7.2 向量操作

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  矢量操作

  When two or more forces are acting on the same object, they combine to create a new force.  A bird flying due south at 10 miles an hour in a headwind of 2 miles an hour only makes headway at a rate of 8 miles per hour.  These forces directly oppose each other.  In real life, most forces are not parallel.  What will happen when the headwind has a slight crosswind as well, blowing NE at 2 miles per hour.  How far does the bird get in one hour? 

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  Basic Vector Operations
  ::基本矢量操作

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  Scalar multiplication means to multiply a vector by a number.  This changes the magnitude of the vector, but not its direction.  If $\begin{array}{r}\stackrel{\to }{v}=<3,4>\end{array}$ , then $\begin{array}{r}2\stackrel{\to }{v}=<6,8>\end{array}$ . Scalar multiplication is fairly simple. 
  ::计算乘法乘以一个矢量乘以数。这会改变矢量的大小,但不会改变其方向。Ifv34.>, then2v6,8>。 计算乘法很简单。

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  Adding and subtracting vectors is slightly more difficult. When adding vectors, place the tail of one vector at the head of the other.  This is called the tail-to-head rule .  The vector that is formed by joining the tail of the first vector with the head of the second is called the vector.
  ::添加和减去矢量比较困难。在添加矢量时,将一个矢量的尾部置于另一个矢量的顶部。这称为尾对头规则。通过连接第一个矢量的尾部和第二个矢量的头而形成的矢量被称为矢量。

  

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  Vector subtraction reverses the direction of the second vector. $\begin{array}{r}\stackrel{\to }{a}-\stackrel{\to }{b}=\stackrel{\to }{a}+(-\stackrel{\to }{b})\end{array}$ :
  ::矢量减法逆向第二个矢量的方向。 aba(-b):

  

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  Adding vectors can be done in either order (just like with regular numbers).  Subtracting vectors must be done in a specific order or else the vector will be negative (just like with regular numbers). 
  ::添加矢量可以按任一顺序进行(就像正常数字一样 ) 。 减量矢量必须按特定顺序进行, 否则矢量将是负的( 就像正常数字一样 ) 。

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  To find the length or magnitude of a resultant vector, you can use the law of cosines. To do this, you also need to know the angle between the two vectors. Say  you were given two vectors $\begin{array}{r}\stackrel{\to }{a}\end{array}$  and $\begin{array}{r}\stackrel{\to }{b}\end{array}$ , have magnitudes of 5 and 9 respectively and that the angle between the vectors is $\begin{array}{r}{53}^{\circ }\end{array}$ . To find the magnitude of $\begin{array}{r}\stackrel{\to }{a}+\stackrel{\to }{b}\end{array}$  , which is written as  $\begin{array}{r}|\stackrel{\to }{a}+\stackrel{\to }{b}|\end{array}$ , notice that you have a parallelogram.
  ::要找到结果矢量的长度或大小, 您可以使用 resine 法则。 要这样做, 您还需要知道两个矢量之间的角 。 假设给您两个矢量 a 和 b , 大小分别为 5 和 9 , 矢量之间的角为 53 。 要找到 a {b 的大小, 以 {a {b} 写成 {a {b} , 提醒您您有一个平行的图形 。

  

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  In order to fine the magnitude of the resulting vector in red, note that the triangle on the bottom has sides 9 and 5 with included angle $\begin{array}{r}{127}^{\circ }\end{array}$  due to the properties of parallelograms. And, so applying the Law of Cosines, you get:
  ::为了微小所生成的红色矢量的大小, 请注意底部的三角形有侧面 9 和 5 , 包括角度 127 , 这是因为平行图的特性。 因此, 应用科辛斯定律, 您可以得到 :

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  $$\begin{array}{rl}{x}^2& =9^2+5^2-2\cdot 9\cdot 5\cdot \cos {127}^{\circ }\\ x& \approx 12.66\end{array}$$

  
  ::x2=92+52-2955cos127x12.66

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  For this video, focus on scalar multiplication and adding and subtracting vectors:
  ::对于此视频, 聚焦于天平乘法的乘法, 并添加和减去矢量 :

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked about how fast a bird was flying. A bird flying due south at 10 miles an hour with a cross headwind of 2 mph heading NE would have a force diagram that looks like this:
  ::早些时候,有人问到一只鸟飞行的速度有多快。 一只鸟以每小时10英里的速度向南飞行,其横风2米长,向NE方向飞来,就会有一个像这样的力量图:

  

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  The angle between the bird’s vector and the wind vector is $\begin{array}{r}{45}^{\circ }\end{array}$  which means this is a perfect situation for the Law of Cosines.  Let $\begin{array}{r}x=\end{array}$  the red vector.
  ::鸟的矢量和风向之间的角是45,这意味着这是《科辛斯定律》中完美的状况。让我们 x = 红向量。

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  $$\begin{array}{rl}{x}^2& ={10}^2+2^2-2\cdot 10\cdot 2\cdot \cos {45}^{\circ }\\ x& \approx 8.7\end{array}$$

  
  ::x2=102+22-21022245x8.7

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  The bird is blown slightly off track and travels only about 8.7 miles in one hour. 
  ::这只鸟被稍稍吹离轨道,一小时内只行驶约8.7英里。

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  Example 2
  ::例2

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  Using the vectors with magnitude 5 and 9 and angle of   $\begin{array}{r}{53}^{\circ }\end{array}$  from the main section, what is the angle that the sum $\begin{array}{r}\stackrel{\to }{a}+\stackrel{\to }{b}\end{array}$  makes with $\begin{array}{r}\stackrel{\to }{a}\end{array}$ ? 
  ::使用主段5级和9级的矢量以及53°Z角的矢量, 和aQQQQ的角是什么?

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  Start by drawing a good picture and labeling what you know.  $\begin{array}{r}|\stackrel{\to }{a}|=5,|\stackrel{\to }{b}|=9,|\stackrel{\to }{a}+\stackrel{\to }{b}|\approx 12.66\end{array}$ .  Since you know three sides of the triangle and you need to find one angle, this is the SSS application of the Law of Cosines.
  ::首先绘制一张好的图片, 并标出您所知道的标签 。 @ @ aç5, @ b9, @ ab12.66。 既然您知道三角形的三边, 您需要找到一个角度, 这是 SSS 应用的科辛定律 。

  

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  $$\begin{array}{rl}{9}^2& ={12.66}^2+5^2-2\cdot 12.66\cdot 5\cdot \cos \theta \\ \theta & ={34.6}^{\circ }\end{array}$$

  
  ::92=12.662+52-212.66=5cos34.6___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Example 3
  ::例3

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  Elaine started a dog walking business.  She walks two dogs at a time named Elvis and Ruby.  They each pull her in different directions at a $\begin{array}{r}{45}^{\circ }\end{array}$  angle with different forces.  Elvis pulls at a force of  $\begin{array}{r}25N\end{array}$ and Ruby pulls at a force of $\begin{array}{r}49N\end{array}$ .  How hard does Elaine need to pull so that she can stay balanced?  Note that N stands for Newtons which is the standard unit of force.
  ::Elaine开始做一条狗行走的生意。她一次走两条狗叫Elvis和Ruby。他们每个人都用不同的力量将她拉向不同方向。Elvis在25N和Ruby的部队拉向49N的部队。Elaine需要多大的拉动才能保持平衡?请注意,N代表牛顿,牛顿是标准的力量单位。

  

  
  

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  Even though the two vectors are centered at Elaine, the forces are added which means that you need to use the tail-to-head rule to add the vectors together.  Finding the angle between each requires logical use of supplement angles. 
  ::尽管两个矢量都以伊莱因为中心, 但仍添加了力量, 这意味着您需要使用尾巴对头规则来将矢量加在一起。 查找每种矢量之间的角需要逻辑使用补充角度 。

  

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  $$\begin{array}{rl}{x}^2& ={49}^2+{25}^2-2\cdot 49\cdot 25\cdot \cos {135}^{\circ }\\ x& \approx 68.98N\end{array}$$

  
  ::x2=492+252-24925cos135x_68.98

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  In order for Elaine to stay balanced, she will need to counteract this force with an equivalent force of her own in the exact opposite direction. 
  ::为了让Elaine保持平衡,她需要用她自己的同等力量以完全相反的方向对抗这支力量。

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  Example 4
  ::例4

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  Consider vector $\begin{array}{r}\stackrel{\to }{v}=<2,5>\end{array}$  and vector $\begin{array}{r}\stackrel{\to }{u}=<-1,9>\end{array}$ .  Determine the component form of the following: $\begin{array}{r}3\stackrel{\to }{v}-2\stackrel{\to }{u}\end{array}$ . 
  ::考虑矢量 v2,5 > 和矢量 u1,9>。确定以下元素的构成形式 : 3v2u。

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  Do multiplication first for each term , followed by vector subtraction.
  ::每个术语首先乘以乘法,然后是矢量减法。

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  $$\begin{array}{rl}3\cdot \stackrel{\to }{v}-2\cdot \stackrel{\to }{u}& =3\cdot <2,5>-2\cdot <-1,9>\\ & =<6,15>-<-2,18>\\ & =<8,-3>\end{array}$$

  
  ::32232221,96,152,188~3>

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  Example 5
  ::例5

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  An airplane is flying at a bearing of $\begin{array}{r}{270}^{\circ }\end{array}$  at 400 mph.  A wind is blowing due south at 30 mph.  Does this cross wind affect the plane’s speed? 
  ::飞机在以400米的方位飞行270°C。 风在以30米的方位向南吹。 横风会影响飞机的速度吗?

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  Since the cross wind is perpendicular to the plane, it pushes the plane south as the plane tries to go directly east.  As a result the plane still has an airspeed of 400 mph but the groundspeed (true speed) needs to be calculated.
  ::由于横风与飞机垂直,在飞机试图直接向东飞行时,横风将飞机推向南面,因此飞机的飞行速度仍为400平方英尺,但需要计算地面速度(实际速度)。

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  $$\begin{array}{rl}{400}^2+{30}^2& =x^2\\ x& \approx 401\end{array}$$

  
  ::4002+302=x2x401

  Summary

  播放段落 Scalar multiplication involves multiplying a vector by a number, changing its magnitude but not its direction. ::Scalar 乘法涉及将矢量乘以数,改变其数量,但不改变方向。 播放段落 Adding and subtracting vectors follows the tail-to-head rule, where the tail of one vector is placed at the head of the other, forming a resultant vector. ::添加和减去矢量遵循尾巴对头规则,将一个矢量的尾部置于另一个矢量的顶部,形成一个由此产生的矢量。 播放段落 Vector subtraction involves reversing the direction of the second vector before adding them. ::矢量减法是指在添加第二个矢量之前颠倒方向。 播放段落 The order of adding vectors does not matter, but the order of subtracting vectors does, as it can result in a negative vector. ::添加矢量的顺序并不重要,但减去矢量的顺序则重要,因为它可能导致负矢量。 播放段落 To find the magnitude of a resultant vector, use the law of cosines. ::为了找到结果的矢量的大小, 使用 共生法则。

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  Review
  ::回顾

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  Consider vector $\begin{array}{r}\stackrel{\to }{v}=<1,3>\end{array}$  and vector $\begin{array}{r}\stackrel{\to }{u}=<-2,4>\end{array}$ . 
  ::考虑矢量 v1,3 > 和矢量 u2,4> 。

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  1. Determine the component form of $\begin{array}{r}5\stackrel{\to }{v}-2\stackrel{\to }{u}\end{array}$ . 
     ::确定 5v+2u的构成表 。
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  2. Determine the component form of $\begin{array}{r}-2\stackrel{\to }{v}+4\stackrel{\to }{u}\end{array}$ . 
     ::确定 - 2v4u的构成形式。
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  3. Determine the component form of $\begin{array}{r}6\stackrel{\to }{v}+\stackrel{\to }{u}\end{array}$ . 
     ::确定 6vu的构成形式。
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  4. Determine the component form of $\begin{array}{r}3\stackrel{\to }{v}-6\stackrel{\to }{u}\end{array}$ . 
     ::确定 3v6u的构成形式。
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  5. Find the magnitude of the resultant vector from #1.
     ::从 # 1 中查找结果矢量的大小 。
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  6. Find the magnitude of the resultant vector from #2.
     ::查找由 # 2 生成的矢量的大小 。
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  7. Find the magnitude of the resultant vector from #3.
     ::从 # 3 查找由此产生的矢量的大小 。
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  8. Find the magnitude of the resultant vector from #4.
     ::查找从 # 4 中生成的矢量的大小 。
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  9. The vector $\begin{array}{r}<3,4>\end{array}$  starts at the origin.  What is the direction of the vector?
     ::矢量 < 3,4> 从源开始。矢量的方向是什么 ?
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  10. The vector $\begin{array}{r}<-1,2>\end{array}$  starts at the origin.  What is the direction of the vector?
      ::矢量 #% 1, 2> 从源开始。 矢量的方向是什么 ?
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  11. The vector $\begin{array}{r}<3,-4>\end{array}$  starts at the origin.  What is the direction of the vector?
      ::矢量 < 3 - 4> 从源开始。矢量的方向是什么?
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  12. A bird flies due south at 8 miles an hour with a cross headwind blowing due east at 15 miles per hour.  How far does the bird get in one hour? 
      ::一只鸟以每小时8英里的速度向南飞行, 横风向东飞行,每小时15英里。 鸟在1小时内能飞多远?
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  13. What direction is the bird in the previous problem actually moving?
      ::上一个问题中鸟儿真正移动的方向是什么?
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  14. A football is thrown at 50 miles per hour due north.  There is a wind blowing due east at 8 miles per hour.  What is the actual speed of the football?
      ::向北以每小时50英里的速度投掷足球,东风以每小时8英里向东吹风。
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  15. What direction is the football in the previous problem actually moving?
      ::足球在前一个问题中 究竟向哪个方向移动?

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  Review (Answers)
  ::回顾(答复)

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  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 。