11.2 矢量操作:点产品

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
• 选择节矢量操作: 点产品

  折叠 展开

  矢量操作: 点产品

  播放段落

  Will is moving into a new apartment on Friday. His friends have bailed on him and he’ll have to move a heavy chest of drawers by himself. He wants to do the least work possible to move the dresser into the house. Should he hoist it through the window, push it up a ramp through the front door, or push it up a hill and into the back door of the apartment?
  ::威尔在周五将搬进新公寓。 他的朋友已经保释了他,他不得不自己搬一个沉重的抽屉箱。 他想做尽可能少的工作才能把梳妆台搬进房子里。 他应该把梳妆台从窗户拉上,从前门推上坡道,还是把车推上山顶,然后推进公寓的后门?

  播放段落

  The Dot Product
  ::点产品

  播放段落

  The dot product is an operation that allows you to multiply two vectors and get a scalar as a result. It works for vectors in two, three, or even four or more dimensions. You can use the dot product to find out if two vectors are perpendicular to each other, to find the angle between two vectors, and to solve physics problems involving concepts like .
  ::点产品是一种操作, 使您能够乘以两个矢量, 并因此获得一个星标。 它适用于两个、 三个甚至四个或更多维的矢量。 您可以使用点产品来找出两个矢量是否相互垂直, 找到两个矢量之间的角, 并解决诸如 . 等概念的物理问题 。

  播放段落

  There are two equivalent methods of calculating the dot product of two vectors:
  ::计算两个矢量的点产物有两种等同的方法:

  播放段落
  1. You can describe the dot product of the vectors $\begin{array}{r}\stackrel{\to }{a}\end{array}$  and $\begin{array}{r}\stackrel{\to }{b}\end{array}$  as the product of the magnitudes of the vectors and the cosine of the angle between the vectors, that is:  $\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\Vert \stackrel{\to }{a}\Vert \Vert \stackrel{\to }{b}\Vert \cos \theta \end{array}$ , where  $\begin{array}{r}\Vert \stackrel{\to }{v}\Vert \end{array}$ is the magnitude of vector  $\begin{array}{r}\stackrel{\to }{v}\end{array}$ . This formula is easiest to use when your vectors are defined in terms of their magnitudes and angles.
     ::您可以将矢量 a 和 b 的点产物描述为矢量的大小和矢量之间角的余弦的产物, 即: aba b bcos , 其中v 是矢量 v 的大小。 当矢量按其大小和角度来定义时, 此公式最容易使用。
  播放段落
  2. For vectors defined in terms of their endpoints, use the formula $\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=x_a{x}_b+y_a{y}_b\end{array}$ , where $\begin{array}{r}(x_a,y_a)\end{array}$  is the endpoint of $\begin{array}{r}\stackrel{\to }{a}\end{array}$  and $\begin{array}{r}(x_b,y_b)\end{array}$  is the endpoint of $\begin{array}{r}\stackrel{\to }{b}\end{array}$ . If your vectors are in three dimensions, the formula just adds another term : $\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=x_a{x}_b+y_a{y}_b+z_a{z}_b\end{array}$ .
     ::对于根据其端点定义的矢量,请使用公式abxaxb+yayb,其中(xa,ya)为a的端点,(xb,yb)为b的端点。如果您的矢量分为三个维度,则该公式只是增加了另一个术语:abxaxb+yayb+zazb。

  播放段落

  You can use the dot product to determine whether two vectors are perpendicular. Two vectors are perpendicular if and only if their dot product is equal to zero. Since magnitude is always a positive number, if the dot product is zero, cosine $\begin{array}{r}\theta \end{array}$  must be zero, and the angle must be a 90 degree angle.
  ::您可以使用点值产品来确定两个矢量是否垂直。 两种矢量只有在它们的点值产品等于零时才具有垂直度。 由于星级总是正数, 如果点值产品为零, 余弦 一定为零, 角必须是 90 度角 。

  播放段落

  Are the vectors (3, 3) and (-5, 5) perpendicular? How about the vectors (-3, 5) and (-2, 1)?
  ::矢量(3,3)和(5,5)是否垂直?矢量(3,5)和(2,1)如何?

  播放段落

  To find out if two vectors are perpendicular, you must see if their dot product is 0. Since you’ve been given the vectors in endpoint form, you should use the second form of the dot product formula in order to compute their dot product.
  ::要发现两个矢量是否垂直, 您必须看到它们的点产品是否为 0。 既然您已经获得了端点形式的矢量, 您应该使用点产品公式的第二种形式来计算它们的点产品 。

  播放段落

  $\begin{array}{r}\stackrel{\to }{a}=(3,3)\\ \stackrel{\to }{b}=(-5,5)\end{array}$
  ::a*(3,3,3,b) (-5,5)

  播放段落

  $$\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=(3)(-5)+(3)(5)=-15+15=0\end{array}$$

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

  播放段落

  The dot product for the first set of vectors is 0, so the vectors are perpendicular.
  ::第一组矢量的点产品为 0, 所以矢量是垂直的 。

  播放段落

  For the second set of vectors,
  ::第二组向量

  播放段落

  $\begin{array}{r}\stackrel{\to }{c}=(-3,5)\\ \stackrel{\to }{d}=(-2,1)\end{array}$
  ::c(-35)d(-2,1)

  播放段落

  $$\begin{array}{r}\stackrel{\to }{c}\cdot \stackrel{\to }{d}=(-3)(-2)+(5)(1)=6+5=11\end{array}$$

  
  ::cd(-3)(-2)+(5)(1)=6+5=11

  播放段落

  Since the dot product of the vectors is not 0, they are not perpendicular.
  ::由于矢量的圆点产物不是0,因此它们不是垂直的。

  播放段落

  You can also use to find the angle between two vectors. Consider vectors $\begin{array}{r}\stackrel{\to }{c}\end{array}$  and $\begin{array}{r}\stackrel{\to }{d}\end{array}$  from the example above:
  ::您也可以从上面的示例中找到两个矢量之间的角。从上面的示例中考虑矢量 c 和 d :

  播放段落

  $\begin{array}{r}\stackrel{\to }{c}=(-3,5)\\ \stackrel{\to }{d}=(-2,1)\end{array}$
  ::c(-35)d(-2,1)

  播放段落

  You’ve already found that their dot product is 11. What is the angle between the vectors?
  ::您已经发现他们的点产品是11, 矢量之间的角度是什么?

  播放段落

  If you calculate the magnitude of each vector, you can use the equation  $\begin{array}{r}\stackrel{\to }{c}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{c}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \end{array}$ to find the value of $\begin{array}{r}\theta \end{array}$ .
  ::如果您计算了每个矢量的大小, 您可以使用方程式 cdc dc dc 的值。

  播放段落

  First, calculate the magnitude of each vector:
  ::首先,计算每个矢量的大小 :

  播放段落

  $$\begin{array}{r}\Vert \stackrel{\to }{c}\Vert =\sqrt{(-3{)}^2+5^2}=\sqrt{34}\\ \Vert \stackrel{\to }{d}\Vert =\sqrt{(-2{)}^2+1^2}=\sqrt{5}\end{array}$$

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

  播放段落

  Substituting these values into the dot product equation, you find that:
  ::用这些值来替代点产品方程式, 你会发现:

  播放段落

  $$\begin{array}{rl}11& =\sqrt{34}\sqrt{5}\cos \theta \\ \frac{11\sqrt{170}}{170}& =\cos \theta \\ \theta & ={32.47}^{\circ }\end{array}$$

  
  ::11=345cos11170170=cos32.47

  播放段落

  Work and the Dot Product
  ::工作和点产品

  播放段落

  In physics, work is defined as force times displacement. In order for work to occur, a force must move an object for some distance . Physicists measure work in Newton Meters, or Joules. One Joule is equivalent to one Newton of force used to move an object one meter.
  ::在物理学中,工作被定义为强制时间变换。为了进行工作,必须用一定距离移动物体。物理学家测量牛顿梅特斯(Newton Meters)或朱利斯(Joules)的工作。一个朱勒等于一个牛顿(Newton)用于移动1米的物体。

  播放段落

  You can express work as the dot product of the vector that describes the force acting on the object and of the vector describing the object’s motion. 
  ::您可以将作品表达为矢量的点产物,该矢量描述该物体上的动力,而矢量描述该物体的运动。

  播放段落

  Bob pulls a wagon up a hill. He exerts a force of 2 N at an angle of 45 degrees relative to the horizontal. The hill has a 15 degree grade and is 20 meters long. How much work does Bob do when he moves the wagon?
  ::鲍勃把一辆马车拉上山顶。他力力力2N,角度与水平相比为45度。山的等级为15度,长20米。当鲍勃移动马车时,他做了多少工作?

  

  播放段落

  To find the amount of work Bob does, you’ll need to find the dot product of the force vector and the movement vector. Use the equation $\begin{array}{r}\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \end{array}$ . You know the magnitudes for force and distance, but you need to find the angle between the vectors. If you look at the diagram, you’ll see that one vector is at a 45 degree angle and the other is at a 15 degree angle. This means that the angle between the vectors is 30 degrees. So:
  ::要找到 Bob 的工作量, 您需要找到强制矢量和移动矢量的点产物。 使用方程式 fdff dcos 。 您知道力和距离的大小, 但是您需要找到矢量之间的角。 如果您查看图表, 您可以看到一个矢量在45度角, 另一个在15度角。 这意味着矢量之间的角是 30 度。 因此 :

  播放段落

  $$\begin{array}{r}\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \\ \stackrel{\to }{f}\cdot \stackrel{\to }{d}=2\cdot 20\cdot \cos (30)=40\left(\frac{\sqrt{3}}{2}\right)=20\sqrt{3}\approx 34.64J\end{array}$$

  
  ::{\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}... {\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

  Bob does about 34.64 joules of work.
  ::鲍勃做约34.64焦耳的工作。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked how Will should move his furniture into his new apartment. He has three options to get the heavy chest of drawers into his new apartment, lifting the dresser through the window, using a ramp, or using the hill up to the backdoor of his apartment.
  ::早些时候,有人问你Will应该如何将家具搬进他的新公寓。 他有三个选择,即把大箱子的抽屉搬进他的新公寓,用斜坡把梳妆台从窗户上拉开,或用山顶到他公寓的后门。

  播放段落

  Option 1: Will can simply lift the dresser through the window and into his apartment. To do this, he’ll need to overcome the force of gravity and use a 1400 Newton force applied at a 90 degree angle to move the dresser 1.5 meters at a 90 degree angle.
  ::选择1:威尔可以简单地将梳妆台从窗户抬到公寓里。 为了做到这一点,他需要克服重力,并用90度角度的1400牛顿力将梳妆台移动到90度角度的1.5米。

  播放段落

  For this move,  $\begin{array}{r}\theta \end{array}$ is 0 because there is no difference in angle between the force vector and the movement vector. So:
  ::对于此移动, 为 0, 因为强制矢量与移动矢量的角没有区别 。 所以 :

  播放段落

  $$\begin{array}{r}\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \\ \stackrel{\to }{f}\cdot \stackrel{\to }{d}=1400\cdot 1.5\cdot \cos (0)=2100J\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=2100 J {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=2100

  播放段落

  If Will uses the first option, he will have to do 2100 Joules of work.
  ::如果威尔使用第一种选择,他必须完成2100焦耳的工作。

  播放段落

  Option 2: Will can use a 243 N force applied at a 30 degree angle to the horizontal and move the chest of drawers 5.7 meters up a 10 degree ramp into the front door of the apartment. This method requires less force than a straight lift, because the force to overcome gravity depends on the sine of the angle of the ramp being used to move an object.
  ::选项2: 威尔可以使用243 N 力,从30度角度向水平倾斜,将5.7米的抽屉胸部向10度斜坡向前门移动。这种方法需要的力小于直升,因为克服重力的力取决于用于移动物体的斜坡角的正弦。

  播放段落

  The angle between the force he’s applying and the surface of the ramp is $\begin{array}{r}{30}^{\circ }-{10}^{\circ }={20}^{\circ }\end{array}$ .
  ::他所施用的力量与坡道表面之间的角是 301020。

  播放段落

  So, the work is:
  ::因此,我们的工作是:

  播放段落

  $$\begin{array}{r}\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \\ \stackrel{\to }{f}\cdot \stackrel{\to }{d}=243\cdot 5.7\cdot \cos (20)=1302J\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=1302 J {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=1302 J {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=1302 J {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=1302 J

  播放段落

  He will do 1302 Joules of work if he moves the dresser up the ramp and into the front door.
  ::如果他把梳妆台搬上坡道 开到前门 他就会干1302 Joules的活儿

  播放段落

  Option 3: Will can exert 73 Newtons of force at a 30 degree angle to the horizontal and move the chest of drawers 19 meters up the slope of a hill with a 3 degree grade and into the back door.
  ::备选方案3:威尔可以从30度角度将73牛顿的力力推向水平,并将抽屉的胸部向山坡上19米处移动,高度为3度,然后进入后门。

  播放段落

  The angle between the force and the hill is 27 degrees, so:
  ::力与山之间的角是27度,所以:

  播放段落

  $$\begin{array}{r}\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \\ \stackrel{\to }{f}\cdot \stackrel{\to }{d}=73\cdot 19\cdot \cos (27)=1236J\end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(27)=1236 J {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}(27)=1236

  播放段落

  A ramp allows you to use less force to move an object to the same height as a straight lift. Will realizes that this is why movers use ramps to load and unload trucks.  If Will wants to do the least work possible, he should choose Option 3. He should go up the hill and into the back door.
  ::斜坡允许您使用较少的武力将物体移动到与直升相同的高度。 将会意识到这就是为什么搬运工使用斜坡来装载和卸载卡车的原因。 如果Will想做最少的工作, 他应该选择选择 3 选项 3 。 他应该上山到后门去 。

  播放段落

  Example 2
  ::例2

  播放段落

  Find the dot products for the following sets of vectors.
  ::为以下几组矢量寻找点产品。

  播放段落

  $\begin{array}{r}\stackrel{\to }{a}=(5,4),\stackrel{\to }{b}=(3,7)\end{array}$
  ::a(5,4,4),b(3,7)

  播放段落

  $\begin{array}{r}\Vert \stackrel{\to }{c}\Vert =5,{\theta }_c={30}^{\circ },\Vert \stackrel{\to }{d}\Vert =4,{\theta }_d={45}^{\circ }\end{array}$
  ::五,三,三,四,四,四,四,四

  播放段落

  $\begin{array}{r}\stackrel{\to }{e}=(12,2),\stackrel{\to }{f}=(5,1)\end{array}$
  ::e(12,2),f(5,1)

  播放段落

  $\begin{array}{r}\Vert \stackrel{\to }{g}\Vert =10,{\theta }_g={90}^{\circ },\Vert \stackrel{\to }{h}\Vert =10,{\theta }_h={30}^{\circ }\end{array}$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}10 10 30 = 30 \fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

  播放段落

  $$\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=x_a{x}_b+y_a{y}_b=(5)(3)+(4)(7)=43\\ \\ \stackrel{\to }{c}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{c}\Vert \Vert \stackrel{\to }{d}\Vert \cos ({\theta }_d-{\theta }_c)\\ \stackrel{\to }{c}\cdot \stackrel{\to }{d}=5\cdot 4\cdot \cos (15)=20\cos 15=19.3\\ \\ \stackrel{\to }{e}\cdot \stackrel{\to }{f}=x_e{x}_f+y_e{y}_f=(12)(5)+(2)(1)=62\\ \\ \stackrel{\to }{g}\cdot \stackrel{\to }{h}=\Vert \stackrel{\to }{g}\Vert \Vert \stackrel{\to }{h}\Vert \cos ({\theta }_g-{\theta }_h)\\ \stackrel{\to }{g}\cdot \stackrel{\to }{h}=10\cdot 10\cdot \cos (90-30)=100\cos 60=50\end{array}$$

  
  ::{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}}}}}}}}}}}}{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{

  播放段落

  Example 3
  ::例3

  播放段落

  Find the angle between the vectors  $\begin{array}{r}\stackrel{\to }{a}=(5,2)\end{array}$ and $\begin{array}{r}\stackrel{\to }{b}=(9,2)\end{array}$ .
  ::查找矢量a(5,2)和b(9,2)之间的角。

  播放段落

  $$\begin{array}{r}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=(5)(9)+(2)(2)=49\\ \Vert \stackrel{\to }{a}\Vert =\sqrt{5^2+2^2}=\sqrt{29}\\ \Vert \stackrel{\to }{b}\Vert =\sqrt{9^2+2^2}=\sqrt{85}\\ \\ \stackrel{\to }{a}\cdot \stackrel{\to }{b}=\Vert \stackrel{\to }{a}\Vert \Vert \stackrel{\to }{b}\Vert \cos \theta \\ 49=\sqrt{29}\sqrt{85}\cos \theta \\ \frac{49}{\sqrt{2465}}=\cos \theta \\ \theta ={9.3}^{\circ }\end{array}$$

  
  ::ab(5)(9)+(2)(2)=49aa52+22=29b92+22=85abaa bbcos49=2985cos492465=cos9.3

  播放段落

  Example 4
  ::例4

  播放段落

  If Bob applies a force of 3 N at an angle of 50 degrees to the horizontal and moves an object 35 m along a 5 degree slope, how much work does he do?
  ::如果Bob在50度角的角向水平上施以3N的力,并在5度斜坡上移动一个物体35米,那么他要工作多少?

  播放段落

  $$\begin{array}{r}\theta ={50}^{\circ }-5^{\circ }={45}^{\circ }\\ W=\stackrel{\to }{f}\cdot \stackrel{\to }{d}=\Vert \stackrel{\to }{f}\Vert \Vert \stackrel{\to }{d}\Vert \cos \theta \\ W=3(35)\left(\frac{\sqrt{2}}{2}\right)=\frac{105\sqrt{2}}{2}\approx 74.25J\end{array}$$

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

  播放段落

  Review
  ::回顾

  播放段落

  For #1-8, find the dot products for the following sets of vectors. Are the vectors perpendicular?
  ::对于 # 1-8, 找到以下矢量组的点产品。 矢量是否垂直?

  播放段落
  1.   $\begin{array}{r}\stackrel{\to }{a}=(4,7),\stackrel{\to }{b}=(-2,5)\end{array}$
     ::a(4,7,b(-2,5))
  播放段落
  2.   $\begin{array}{r}\Vert \stackrel{\to }{c}\Vert =5,{\theta }_c={120}^{\circ },\Vert \stackrel{\to }{d}\Vert =12,{\theta }_d={45}^{\circ }\end{array}$
     ::======================================================================================================================================================================================================================================= =====================================================================================================================================================================================================================================================================================
  播放段落
  3.   $\begin{array}{r}\stackrel{\to }{e}=(10,1),\stackrel{\to }{f}=(5,-50)\end{array}$
     ::e*(10,1),f(5,-50)
  播放段落
  4.   $\begin{array}{r}\stackrel{\to }{g}=(-4,2),\stackrel{\to }{h}=(6,8)\end{array}$
     ::g( - 4, 2, h( 6, 8) 。
  播放段落
  5.   $\begin{array}{r}\Vert \stackrel{\to }{i}\Vert =8,{\theta }_i={10}^{\circ },\Vert \stackrel{\to }{j}\Vert =10,{\theta }_j={100}^{\circ }\end{array}$
     ::8,i=10,j10,j=100
  播放段落
  6.   $\begin{array}{r}\stackrel{\to }{k}=(4,14),\stackrel{\to }{l}=(-2,-9)\end{array}$
     ::k(4,14,l(-2,-9))
  播放段落
  7.   $\begin{array}{r}\Vert \stackrel{\to }{m}\Vert =15,{\theta }_m={35}^{\circ },\Vert \stackrel{\to }{n}\Vert =2,{\theta }_n={55}^{\circ }\end{array}$
     ::========================================================================================================================================================================================================================================================
  播放段落
  8.   $\begin{array}{r}\Vert \stackrel{\to }{p}\Vert =9,{\theta }_p={50}^{\circ },\Vert \stackrel{\to }{q}\Vert =16,{\theta }_q={75}^{\circ }\end{array}$
     ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

  播放段落

  For #9-12, find the angle between each pair of vectors.
  ::对于 # 9-12, 找到每一对矢量之间的角 。

  播放段落
  9.   $\begin{array}{r}\stackrel{\to }{a}=(4,7),\stackrel{\to }{b}=(-2,5)\end{array}$
     ::a(4,7,b(-2,5))
  播放段落
  10.   $\begin{array}{r}\stackrel{\to }{e}=(10,1),\stackrel{\to }{f}=(5,-50)\end{array}$
      ::e*(10,1),f(5,-50)
  播放段落
  11.   $\begin{array}{r}\stackrel{\to }{g}=(-4,2),\stackrel{\to }{h}=(6,8)\end{array}$
      ::g( - 4, 2, h( 6, 8) 。
  播放段落
  12.   $\begin{array}{r}\stackrel{\to }{k}=(4,14),\stackrel{\to }{l}=(-2,-9)\end{array}$
      ::k(4,14,l(-2,-9))
  播放段落
  13.  If Julie applies a force of 5 N at an angle of 14 degrees to the horizontal and moves an object 12 m along a 13 upward degree slope, how much work does she do?
      ::如果Julie在14度角向水平倾角处施以5N的力,然后沿着13度向上斜坡移动一个物体12米,那么她做了多少工作?
  播放段落
  14.  If Michelle applies a force of 8 N at an angle of 6 degrees to the horizontal and moves an object 18 m along a 2 degree upward slope, how much work does she do?
      ::如果米歇尔在水平6度的角度上施用8N的力,然后沿着2度的向上斜坡上移物体18米,她能工作多少?
  播放段落
  15.  If John applies a force of 10 N at an angle of 10 degrees to the horizontal and moves an object 25 m along a flat surface, how much work does he do?
      ::如果约翰在10度角的角向水平上施以10N的力,并在平面上移动一个物体25米,那么他做了多少工作?

  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 。