9.3 向量操作

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  矢量操作

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  Introduction
  ::导言

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  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 mph in a headwind of 2 mph makes headway at a rate of only 8 mph. 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 northeast at 2 mph? How far will the bird get in one hour?
  ::当两个或两个以上部队在同一物体上行动时,它们联合起来形成一支新的力量。在2米长的风中以10mph向南飞行的一只鸟,在2米长的风中以10mph向南飞行,以只有8mph的速度前进。这些力量直接相互对立。在现实生活中,大多数力量并不平行。当风向有轻微的横风,又在2mph向东北吹风时,会发生什么情况?1小时后鸟能飞多远?

  

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  Operations with Vectors
  ::矢量操作

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  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 formed by joining the tail of the 1st with the head of the 2nd is called the  vector . The order in which we add vectors is irrelevant, because we will obtain the same result regardless of the order.
  ::当添加矢量时, 将一个矢量的尾巴置于另一个矢量的顶部。 这被称为尾对头规则。 通过连接第一尾和第二头的尾部而形成的矢量被称为矢量。 我们添加矢量的顺序无关紧要, 因为无论顺序如何, 我们都会获得同样的结果 。

  

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  To determine the sum of two vectors, add the corresponding components of the vectors. Suppose $\begin{array}{r}\stackrel{\to }{c}\end{array}$ is the sum of  $\begin{array}{r}\stackrel{\to }{a}\end{array}$  and  $\begin{array}{r}\stackrel{\to }{b}\end{array}$ :
  ::为确定两个矢量的总和,加上矢量的相应组成部分。

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  $$\begin{array}{rl}\stackrel{\to }{a}& =<5,12>\\ \stackrel{\to }{b}& =<3,8>\\ \stackrel{\to }{c}& =<(5+3),(12+8)>=<8,20>.\end{array}$$

  
  ::=================================================================================================================================

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  Vector subtraction reverses the direction of the 2nd vector:  $\begin{array}{r}\stackrel{\to }{a}-\stackrel{\to }{b}=\stackrel{\to }{a}+(-\stackrel{\to }{b})\end{array}$ .  Subtracting two vectors is equivalent to putting the head of the vectors together. 
  ::矢量减法颠倒第2矢量的方向:aba(-b)。减法中两个矢量等于将矢量头组合在一起。

  

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  To determine the difference of two vectors, reverse the direction of the vector we would like to subtract, and then add the two vectors. Suppose $\begin{array}{r}\stackrel{\to }{t}\end{array}$ is the difference of $\begin{array}{r}\stackrel{\to }{r}\end{array}$  and vector $\begin{array}{r}\stackrel{\to }{s}\end{array}$ :
  ::为了确定两个矢量的差数, 请将矢量的方向反转, 然后添加两个矢量。 假设 t... 是 r 和 矢量 s 的差数 :

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  $$\begin{array}{rl}A=(1,3),\stackrel{\to }{v}& =<4,8>,\stackrel{\to }{u}=<-1,-5>\\ A+\stackrel{\to }{v}+\stackrel{\to }{u}& =(4,6).\end{array}$$

       
  ::A=(1,3), v4,8>, u1, -5>A+vu[4,6]。

   

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  Adding vectors can be done in either order (just as with regular numbers). Subtracting vectors must be done in a specific order or else the vector will be negative (just as with regular numbers). In either case, use geometric reasoning and the Law of Cosines with the parallelogram that is formed to find the magnitude of the resultant vector. 
  ::添加矢量可以按任一顺序(和正常数量一样)进行。 减量矢量必须按特定顺序进行,否则矢量将是负的(和正常数量一样 ) 。 在这两种情况下,使用几何推理法和科辛斯定律,并使用为查找结果矢量而形成的平行图。

  

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  Scalar multiplication means to multiply a vector's components by a scalar number.
  ::计算乘法乘以星标数乘以矢量的元件。

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  $$\begin{array}{rl}\stackrel{\to }{a}& =<x,y>\\ k\cdot \stackrel{\to }{a}& =k\cdot <x,y>=<k\cdot x,k\cdot y>\end{array}$$

   
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊

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  Scalar multiplication often changes the magnitude of the vector, and can potentially result in a change of direction. The effect of the scalar on the magnitude of the vector depends on the absolute value of the scalar. An absolute value of the scalar that is greater than one will increase the magnitude of the vector, that is less than one will decrease the magnitude of the vector, and that equals one will have no effect on the magnitude of the vector. Multiplying a vector by a positive scalar will have no impact on direction, and multiplying a vector by a negative scalar will reverse the direction of the vector.
  ::计算倍增往往会改变矢量的大小,并可能导致方向的改变。标量对矢量的大小的影响取决于标量的绝对值。标量的绝对值如果大于一个,就会增加矢量的大小,即小于一个,就会降低矢量的大小,而等值则不会对矢量的大小产生影响。正标的乘以一个矢量不会对方向产生影响,而负标的乘以一个矢量的绝对值将会逆转矢量的方向。

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  Suppose  $\begin{array}{r}\stackrel{\to }{v}=<3,4>\end{array}$ . T hen $\begin{array}{r}2\stackrel{\to }{v}=<6,8>\end{array}$ .
  ::假设:v3,4>。然后是2v,6,8>。

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  The following video further  explains basic vector operations, such as vector addition, vector subtraction, and scalar multiplication: 
  ::以下视频进一步解释了病媒基本操作,如病媒添加、病媒减法和计算乘法:

    

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  Examples
  ::实例

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  Example 1
  ::例1

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  Given the following vectors, compute the sum:
  ::根据下列矢量计算总和:

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  $\begin{array}{r}\stackrel{\to }{w}=<1,3>,\stackrel{\to }{v}=<4,8>,\stackrel{\to }{u}=<-1,-5>\end{array}$
  ::1,3>,v4,8>,u1,-5>

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  $\begin{array}{r}\stackrel{\to }{w}+\stackrel{\to }{v}+\stackrel{\to }{u}=?\end{array}$
  ::怎么样?

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  Solution:
  ::解决方案 :

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  $$\begin{array}{rl}\stackrel{\to }{w}=<1,3>,\stackrel{\to }{v}& =<4,8>,\stackrel{\to }{u}=<-1,-5>\\ \stackrel{\to }{w}+\stackrel{\to }{v}+\stackrel{\to }{u}& =<4,6>\end{array}$$

  
  ::1,3>,v4,8>,u1,-5>

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  Example 2 
  ::例2

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  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. The angle between the vectors is $\begin{array}{r}{53}^{\circ }\end{array}$ . Determine the magnitude of the resultant vector,   $\begin{array}{r}|\stackrel{\to }{a}+\stackrel{\to }{b}|\end{array}$ .   
  ::两个矢量, a和 b, 的大小分别为 5 和 9 。 矢量之间的角为 53 。 确定由此产生的矢量的大小 , ab 。

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  Solution:
  ::解决方案 :

  

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  To find the magnitude of the resulting vector $\begin{array}{r}(x)\end{array}$ , note the triangle on the bottom that has sides 9 and 5, with included angle $\begin{array}{r}{127}^{\circ }\end{array}$ .  Since you know two sides of the triangle and its included angle, use  the Law of Cosines to calculate the 3rd side of the triangle.
  ::要找到结果矢量(x) 的大小, 请注意底部有侧面 9 和 5 的三角形, 包括角度 127 。 既然您知道三角形的两侧及其角度, 请使用 科辛斯 定律来计算三角形的第三侧 。

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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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  Example 3
  ::例3

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  Using the picture from Example 2, what is the angle that the sum $\begin{array}{r}\stackrel{\to }{a}+\stackrel{\to }{b}\end{array}$  makes with $\begin{array}{r}\stackrel{\to }{a}\end{array}$ ? 
  ::使用例2中的图片,a_b和a_之间的角是什么?

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  Solution:
  ::解决方案 :

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  Start by drawing a picture and labeling what you know: 

  $$\begin{array}{r}|\stackrel{\to }{a}|=5,|\stackrel{\to }{b}|=9,|\stackrel{\to }{a}+\stackrel{\to }{b}|\approx \mathrm{12.66.}\end{array}$$

  
  ::首先绘制一张图片并标注你所知道的: @aa_5, @b_9, @a_b_12.66。

  

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  Since you know three sides of the triangle and you need to find one angle, use the Law of Cosines.
  

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

  
  ::既然您知道三角形的三边, 您需要找到一个角度, 请使用 Cosines 法则 92= 12. 662+52-212. 665534. 634. {}\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Example 4
  ::例4

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  Return to the problem from the Introduction, in which a bird is flying due south at 10 mph, with a cross headwind of 2 mph heading northeast.
  ::回到引言中的问题,其中一只鸟在10mph向南飞行,向东北飞来,横风2mph。

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  Solution:
  ::解决方案 :

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  The force diagram looks like this:
  ::力图看起来是这样的:

  

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  Given the directions of the vectors are based on a compass, 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.
  ::鉴于矢量的方向是基于一个罗盘,鸟类矢量和风向之间的角是45,这意味着这是《科辛定律》的完美情况。让 x = 红矢量。

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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 mph. 
  ::这只鸟被稍稍吹离轨道,只飞行约8.7米。

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  Example 5
  ::例5

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  Elaine started a dog-walking business. She walks two dogs at a time. The dogs, Elvis and Ruby, each pull her in a different direction at a $\begin{array}{r}{45}^{\circ }\end{array}$  angle, with different forces. 
  ::Elaine开始做一条狗行走的生意,她一次走两条狗, 狗、猫王和Ruby, 都用不同的力量把她拉向不同方向,

  

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  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 with a constant force so she can stay in place? (Note: N stands for Newtons, which is the standard unit of force.)
  ::猫王在25N的军队中拉力,鲁比在49N的军队中拉力。 Elaine需要如何用恒定的武力拉力才能保持原位? (注:N代表牛顿,这是标准的力量单位。 )

  

  
  

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  Solution:
  ::解决方案 :

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  Even though the two vectors are centered at Elaine, the forces are added, which means 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, 但力量还是被添加了, 这意味着您需要使用尾巴对头规则来将矢量加在一起。 在每个矢量之间寻找角度需要逻辑使用补充角度 。

  

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  Evaluate the Law of Cosines with the given information.

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

  
  ::利用所提供的信息评估科辛斯定律.x2=492+252-24925cos135x68.98

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  In order for Elaine to stay in place, she will need to counteract this force with an equivalent force of her own in the exact opposite direction. 
  ::为了让Elaine留在原地,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 6
  ::例6

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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   $\begin{array}{r}3\stackrel{\to }{v}-2\stackrel{\to }{u}.\end{array}$
  ::考虑矢量 v2,5 > 和矢量 u1,9>。确定 3v2u的构成形式 。

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  Solution:
  ::解决方案 :

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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 7
  ::例7

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  An airplane is flying at a bearing of  270°  at 400 mph. A wind is blowing due south at 30 mph. Does this crosswind affect the plane's speed? 
  ::飞机在以400米方位以270度的方位飞行,风向以30米方位向南吹,横风影响飞机速度吗?

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  Solution:
  ::解决方案 :

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  Bearing is based on cardinal directions, or clockwise directions from due north, so a bearing of 270° means the plane is heading due west. Since the crosswind is pushing the plane due south as the plane tries to go directly west, the crosswind is perpendicular to the plane . As a result, the plane still has an airspeed of 400 mph, but the groundspeed (true speed) needs to be calculated.
  ::由于横风将飞机往南推,横风与飞机垂直。因此,飞机的飞行速度仍为400米,但需要计算地面速度(真正的速度)。

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

  
  ::4002+302=x2x401

  

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  Summary
  ::摘要

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  • A resultant vector is the vector that is produced when two or more vectors are summed or subtracted. It is also what is produced when a single vector is scaled by a constant. 
    ::由此产生的矢量是指在对两个或两个以上矢量进行总和或减去时产生的矢量,也是指单个矢量用常数缩放时产生的矢量。
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  • To determine the sum of two vectors , add the corresponding components of the vectors. 
    ::为确定两个矢量的总和,添加矢量的相应组成部分。
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  • To determine the difference of two vectors ,  reverse the direction of the vector that we would like to subtract, and then add the two vectors.
    ::为了确定两个矢量的差数, 将我们想要减去的矢量方向反转, 然后添加两个矢量 。
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  • Scalar multiplication  means to multiply a vector's components by a scalar number.
    ::计算乘法乘以星标数乘以矢量的元件。
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  • Calculations involving vectors can be solved using formulas from trigonometry such as the Law of Sines and the Law of Cosines.
    ::涉及矢量的计算可以通过三角测量公式来解决,例如《辛烷法》和《科辛斯法》。

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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 number 1.
     ::从 1 中查找由此产生的矢量的大小。
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  6. Find the magnitude of the resultant vector from number 2.
     ::从 2 中查找由此产生的矢量的大小。
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  7. Find the magnitude of the resultant vector from number 3.
     ::从第3号中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 number 4.
     ::从第4号中找出由此产生的矢量的大小。
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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 mph with a cross headwind blowing due east at 15 mph. How far does the bird get in one hour? 
      ::一只鸟从南面向南8英里处飞来 横风向东吹来 15米处
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  13. In what direction is the bird in the previous problem actually moving?
      ::上一个问题中的鸟到底在向什么方向移动?
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  14. A football is thrown at 50 mph due north. A wind is blowing due east at 8 mph. What is the actual speed of the football?
      ::向北50米处投掷足球,风向向东面8米处。 足球的实际速度是多少?
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  15. In what direction is the football in the previous problem actually moving?
      ::足球在前一个问题中 究竟向什么方向移动?

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  Review (Answers)
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。