11.1 平面中的矢量

章节大纲

• 选择节常规

  折叠 展开

  常规

  全部折叠 展开全部

  • 选择活动新闻通告

    

    新闻通告 讨论区
• 选择节平面中的矢量

  折叠 展开

  平面中的矢量

  播放段落

  Meghan has planned a Saturday zip line adventure for her family. Her grandmother is a little nervous about one part of the course. As the riders plummet toward the surface of a lake, they reach speeds of 40 miles an hour. Meghan argues that the speed isn’t as scary as it sounds. Can she break it into its horizontal and vertical components so that her grandmother can enjoy the course with the rest of the family?
  ::Meghan计划为她的家人做一个周六的拉链冒险。她祖母对其中一部分感到有点紧张。 当骑手们向湖面倾斜时,他们的速度达到每小时40英里的速度。 Meghan说,速度没有听起来那么可怕。 她能打破速度的横向和纵向部分,以便她奶奶能够和家人一起享受这个过程吗?

  播放段落

  Introduction to Vectors
  ::矢量介绍

  播放段落

  Some numbers simply describe how large something is. These numbers, which express magnitude , are called scalars. Speed limits are examples of scalars. They tell you how fast you can drive. Air temperature is a scalar. It tells you how warm the air is. Your height is a scalar. It describes how tall you are.
  ::一些数字简单描述某物的大小。 这些数字, 表示音量, 被称为星标。 速度限制是星标的示例。 它们告诉您可以开多快。 空气温度是一个星标。 它能说明空气的温度。 您的高度是星标。 它能说明您有多高 。

  播放段落

  Sometimes, you need more information. When you need a number that can express both magnitude and direction, you can use a vector. Velocity is an example of an idea best expressed by a vector. While speed simply tells you how fast you’re going, velocity tells you your speed and your direction. You can express vectors in one, two, three, or even more dimensions.
  ::有时,您需要更多信息。当您需要能够表达大小和方向的数字时,您可以使用矢量。速度是矢量表达最佳想法的一个实例。虽然速度简单告诉你你行进的速度,速度可以告诉你速度和方向。您可以在一、二、三或更多维中表达矢量。

  播放段落

  When you work with a vector on the coordinate plane, you can express it in terms of its location with respect to the origin. For example,  $\begin{array}{r}\stackrel{\to }{a}=(5,8)\end{array}$ $\begin{array}{r}\stackrel{\to }{a}\end{array}$ begins at the origin and ends at point $\begin{array}{r}(5,8)\end{array}$ . You can express the vector’s direction by finding its angle with respect to the $\begin{array}{r}x\end{array}$ -axis, and you can find its magnitude by using the distance formula for points on the coordinate plane.
  ::当您在坐标平面上与矢量一起工作时,您可以用其来源位置表示它。例如,a(5,8)a(以原为起点)和终点(5,8)为终点。您可以通过在X轴上找到矢量的角度来表达矢量的方向,并通过使用坐标平面上的点的距离公式来发现其大小。

  播放段落

  Let's find the direction and magnitude of $\begin{array}{r}\stackrel{\to }{a}=(5,8)\end{array}$ .
  ::让我们找出A(5,8)的方向和范围。

  

  播放段落

  To find the direction, you’ll need to find $\begin{array}{r}\theta \end{array}$ , the angle that the vector makes with the $\begin{array}{r}x\end{array}$ -axis. You can use tangent to do this:
  ::要找到方向, 您需要找到 °, 即矢量与 x 轴的角。 您可以使用切线来做到这一点 :

  播放段落

  $$\begin{array}{r}\tan \theta =\frac{8}{5}\\ \theta ={\tan }^{-1}\left(\frac{8}{5}\right)\\ \theta \approx {58}^{\circ }\end{array}$$

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

  播放段落

  The vector’s direction is $\begin{array}{r}{58}^{\circ }\end{array}$
  ::矢量的方向是58

  播放段落

  $$\begin{array}{r}d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\\ d=\sqrt{25+64}=\sqrt{89}\end{array}$$

  
  ::d= (x2-x1) 2+ (y2-y1) 2d=25+64=89

  播放段落

  The magnitude of the vector is $\begin{array}{r}\sqrt{89}\end{array}$ .
  ::矢量的大小是89

  播放段落

  Unit Vectors
  ::单位矢量

  播放段落

  You can also describe vectors in terms of unit vectors and scalars. A unit vector is a vector with a magnitude of 1. Any vector can be described as a multiple of a unit vector. For instance, the vector $\begin{array}{r}\stackrel{\to }{a}=(5,8)\end{array}$  from above be described as  $\begin{array}{r}\sqrt{89}\cdot \stackrel{\to }{v}\end{array}$ where  $\begin{array}{r}\stackrel{\to }{v}\end{array}$ is a vector with a magnitude of 1 and an angle of $\begin{array}{r}{58}^{\circ }\end{array}$ .
  ::您也可以用单位矢量和弧度来描述矢量。单位矢量是一个矢量,大小为1。任何矢量都可以描述为单位矢量的倍数。例如,上面的矢量a(5,8)被描述为 89v,其中 v是1 的矢量,角为 58。

  播放段落

  Two important unit vectors are  $\begin{array}{r}\stackrel{\to }{i}=(1,0)\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}=(0,1)\end{array}$ . They are the unit vectors that describe vectors at the angles  $\begin{array}{r}{0}^{\circ }\end{array}$ and $\begin{array}{r}{90}^{\circ }\end{array}$ .  You can describe any vector as the sum of multiples of these vectors. For instance, the vector from above,  $\begin{array}{r}\stackrel{\to }{a}=(5,8)\end{array}$ , could also be written as $\begin{array}{r}\stackrel{\to }{a}=5\stackrel{\to }{i}+8\stackrel{\to }{j}\end{array}$ . When you write vectors as a sum of multiples of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ , it becomes easy to add them and to multiply them by scalars.  To add two vectors, first add the horizontal components together (the multiples of $\begin{array}{r}\stackrel{\to }{i}\end{array}$ ). Then add the vertical components (the multiples of $\begin{array}{r}\stackrel{\to }{j}\end{array}$ ),  and express the new vector as a sum of multiples of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
  ::两个重要的单位矢量是 i( 1, 0) 和 j( 0. 1) 。 它们是描述角度 0 和 90 的矢量的单位矢量。 您可以将任何矢量描述为这些矢量的倍数之和。 例如, 上面的矢量a( 5, 8) 也可以写成 a5i8j。 当您将矢量写成 i 和 j 的倍数之和时, 添加它们并乘以 scalars 很容易。 要添加两个矢量, 首先将水平组件加在一起( i的倍数) 。 然后添加垂直组件( j的倍数) , 并将新的矢量表示为 i 和 j 的倍数之和 。

  播放段落

  Now, let's find the magnitude and direction of vector  $\begin{array}{r}\stackrel{\to }{b}=(3,2)\end{array}$ and vector $\begin{array}{r}\stackrel{\to }{c}=(1,4)\end{array}$  and then rewrite both vectors in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ . 
  ::现在,让我们找到矢量 b( 3, 2) 和矢量 c( 1, 4) 的大小和方向, 然后重写 i 和 j 的矢量和方向 。

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{b}& =(3,2)\\ \tan \theta & =\frac{2}{3}\\ \theta & ={34}^{\circ }\\ d& =\sqrt{a^2+b^2}\\ d& =\sqrt{3^2+2^2}=\sqrt{13}\end{array}$$

  
  ::==a2+b2d=32+22=13

  播放段落

  $\begin{array}{r}\stackrel{\to }{b}\end{array}$ has a direction of  $\begin{array}{r}{34}^{\circ }\end{array}$ and a magnitude of $\begin{array}{r}\sqrt{13}\end{array}$ .
  ::b 方向为34,大小为13。

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{c}& =(1,4)\\ \tan \theta & =\frac{4}{1}\\ \theta & ={76}^{\circ }\\ d& =\sqrt{a^2+b^2}\\ d& =\sqrt{1^2+4^2}=\sqrt{17}\end{array}$$

  
  ::c(1,4)tan4176d=a2+b2d=12+42=17

  播放段落

  $\begin{array}{r}\stackrel{\to }{c}\end{array}$ has a direction of  $\begin{array}{r}{76}^{\circ }\end{array}$ and a magnitude of $\begin{array}{r}\sqrt{17}\end{array}$ .
  ::c 方向为 76 和 17 , 范围为 76 和 17 。

  播放段落

  To rewrite in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ , use the  $\begin{array}{r}x\end{array}$ and  $\begin{array}{r}y\end{array}$ coordinates of each vector’s endpoint. Then you’ll see that  $\begin{array}{r}\stackrel{\to }{b}=3\stackrel{\to }{i}+2\stackrel{\to }{j}\end{array}$ and $\begin{array}{r}\stackrel{\to }{c}=\stackrel{\to }{i}+4\stackrel{\to }{j}\end{array}$ .  
  ::重写 i 和 j , 请使用每个矢量端点的 x 和 y 坐标 。 然后您可以看到 b 3 i 2j 和 c i 4j 。

  播放段落

  As one last step, let's find the sum of the two vectors and find the direction and magnitude of the resulting vector. When you add vectors, you add the horizontal components together and the vertical components together. So:
  ::作为最后一步,让我们找到两个矢量的总和, 并找到结果矢量的方向和大小。 当您添加矢量时, 将水平组件和垂直组件合并在一起。 所以 :

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{d}& =\stackrel{\to }{b}+\stackrel{\to }{c}\\ \stackrel{\to }{d}& =3\stackrel{\to }{i}+2\stackrel{\to }{j}+\stackrel{\to }{i}+4\stackrel{\to }{j}\\ \stackrel{\to }{d}& =4\stackrel{\to }{i}+6\stackrel{\to }{j}\end{array}$$

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

  

  播放段落

  Now, find the direction and magnitude of the new vector.
  ::现在,找到新矢量的方向和范围。

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{d}& =4\stackrel{\to }{i}+6\stackrel{\to }{j}\\ \tan \theta & =\frac{6}{4}\\ \theta & ={56}^{\circ }\\ d& =\sqrt{a^2+b^2}\\ d& =\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}\end{array}$$

  
  ::46j46456d=a2+b2d=42+62=52=213

  播放段落

  $\begin{array}{r}\stackrel{\to }{d}\end{array}$  has a direction of  $\begin{array}{r}{56}^{\circ }\end{array}$ and a magnitude of $\begin{array}{r}2\sqrt{13}\end{array}$ .
  ::d 具有56 和213的高度方向。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked about a zip line and breaking the speed down into components. At the point where the zip line hits 40 miles per hour, the angle between the zip line’s path and the water below is $\begin{array}{r}{25}^{\circ }\end{array}$ .
  ::早些时候,有人问过你一条拉链线,然后将速度折成部件。 当拉链线达到每小时40英里时,拉链线路径与下方水之间的角是25英里。

  

  播放段落

  Meghan plans to put the vector in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and  $\begin{array}{r}\stackrel{\to }{j}\end{array}$ so that her grandmother can see that the vertical and horizontal speeds aren’t nearly as scary as the combined speed. 
  ::Meghan计划将矢量放在i和j上, 这样她的祖母就能看到垂直和水平速度不像综合速度那样可怕。

  播放段落

  Keep in mind that the actual velocity of the travelers will be -40, because they’ll be moving in a downward direction. Meghan will find the vertical velocity first, and then the horizontal velocity.
  ::记住旅行者的实际速度将是 - 40 , 因为他们会向下移动。 Meghan将首先找到垂直速度,然后找到水平速度。

  播放段落

  $$\begin{array}{rl}\sin 25& =\frac{y}{40}\\ 40\sin (25)& =y\\ y& =16.9\\ \cos 25& =\frac{x}{40}\\ 40\cos 25& =x\\ x& =36.3\\ \stackrel{\to }{v}& =36.3\stackrel{\to }{i}+16.9\stackrel{\to }{j}\end{array}$$

  
  :25)=yy=16.9cos25=x4040cos25=xxx=36.3v36.3i16.9j

  播放段落

  After looking at the vectors, Meghan decides to tell her grandmother that she’ll only be falling toward the lake at about 17 mph - a little faster than a leisurely bike ride!
  ::Meghan在观察矢量后决定告诉她的祖母,

  播放段落

  Example 2
  ::例2

  播放段落

  If you’re given a vector’s magnitude and direction, you can break the vector into its component parts and express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and  $\begin{array}{r}\stackrel{\to }{j}\end{array}$ . For instance, consider the vector $\begin{array}{r}\stackrel{\to }{e}\end{array}$ , which has a magnitude of 8 units and makes an angle of  $\begin{array}{r}{30}^{\circ }\end{array}$ with the $\begin{array}{r}x\end{array}$ -axis. Find the component parts of $\begin{array}{r}\stackrel{\to }{e}\end{array}$ .
  ::如果给定矢量的大小和方向,您可以将矢量分解成其组成部分,并以 i 和 j 表示。例如,考虑矢量 e ,该矢量的大小为 8 个单位,并用 x 轴的角为 30 。找到 e 的构成部分。

  播放段落

  This is what  $\begin{array}{r}\stackrel{\to }{e}\end{array}$ looks like:
  ::这就是我们看起来的样子:

  

  播放段落

  You can use trigonometry to find the component parts of $\begin{array}{r}\stackrel{\to }{e}\end{array}$ :
  ::您可以使用三角测量来找到 e__ 的组件 :

  播放段落

  $$\begin{array}{rl}\sin (30)& =\frac{y}{8}\\ 8\sin (30)& =y\\ y& =4\\ \cos (30)& =\frac{x}{8}\\ 8\cos (30)& =x\\ x& =4\sqrt{3}\end{array}$$

  
  :30)=y88sin(30)=yy=4cos*(30)=x88cos*(30)=xxx=43

  播放段落

  So, $\begin{array}{r}\stackrel{\to }{e}=4\sqrt{3}\stackrel{\to }{i}+4\stackrel{\to }{j}\end{array}$ .
  ::那么,我们43i4j。

  播放段落

  Example 3
  ::例3

  播放段落

  Find the magnitude and direction of the vector that ends at (4, 7).
  ::查找矢量的大小和方向,以(4, 7)为终点。

  播放段落

  Use tangent to find the direction of the vector and the distance formula to find the magnitude of the vector:
  ::使用切线查找矢量的方向和距离公式查找矢量的大小:

  播放段落

  $$\begin{array}{rl}\tan \theta & =\frac{7}{4}\\ \theta & ={60.3}^{\circ }\\ d& =\sqrt{x^2+y^2}=\sqrt{4^2+7^2}=\sqrt{65}\end{array}$$

  
  ::=x2+y2=42+72=65

  播放段落

  The magnitude of the vector is  $\begin{array}{r}\sqrt{65}\end{array}$ and the direction of the vector is $\begin{array}{r}{60.3}^{\circ }\end{array}$ .
  ::向量的大小为65,向量的方向为60.3。

  播放段落

  Example 4
  ::例4

  播放段落

  Put the following 3 vectors in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ . Add them together, then find the magnitude and angle of direction for the resulting vector.
  ::将以下 3 个矢量以 i 和 j 和 i 和 j 相加。 将它们加在一起, 然后找到由此生成的矢量方向的大小和角度 。

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{a}& =(-2,3)\\ \stackrel{\to }{b}& =(5,-1)\\ \stackrel{\to }{c}& =(4,4)\end{array}$$

  
  ::a(-2,3)b(5,-1)c(4,4)

  播放段落

  First write each vector in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ :
  ::以 i_ 和 j_ 表示的 i_ 和 j_ 首写每个矢量 :

  播放段落

  $$\begin{array}{r}\stackrel{\to }{a}=-2\stackrel{\to }{i}+3\stackrel{\to }{j}\\ \stackrel{\to }{b}=5\stackrel{\to }{i}+-\stackrel{\to }{j}\\ \stackrel{\to }{c}=4\stackrel{\to }{i}+4\stackrel{\to }{j}\end{array}$$

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

  播放段落

  Add the vectors to find the resulting vector:
  ::添加矢量以查找由此生成的矢量 :

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{d}& =-2\stackrel{\to }{i}+3\stackrel{\to }{j}+5\stackrel{\to }{i}+-\stackrel{\to }{j}+4\stackrel{\to }{i}+4\stackrel{\to }{j}\\ \stackrel{\to }{d}& =5\stackrel{\to }{i}+6\stackrel{\to }{j}\end{array}$$

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

  播放段落

  Find the magnitude and direction of the resulting vector:
  ::查找由此生成的矢量的大小和方向 :

  播放段落

  $$\begin{array}{rl}\tan \theta & =\frac{6}{5}\\ \theta & ={50.2}^{\circ }\\ d& =\sqrt{x^2+y^2}=6^2+5^2=\sqrt{61}\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=x2+y2=62+52=61 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=62+52=61

  播放段落

  The magnitude of the resulting vector is  $\begin{array}{r}\sqrt{61}\end{array}$ and the direction of the resulting vector is $\begin{array}{r}{50.2}^{\circ }\end{array}$ .
  ::由此形成的矢量规模为61,而由此形成的矢量方向为50.2。

  播放段落

  Example 5
  ::例5

  播放段落

  A vector has a magnitude of 12 and an angle of $\begin{array}{r}{40}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
  ::矢量的大小为 12 , 角度为 40 。 用 i 和 j 表示 。

  播放段落

  Use trigonometry to find the horizontal and vertical components of the vector:
  ::使用三角测量法查找矢量的水平和垂直成分:

  播放段落

  $$\begin{array}{rl}\sin (40)& =\frac{y}{12}\\ 12\sin 40& =y\\ y& =7.7\\ \cos (40)& =\frac{x}{12}\\ 12\cos (40)& =x\\ x& =9.2\end{array}$$

  
  :40)=y1212sin40=yy=7.7cos(40)=x1212cos(40)=xxx=9.2

  播放段落

  The resulting vector is $\begin{array}{r}\stackrel{\to }{v}=9.2\stackrel{\to }{i}+7.7\stackrel{\to }{j}\end{array}$ .
  ::产生的矢量为 v9.2i7.7j。

  播放段落

  Review
  ::回顾

  播放段落

  For #1-5, find the magnitude and direction of the vector that ends at each of the point.
  ::对于 # 1 5, 找到矢量的大小和方向, 以每个点结束 。

  1.   $\begin{array}{r}(2,5)\end{array}$
  2.   $\begin{array}{r}(-2,-3)\end{array}$
  3.   $\begin{array}{r}(6,4)\end{array}$
  4.   $\begin{array}{r}(7,-5)\end{array}$
  5.   $\begin{array}{r}(-1,4)\end{array}$
  播放段落
  6. Put the following 5 vectors in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ $\begin{array}{r}\stackrel{\to }{j}\end{array}$
     ::将以下5个矢量( ij) 设置为 5 个矢量( ij) 。

  播放段落

  $$\begin{array}{rl}\stackrel{\to }{a}& =(1,2)\\ \stackrel{\to }{b}& =(-4,8)\\ \stackrel{\to }{c}& =(-3,-6)\\ \stackrel{\to }{d}& =(2,-4)\\ \stackrel{\to }{e}& =(1,9)\end{array}$$

  
  ::a(1,2)b(-4,8)c(-3,-6)d(2,-4)e(1,9)

  播放段落
  7. Find the magnitude and direction of the resulting vector of $\begin{array}{r}\stackrel{\to }{a}+\stackrel{\to }{e}\end{array}$ .
     ::查找 ae 所生成矢量的大小和方向 。
  播放段落
  8. Find the magnitude and direction of the resulting vector of $\begin{array}{r}\stackrel{\to }{c}+\stackrel{\to }{d}\end{array}$ .
     ::查找 Cd 矢量的大小和方向 。
  播放段落
  9. Find the magnitude and direction of the resulting vector of $\begin{array}{r}\stackrel{\to }{b}+\stackrel{\to }{d}\end{array}$ .
     ::查找 bd 矢量的大小和方向 。
  播放段落
  10. Find the magnitude and direction of the resulting vector of $\begin{array}{r}\stackrel{\to }{a}+\stackrel{\to }{d}+\stackrel{\to }{e}\end{array}$ .
      ::查找 ade 所生成矢量的大小和方向 。
  播放段落
  11. A vector has a magnitude of 10 and an angle of $\begin{array}{r}{140}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
      ::矢量的大小为 10 , 角度为 140 。 用 i 和 j 表示 。
  播放段落
  12. A vector has a magnitude of 8 and an angle of $\begin{array}{r}{80}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
      ::矢量的大小为 8,角度为 80 。 用 i 和 j 表示 。
  播放段落
  13. A vector has a magnitude of 2 and an angle of $\begin{array}{r}{10}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
      ::矢量的大小为 2 , 角度为 10 。 用 i 和 j 表示 。
  播放段落
  14. A vector has a magnitude of 6 and an angle of $\begin{array}{r}{210}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
      ::矢量的大小为 6 , 角度为 210 。 用 i 和 j 表示 。
  播放段落
  15. A vector has a magnitude of 4 and an angle of $\begin{array}{r}{300}^{\circ }\end{array}$ . Express it in terms of  $\begin{array}{r}\stackrel{\to }{i}\end{array}$ and $\begin{array}{r}\stackrel{\to }{j}\end{array}$ .
      ::矢量的大小为 4 , 角度为 300 。 用 i 和 j 表示 。

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。