $\begin{aligned} \vec{R} = \vec{p} + \vec{v} \end{aligned}$
::位于 P 和 Q 之间的一点的矢量 R 位置由 Rpv提供。

All other points on this line can be reached by traveling along the line from point P , therefore the position vector for any point on the line is given by $\begin{aligned} \vec{r} = \vec{p} + k \vec{v} \end{aligned}$ .
::这条线上的其他所有点都可以通过从P点沿这条线行走达到,因此线上任何点的位置矢量由 rpkv提供。

If we know the locations of two points on a line, we can determine the equation of the line. All points on a line fulfill the equation $\begin{aligned} \vec{r} = \vec{p} + k \vec{v} \end{aligned}$ , where k is a scalar that varies from -∞ to ∞. If we already know the position vectors for two points on the line, $\begin{aligned} \vec{p} \end{aligned}$ and $\begin{aligned} \vec{q} \end{aligned}$ , we can use the method of vector subtraction to determine the equation of the vector, $\begin{aligned} \vec{v} = \vec{p} - \vec{q} \end{aligned}$ . Therefore, $\begin{aligned} \vec{r} = \vec{p} + k (\vec{p} - \vec{q}) \end{aligned}$ , where k varies from -∞ to ∞.
::如果我们知道线上两个点的位置, 我们可以确定线的方程。线上的所有点都符合 rpkv 的方程。 k 是一个从\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

General Vector Equations
::一般矢量等量

In general, a vector equation is any function that takes any one or more variables and returns a vector. The vector equation of a line is an equation that identifies the position vector of every point along the line. This works for straight lines and for curves. For example, the vector equation $\begin{aligned} \vec{p} = ⟨ 3 cos θ, 3 sin θ, 2 ⟩ \end{aligned}$ defines a circle having a radius of 3 units which sits parallel to the x-y plane at a distance of 2 units along the z-axis as shown below. The orange line is the set of points that satisfy the vector equation $\begin{aligned} \vec{p} = ⟨ 3 cos θ, 3 sin θ, 2 ⟩ \end{aligned}$ while the grey lines are various individual position vectors which satisfy the vector equation.
::一般而言,矢量方程式是指需要任何一个或多个变量并返回矢量的任何函数。一线的矢量方程式是一种公式,用以识别线上每个点的位置矢量。这适用于直线和曲线。例如,矢量方程式p3 cos ,3 sin ,2 定义了一个圆形,其半径为3个单位,与X-y平面平行,其距离为Z-轴两单位,如下所示。橙色线是满足矢量方程式p3 cos 3 sin 2的一组点,而灰线则是满足矢量方方方形的各种单个位置矢量。

Examples
::实例

Example 1
::例1

Earlier, you were asked a question about two aircrafts.
::早些时候,有人问了你两架飞机的问题。

Air traffic control is tracking two planes in the vicinity of their airport. At a given moment, one plane is at a location 45 km east and 120 km north of the airport at an altitude of 7.5 km. The second plane is located 63 km east and 96 km south of the airport at an altitude of 6.0 km. The first plane is flying directly toward the airport while the second plane is continuing at a constant altitude with a heading defined by the vector $\begin{aligned} \vec{h_{2}} = ⟨ 3, 4, 0 ⟩ \end{aligned}$ to land eventually at another airport to the northwest of our air traffic controllers. Do the paths of these two aircraft cross?
::在某一时刻,一架飞机位于机场以东45公里和以北120公里处,高度为7.5公里;第二架飞机位于机场以东63公里和以南96公里处,高度为6.0公里;第一架飞机正直接飞往机场,而第二架飞机正以恒定高度飞行,航向由矢量h2340确定,最后降落在我国空中交通管制员西北的另一机场。

The first thing we need to do is to determine the position vectors of the two planes. Define our airport as the origin of coordinates and define $\begin{aligned} \hat{x} \end{aligned}$ = east, $\begin{aligned} \hat{y} \end{aligned}$ = north, and $\begin{aligned} \hat{z} \end{aligned}$ = upward. Call the position of plane #1, P , and the position of plane #2, Q . This gives $\begin{aligned} \vec{p} = ⟨ 45, 120, 7.5 ⟩ \end{aligned}$ and $\begin{aligned} \vec{q} = ⟨ 63, - 96, 6.0 ⟩ \end{aligned}$ .
::我们首先需要做的是确定两架飞机的位置矢量。定义我们的机场为坐标源, 定义坐标为 x = 东, y = 北, z = 上方。调用第 1 、 P 和 2 、 Q 的方位。这给出了 p 45 、 120 、 75 和 q 63 、 96 、 6. 0 的方位。

Plane #1 is heading directly toward the airport. The vector from the position of plane #1 to the origin is given by
::第1飞机正直接朝机场方向飞去。从第1飞机的位置到原点的矢量由

$\begin{aligned} \vec{v} = \vec{o} - \vec{p} = ⟨ (0 - 45), (0 - 120), (0 - 7.5) ⟩ \end{aligned}$
:0-45),(0-120),(0-7.5/5)

$\begin{aligned} \vec{v} = \vec{o} - \vec{q} = ⟨ - 45, - 120, - 7.5 ⟩ \end{aligned}$
::45,-120,-7.5

The equation of the line representing plane #1’s motion then becomes
::代表 1 平面运动的直线等式变成

$\begin{aligned} \vec{r_{1}} = ⟨ 45, 120, 7.5 ⟩ + k_{1} ⟨ - 45, - 120, - 7.5 ⟩ \end{aligned}$
::-45,120,7.5,k1 -45,-120,-7.5

Plane #2 continues at a constant altitude with a heading $\begin{aligned} \vec{h_{2}} = ⟨ 3, 4, 0 ⟩ \end{aligned}$ . The equation of the line representing plane #2’s motion then becomes
::平面# 2 继续以恒定高度飞行, 标题为 h23,4,0。代表平面运动的直线的方程式随后变成

$\begin{aligned} \vec{r_{2}} = ⟨ - 63, 96, 6.0 ⟩ + k_{2} ⟨ 3, 4, 0 ⟩ \end{aligned}$
::r263,96,6.0k23,4,0

If the paths of the two planes intersect, the position of that intersection must have coordinates which satisfy both equations.
::如果两架飞机的路径交叉,该十字路口的位置必须具有能满足两个方程的坐标。

$\begin{aligned} \vec{r_{2}} = ⟨ - 63, 96, 6.0 ⟩ + k_{2} ⟨ 3, 4, 0 ⟩ \\ \vec{r_{1}} = ⟨ 45, 120, 7.5 ⟩ + k_{1} ⟨ - 45, - 120, - 7.5 ⟩ \end{aligned}$
::r263,96,6.0k23,4,0r145,120,7.5k145,-120,-7.55

To satisfy both equations, r ₁ _x must equal r ₂ _x , r ₁ _y must equal r ₂ _y , and r ₁ _z must equal r ₂ _z . This means that 45 - 45 k ₁ = 63 + 3 k ₂ , 120 - 120 k ₁ = -96 + 4 k ₂ , and 7.5 - 7.5 k ₁ = 6.0 + 0 k ₂ must all be simultaneously true.
::为了满足两种方程, r1x必须等值 r2x, r1y必须等值 r2y, r1z必须等值 r2z。这意味着45 - 45k1 = 63 + 3k2, 120 - 120k1 = - 96 + 4k2, 7.5 - 7.5k1 = 6.0 + 0k2 必须同时真实。

The first equation simplifies to 15 - 15 k ₁ = 21 + k ₂ or k ₂ = -6 - 15 k ₁ . The second simplifies to 30 - 30 k ₁ = -24 + k ₂ or k ₂ = 54 - 30 k ₁ . The third equation simplifies to 7.5 - 7.5 k ₁ = 6 or $\begin{aligned} k_{1} = \frac{1}{5} \end{aligned}$ .
::第一个方程式简化为 15 - 15k1 = 21 + k2 或 k2 = - 6 - 15k1. 第二个简化为 30 - 30k1 = - 24 + k2 或 k2 = 54 - 30k1. 第三个方程式简化为 7.5 - 7.5k1 = 6 或 k1= 15。

Inserting the value of k ₁ from the third equation into the first two equations, we find $\begin{aligned} k_{2} = - 6 - 15 (\frac{1}{5}) = - 6 - 3 = - 9 \end{aligned}$ and $\begin{aligned} k_{2} = 54 - 30 k_{1} = 54 - 30 (\frac{1}{5}) = 54 - 6 = 48 \end{aligned}$ Since k ₂ = 33 and k ₂ = 0 cannot be simultaneously true, the two planes’ paths do not cross.
::在前两个方程中插入从第三个方程中的 k1 值时, 我们发现 k2\\ 6- 15( 15)\\ 6- 39 和 k2= 54- 30k1= 54- 30( 15)= 54- 6= 48 因为 k2 = 33 和 k2 = 0 不能同时真实, 两平面的路径无法交叉。

Note: There are other ways to reason through this problem. Can you find some? (Hint: Think of the geometry in 2-dimensions, or focus on the altitude as a way to simplify this problem).
::注意: 有其他方法可以解释这个问题。您能找到一些吗? (提示: 想象二维的几何, 或者关注高度, 以此来简化这个问题 ) 。

Example 2
::例2

Write the equation of $\begin{aligned} y = - \frac{5}{3} x + 5 \end{aligned}$ as a vector equation.
::将 y53x+5 的方程式写成矢量方程式。

Start by choosing two points on the line, say (3,0) and (6,-5). Then $\begin{aligned} \vec{p} = ⟨ 3, 0 ⟩ \end{aligned}$ and $\begin{aligned} \vec{q} = ⟨ 6, - 5 ⟩ \end{aligned}$ . So,
::首先在线上选择两点, 例如( 3, 0) 和 (6, 5) 。然后 p 3, 0 和 q 6, 5 。所以,

$\begin{aligned} \vec{p} - \vec{q} = ⟨ 3, 0 ⟩ - ⟨ 6, - 5 ⟩ = ⟨ 3 - 6, 0 - (- 5) ⟩ = ⟨ - 3, 5 ⟩ \end{aligned}$
::-=YTET -伊甸园字幕组=-伊甸园字幕组=- 翻译:

Finally, the vector equation of the line is
::最后,线的矢量方程式是

$\begin{aligned} \vec{r} = ⟨ 3, 0 ⟩ + k ⟨ - 3, 5 ⟩ = ⟨ 3 - 3 k, 5 k ⟩ \end{aligned}$
::3,0,3,5,3,3,3,3,3,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,3,3,3,5,3,3,5,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5

Example 3
::例3

Write the vector equation of the line defined by the points (2, 3, 4) and (-5, 2, -1).
::写入由点(2、3、4)和点(-5、2、-1)定义的线条的矢量方程式。

These two points have position vectors $\begin{aligned} ⟨ 2, 3, 4 ⟩ \end{aligned}$ and $\begin{aligned} ⟨ - 5, 2, - 1 ⟩ \end{aligned}$ . The vector of the line connecting the two points is given by
::这两个点都有位置矢量 2,3,444 和5,2,-11。连接这两个点的线的矢量由

$\begin{aligned} \vec{v} = \vec{p} - \vec{q} = ⟨ (2 - (- 5)), (3 - 2), (4 - (- 1)) ⟩ \end{aligned}$
:2 -(5)),(3 - 2),(4 -(1))

$\begin{aligned} \hat{v} = \vec{p} - \vec{q} = ⟨ 7, 1, 5 ⟩ \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}七,一,五

The equation of the line then becomes
::然后直线的方形变成

$\begin{aligned} \vec{r} = ⟨ 2, 3, 4 ⟩ + k ⟨ 7, 1, 5 ⟩ \vec{r} = ⟨ 2, 3, 4 ⟩ + k ⟨ 7, 1, 5 ⟩ \end{aligned}$
::2,3,447,1,552,347,1,55

Example 4
::例4

Determine the equation for the line defined by the points P = (6, 7, 5) and Q = (3, 2, 11). Then find the position vector for a point, R, half-way between these two points.
::确定由 P = (6, 7, 5) 和 Q = (3, 2, 11) 所定义的线条的方程,然后为这两个点之间的一个点( R) 找到位置矢量。

$\begin{aligned} \vec{v} = \vec{p} - \vec{q} = ⟨ (6 - 3), (7 - 2), (5 - 11) ⟩ \end{aligned}$
:6-3),(7)-2,(5)-(11)_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} \hat{v} = \vec{p} - \vec{q} = ⟨ 3, 5, - 6 ⟩ \end{aligned}$
::QQ3,5,-665,6666

The equation of the line then becomes
::然后直线的方形变成

$\begin{aligned} \vec{r} = ⟨ 6, 7, 5 ⟩ + k ⟨ 3, 5, - 6 ⟩ \vec{r} = ⟨ 6, 7, 5 ⟩ + k ⟨ 3, 5, - 6 ⟩ \end{aligned}$
::6,7,53,56,66,55,66

Since the vector $\begin{aligned} \vec{v} \end{aligned}$ points from P , the value of k for a particular point gives us some information about the location of that point. If 0 < k < 1, the point lies on the line between P and Q . If k < 0, the point is not between P and Q and is closer to P than Q . If k > 1, the point lies on the line between P and Q . If k < 0, the point is not between P and Q and is closer to P than Q . The point halfway between the two points has $\begin{aligned} k = \frac{1}{2} \end{aligned}$ .
::由于矢量 v 点来自 P, k 值对某个点的数值为我们提供了关于该点位置的一些信息。如果 0 < k < 1, 点位于 P 和 Q 之间的线条上。如果 k < 0, 点在 P 和 Q 之间, 点在 P 和 Q 之间, 点在 Q 之间。如果 k > 1, 点在 P 和 Q 之间的线条上。如果 k < 0, 点在 P 和 Q 之间, 点在 P 之间, 点在 Q 之间。两个点之间的中间点是 k=12 。

$\begin{aligned} \vec{R} = ⟨ 6, 7, 5 ⟩ + \frac{1}{2} ⟨ 3, 5, - 6 ⟩ \end{aligned}$
::R6,7,512,3,5,-66

$\begin{aligned} \vec{R} = ⟨ (6 + 1.5), (7 + 2.5), (5 + (- 3) ⟩ = ⟨ 7.5, 9.5, 2 ⟩ \end{aligned}$
::R*(6+1.5),(7+2.5),(5+(3))

Example 5
::例5

Are the two vectors $\begin{aligned} \vec{D} = ⟨ 1, - 1, 4 ⟩ + d ⟨ 1, - 1, 1 ⟩ \end{aligned}$ and $\begin{aligned} \vec{F} = ⟨ 2, 4, 7 ⟩ + f ⟨ 2, 1, 3 ⟩ \end{aligned}$ skew lines or do they intersect one another?
::两种矢量是D1, -1,44,44,1,1,1 和F2,4,4,7f,2,1,3,3Skew线,还是它们相互交叉?

If the two vectors intersect, there must be a point identified by position vector $\begin{aligned} \vec{p} \end{aligned}$ which satisfies the equations of both lines. In other words, we must be able to find values for d and f such that $\begin{aligned} \vec{D} = \vec{F} \end{aligned}$ or $\begin{aligned} ⟨ 1, - 1, 4 ⟩ + d ⟨ 1, - 1, 1 ⟩ = ⟨ 2, 4, 7 ⟩ + f ⟨ 2, 1, 3 ⟩ \end{aligned}$ .
::如果两个矢量相互交叉, 则必须有一个由位置矢量 p 确定的点, 满足两行的方程。换句话说, 我们必须能够找到 d 和 f 的值, 如 DF 或 1, - 1, 4d 1, - 1, 1, 1, 12, 4, 7f2, 1,3 。

Each of the three components of vectors $\begin{aligned} \vec{D} \end{aligned}$ and $\begin{aligned} \vec{F} \end{aligned}$ must independently be equal if $\begin{aligned} \vec{D} = \vec{F} \end{aligned}$ . This means that 1 + d = 2 + 2 f , -1 - d = 4 + f , and 4 + d = 7 + 3 f Combining the first two equations gives: 1 + d - 1 - d = 2 + 2 f + 4 + f which when simplified becomes 0 = 6 + 3 f or f = -2. If we substitute this back into the first equation we obtain, 1 + d = 2 + 2(-2) which when simplified becomes d = -3. If the two lines cross, f = -2 and d = -3 should satisfy each of the component equations.
::如果DF,则D和F的三个矢量组成部分的每个组成部分必须独立平等。这意味着1+d=2+2f, -1-d=4+f,4+d=4+f,和4+d=7+3f 合并前两个方程式:1+d-1-d=2+2f+4+f,当简化为0=6+3f或f=2时,它必须独立平等。如果我们将此替换回我们获得的第一个方程式,那么,1+d=2+2(-2),当简化后成为d=3,如果两条线交叉,f=2和d=3应满足每个组成部分方程式。

1 + d = 2 + 2 f becomes 1 + (-3) = 2 + 2(-2) or -2 = -2
::1 + d = 2 + 2 f 改为 1 + (3) = 2 + 2( 2-2) 或 - 2 = - 2

-1 - d = 4 + f becomes -1 - (-3) = 4 + (-2) or +2 = +2
::- 1 - d = 4 + f 改为 - 1 - (3) = 4 + (2) 或 + 2 = +2

4 + d = 7 + 3 f becomes 4 + (-3) = 7 + 3(-2) or +1 = +1
::4 + d = 7 + 3f 改为 4 + (3) = 7 + 3(2)或 + 1 = +1

All three equations are equally satisfied, so the two lines do intersect and the point of intersection is (-2, 2, 1).
::所有三个方程都同样得到满足,因此两条线相互交叉,交叉点是(2、2、1)。

Example 6
::例6

In physics, the motion of an object traveling at a constant speed is described by the equation $\begin{aligned} \vec{s_{t}} = \vec{s_{i}} + \vec{v} t \end{aligned}$ where s _i is the initial position, s is the position at some later time t , and v is the velocity of the object. Write the vector equation which returns the set of position vectors $\begin{aligned} \vec{s} \end{aligned}$ for an object having an initial position $\begin{aligned} \vec{s_{i}} = ⟨ 2, 3, 4 ⟩ \end{aligned}$ and a velocity of $\begin{aligned} \vec{v} = ⟨ 1, 1, - 2 ⟩ \end{aligned}$ and determine the object’s location at t = 10s.
::在物理学中,以恒定速度飞行的物体的动作由公式 stsivt 描述,该方程式的初始位置是 si, s 是稍后时间的方位 t, v 是对象的速度。写入矢量方程, 该矢量方程返回位置矢量 s 的一组位置矢量 s , 该对象的初始位置是 2, 3,4 和 v1, 1, -2 的速度, 并确定天体在 t = 10 的位置。

$\begin{aligned} \vec{s_{t}} = \vec{s_{i}} + \vec{v} t = ⟨ 2, 3, 4 ⟩ + t ⟨ 1, 1, - 2 ⟩ = ⟨ 2 + t, 3 + t, 4 - 2 t ⟩ \end{aligned}$
::2,3,4,4,4,1,1,2-2,2+t,3+t,4 -2

At t = 10s, $\begin{aligned} \vec{s_{10}} = ⟨ 2 + 10, 3 + 10, 4 - 2 (10) ⟩ = ⟨ 12, 13, - 16 ⟩ \end{aligned}$
::at t = 10 = 10s, s102+10,3+10,4-2(10) = 10 = 12,13,-16

Example 7
::例7

An object has a position of $\begin{aligned} \vec{s_{i}} = ⟨ 3, 3, 6 ⟩ \end{aligned}$ at t = 0 and a velocity of $\begin{aligned} \vec{v} = ⟨ 10, 7, 3 ⟩ \end{aligned}$ . Use the vector equation $\begin{aligned} \vec{s} = \vec{s_{i}} + \vec{v} t \end{aligned}$ to determine the distance traveled by the object between t = 3s and t = 5s. Distance measured in meters.
::对象在 t = 0 时的方位为 si 3,3,6 和 v 10,7,3 的速率为 v 10,7,3 。使用矢量方程式 s si vt 来确定天体在 t = 3 和 t = 5 之间的距离。以米计的距离。

The vector equation describing the motion of the object is
::描述对象运动的矢量方程式是

$\begin{aligned} \vec{s} = \vec{s_{i}} + \vec{v} t = ⟨ 3, 3, 6 ⟩ + t ⟨ 10, 7, 3 ⟩ = ⟨ 3 + 10 t, 3 + 7 t, 6 + 3 t ⟩ \end{aligned}$
::=================================================================================================================================================================;===========================================================================================================================================================================================================================================================================================================================================================

The object’s position at t = 3s is obtained from the vector equation:
::对象在 t = 3s 的位置取自矢量方程式:

$\begin{aligned} \vec{s_{3}} = \vec{s_{i}} + \vec{v} t = ⟨ 3, 3, 6 ⟩ + (3) ⟨ 10, 7, 3 ⟩ = ⟨ 3 + 10 (3), 3 + 7 (3), 6 + 3 (3) ⟩ = \end{aligned}$ $\begin{aligned} ⟨ 33, 24, 15 ⟩ \end{aligned}$
::s33,3,63,6(3) 10,7,33+10(3),3+7(3),6+3(3),33,24,15

The object’s position at t = 5s is obtained from the vector equation:
::对象在 t = 5s 的位置取自矢量方程式:

$\begin{aligned} \vec{s_{5}} = \vec{s_{i}} + \vec{v} t = ⟨ 3, 3, 6 ⟩ + (5) ⟨ 10, 7, 3 ⟩ = ⟨ 3 + 10 (5), 3 + 7 (5), 6 + 3 (5) ⟩ = \end{aligned}$ $\begin{aligned} ⟨ 53, 38, 21 ⟩ \end{aligned}$
::=============================================================================================;===============================================================================================================================================================================================================================================================================================================================================================================================================================

The distance traveled between these two points is the magnitude of the vector starting at (33, 24, 15) and ending at (53, 38, 21).
::这两个点之间的距离是从(33、24、15)开始到(53、38、21)结束的矢量的大小。

$\begin{aligned} \vec{△ s} = \vec{s_{5}} - \vec{s_{3}} = ⟨ 53, 38, 21 ⟩ - ⟨ 33, 24, 15 ⟩ = ⟨ 20, 14, 6 ⟩ \end{aligned}$
::5353,38,2133,24,15,20,14,6

Now we can use the Pythagorean theorem to determine the magnitude of the vector.
::现在我们可以使用毕达哥里定理来确定矢量的大小。

$\begin{aligned} | \vec{△ s} | = \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{(20)^{2} + (14)^{2} + (6)^{2}} = 25.1 \end{aligned}$ meters
::a2+b2+c2=(20)2+(14)2+(6)2=25.1米

Example 8
::例8

Determine the vector equation of the straight line defined by the points (2, 2, 2) and (1, 3, 5).
::确定由点(2、2、2)和点(1、3、5)界定的直线的矢量方程式。

These two points have position vectors $\begin{aligned} \vec{p} = ⟨ 2, 2, 2 ⟩ \end{aligned}$ and $\begin{aligned} \vec{q} = ⟨ 1, 3, 5 ⟩ \end{aligned}$ . The vector of the line connecting the two points is given by
::这两个点都有位置矢量 p2,2,2 和 q1,3,5。连接这两个点的线条矢量由

$\begin{aligned} \vec{v} = \vec{p} - \vec{q} = ⟨ (2 - 1), (2 - 3), (2 - 5) ⟩ = ⟨ 1, - 1, - 3 ⟩ \end{aligned}$
:二)-1,(二)-3,(二)-5,(一)-1,(一)-3,(三)

The equation of the line, $\begin{aligned} \vec{r} = \vec{p} + k \vec{v} \end{aligned}$ , then becomes
::直线的方程, rpkv, 然后变成

$\begin{aligned} \vec{r} = ⟨ 2, 2, 2 ⟩ + k ⟨ 1, - 1, - 3 ⟩ \vec{r} = ⟨ 2, 2, 2 ⟩ + k ⟨ 1, - 1, - 3 ⟩ \end{aligned}$
::2,2,221,-1,-1,-32,2,22,21,-1,-33

Example 9
::例9

Do the two lines $\begin{aligned} \vec{R} = ⟨ 4, 4, - 2 ⟩ + r ⟨ - 3, 7, 2 ⟩ \end{aligned}$ and $\begin{aligned} \vec{K} = ⟨ 9, - 8, 7 ⟩ + k ⟨ - 2, 1, 3 ⟩ \end{aligned}$ intersect?
::两条线R4、4、-23、7、29、8、72、1、33交叉吗?

If the two vectors intersect, there must be a point identified by position vector $\begin{aligned} \vec{p} \end{aligned}$ which satisfies the equations of both lines. In other words, we must be able to find values for r and k such that $\begin{aligned} \vec{R} = \vec{K} \end{aligned}$ or $\begin{aligned} ⟨ 4, 4, - 2 ⟩ + r ⟨ - 3, 7, 2 ⟩ = ⟨ 9, - 8, 7 ⟩ + k ⟨ - 2, 1, 3 ⟩ \end{aligned}$ .
::如果两个矢量交叉, 则必须有一个位置矢量 p 表示的点, 满足两行的方程。换句话说, 我们必须能够找到 r 和 k 的值, 以便 R K 或 {4, 4, - 2 r3, 7, 29, 8, 7k2, 1,3 。

Each of the three components of vectors $\begin{aligned} \vec{R} \end{aligned}$ and $\begin{aligned} \vec{K} \end{aligned}$ must independently be equal if $\begin{aligned} \vec{R} = \vec{K} \end{aligned}$ . This means that 4 - 3 r = 9 - 2 k , 4 + 7 r = -8 + k , and -2 + 2 r = 7 + 3 k
::矢量 R 和 K 的三种成分如果 R 和 K , 则必须独立平等。这意味着 4 - 3r = 9 - 2k, 4 + 7r = - 8 + k, 和 - 2 + 2r = 7 + 3k

Solving the second equation for k gives 12 + 7 r = k . Now substitute this into one of the other two equations: 4 - 3 r = 9 - 2(12 + 7 r ) or 4 - 3 r = 9 - 24 - 14 r which simplifies to 19 = -11 r or $\begin{aligned} r = \frac{- 19}{11} = - 1.727. \end{aligned}$ Substitute this value into one of the other equations: -2 + 2(-1.727) = 7 + 3 k which becomes -2 - 3.454 = 7 + 3 k . Solving for k gives k = -4.151. If the two lines cross, r = -1.727 and k = -4.151 should satisfy each of the component equations.
::解决 k 的第二个方程式给出 12 + 7r = k。现在将它替换为另外两个方程式之一 : 4 - 3r = 9 - 2( 12 + 7r ) 或 4 - 3r = 9 - 24 - 14r = 9 - 24 - 14r 简化为 19 = - 11r 或 r 11\\\ 1. 727 。将这个数值替换为其他方程式之一 : - 2 + 2( 1- 727 ) = 7 + 3k, 成为 - 2 - 3. 454 = 7 + 3k 。 k 溶解 k = - 4. 151。如果两条线的r = - 1. 727 和 k = - 4. 151 将满足每个方程式。

4 - 3 r = 9 - 2 k becomes 4 - 3 (-1.727) = 9 - 2(-4.151) or -1.181 = 17.302 Since even this first equation does not hold true, these two lines are skew and do not intersect.
::4 - 3r = 9 - 2k = 4 - 3 (-1.727) = 9 - 2(-4.51) 或 - 1.181 = 17. 302 由于即使第一个方程也不正确,这两条线是斜线,不交叉。

Review
::回顾

Write the vector equation of the line defined by the the following points:
::写入由以下各点定义的线条矢量方程式:

$\begin{aligned} (2, - 2, 5) \end{aligned}$ and $\begin{aligned} (1, 5, 4) \end{aligned}$
:2,2-2,5)和(1,5,4)

$\begin{aligned} (2, - 9, 5) \end{aligned}$ and $\begin{aligned} (8, 4, - 6) \end{aligned}$
:2,-9,5)和(8,4,-6)

$\begin{aligned} (15, 3, - 3) \end{aligned}$ and $\begin{aligned} (4, - 3, 9) \end{aligned}$
:15,3,3-3)和(4,3,9)

$\begin{aligned} (- 1, - 1, 7) \end{aligned}$ and $\begin{aligned} (3, 11, 8) \end{aligned}$
:-1,1,7)和(3,11,8)

$\begin{aligned} (1, - 3, 2) \end{aligned}$ and $\begin{aligned} (- 5, 3, - 1) \end{aligned}$
:1,-3,2)和(-5,3,1)

$\begin{aligned} (25, 17, 42) \end{aligned}$ and $\begin{aligned} (- 16, 12, 23) \end{aligned}$
:25,17,42)和(-16,12,23)

Determine if the two vectors are skew lines or if they intersect each other.
::确定两个矢量是斜线线,还是相互交错。

$\begin{aligned} \vec{D} = ⟨ 2, 3, 1 ⟩ + d ⟨ 5, 3, 6 ⟩ \end{aligned}$ and $\begin{aligned} \vec{F} = ⟨ - 5, - 3, - 5 ⟩ + f ⟨ 15, 3, - 3 ⟩ \end{aligned}$
::D2,3,1,1,5,3,6 和F5,5,3,5,5,5,5,5,15,3,3,3

$\begin{aligned} \vec{D} = ⟨ 3, 4, 7 ⟩ + d ⟨ 3, 3, 2 ⟩ \end{aligned}$ and $\begin{aligned} \vec{F} = ⟨ - 2, 11, 7 ⟩ + f ⟨ - 2, 11, 7 ⟩ \end{aligned}$
::D3,4,7,7,3,3,2 和F2,2,11,7 f,2,11,7

$\begin{aligned} \vec{D} = ⟨ 15, 3, - 3 ⟩ + d ⟨ 3, 11, 8 ⟩ \end{aligned}$ and $\begin{aligned} \vec{F} = ⟨ 5, 3, 6 ⟩ + f ⟨ 1, - 4, 6 ⟩ \end{aligned}$
::D15,3,3,3,3,3,3,11,8 和F5,3,6,6,5,4,6

$\begin{aligned} \vec{D} = ⟨ 13, - 1, 6 ⟩ + d ⟨ - 5, 4, 12 ⟩ \end{aligned}$ and $\begin{aligned} \vec{F} = ⟨ 6, 9, 0 ⟩ + f ⟨ 21, 0, 14 ⟩ \end{aligned}$
::13, -1,65,4,12 和F6,9,0F21,0,1414

Identify the position vector for the midpoint of each line (you already found the vector equations in problems 1 through 4).
::确定每行中点的位置矢量(您已经在问题1至4中找到了矢量方程)。

$\begin{aligned} (2, - 2, 5) \end{aligned}$ and $\begin{aligned} (1, 5, 4) \end{aligned}$
:2,2-2,5)和(1,5,4)

$\begin{aligned} (2, - 9, 5) \end{aligned}$ and $\begin{aligned} (8, 4, - 6) \end{aligned}$
:2,-9,5)和(8,4,-6)

$\begin{aligned} (15, 3, - 3) \end{aligned}$ and $\begin{aligned} (4, - 3, 9) \end{aligned}$
:15,3,3-3)和(4,3,9)

$\begin{aligned} (- 1, - 1, 7) \end{aligned}$ and $\begin{aligned} (3, 11, 8) \end{aligned}$
:-1,1,7)和(3,11,8)

Use the vector equation, and the following values for t: t = 3s and t = 5s. Distance measured in meters.
::使用矢量方程式, t 的数值如下: t = 3s 和 t = 5s. 以米计的距离。

An object has a position of $\begin{aligned} \vec{s_{i}} = ⟨ 3, - 5, 1 ⟩ \end{aligned}$ at t = 0 and a velocity of $\begin{aligned} \vec{v} = ⟨ 2, 7, - 3 ⟩ \end{aligned}$
::对象在 t = 0 时的方位为 si%3, - 5,1 和 vv2, 7, - 3 的速率。

An object has a position of $\begin{aligned} \vec{s_{i}} = ⟨ 1, - 4, - 8 ⟩ \end{aligned}$ at t = 0 and a velocity of $\begin{aligned} \vec{v} = ⟨ 9, - 7, 3 ⟩ \end{aligned}$
::对象在 t = 0 时的方位为 si1,- 4,- 8,且速度为 v9,- 7,3。

An object has a position of $\begin{aligned} \vec{s_{i}} = ⟨ - 1, 2, - 3 ⟩ \end{aligned}$ at t = 0 and a velocity of $\begin{aligned} \vec{v} = ⟨ 3, - 1, - 2 ⟩ \end{aligned}$
::对象在 t = 0 时的方位为 si%1, 2, - 3 , 速度为 v%3, 1, 2- 2 。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

线条矢量方对数

Vector Equation of a Line ::线条矢量方对数

Lines in Space ::空间线

Finding the Vector Equation of a Line ::查找一线的矢量等量

General Vector Equations ::一般矢量等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Example 8 ::例8

Example 9 ::例9

Review ::回顾

Review (Answers) ::回顾(答复)

Vector Equation of a Line
::线条矢量方对数

Lines in Space
::空间线

Finding the Vector Equation of a Line
::查找一线的矢量等量

General Vector Equations
::一般矢量等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

Example 8
::例8

Example 9
::例9

Review
::回顾

Review (Answers)
::回顾(答复)