Section: 点产品 | 9.5 点产品

Section outline

Introduction
::导言

A ramp allows you to use less force than a straight lift to move an object to the same height. Will realizes this is why movers use ramps to load and unload trucks. He has three options to get a heavy chest of drawers onto a moving truck :
::斜坡允许您使用比直升低的强度来将物体移动到同一高度。将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。他有三个选择, 可以把一个大箱子的抽屉放在移动的卡车上:

Option 1: Will can simply lift the dresser onto the moving truck . To do this, he'll need to overcome the force of gravity and use a 540 Newton (N) force applied at a 90 ° angle to move the dresser 1.2 meters at a 90 ° angle onto the bed of the truck.
::选项1:威尔可以简单地将梳妆机抬上移动的卡车。要做到这一点,他需要克服重力, 并用90度角度的540牛顿(N)力, 将梳妆机在90度角度的1.2米移动到卡车的床上。

Option 2: Will can place a 5.7 meter ramp to the truck at a 30 ° angle from the ground.
::备选办法2:Will可以从地面以30度角向卡车倾斜5.7米坡道。

Option 3: Will can also place the 5.7 meter from the base of his front door to the truck , creating a 17° angle to the horizontal.
::备选方案3:Will还可以将5.7米从前门的底部放置到卡车上,将角度从17度提高到水平。

If Will wants to move the dresser with the least amount of force possible , w hich is the best option for moving his heavy chest of drawers ?
::如果Will想用最少的力气移动梳妆台那这是移动他沉重的抽屉箱的最佳选择吗?

Dot Products
::点产品

While two vectors cannot be strictly multiplied like numbers can, there are two different ways to find the product between two vectors. The between two vectors results in a new vector perpendicular to the other two vectors. The 2nd type of product is the dot product between two vectors, which results in a scalar number. This number represents how much of one vector goes in the direction of the other . In one sense, it indicates how much the two vectors agree with each other.
::虽然两个矢量不能严格地像数字一样乘以两个矢量,但在两个矢量之间找到产品有两种不同的方式。两个矢量之间产生与另外两个矢量相对的一个新的矢量。第二种产品是两个矢量之间的圆点产物,结果得出一个星标数。这个数字表示一个矢量朝另一个矢量的方向移动了多少。从一种意义上说,它表明两个矢量之间有多少一致。

The dot product is defined as
::点产品的定义是:

$\begin{aligned} u \cdot v =< u_{1}, u_{2} > \cdot < v_{1}, v_{2} >= u_{1} v_{1} + u_{2} v_{2} . \end{aligned}$

::uvu1,u22v1,v2u1v1+u2v2。

This procedure states that you multiply the corresponding components and then sum the resulting products. It can work with vectors that are more than two dimensions in the same way, as long as the vectors involved have the same number of dimensions.
::此程序规定, 您将相应的组件乘以相应的组件, 然后对产生的产品进行总和。只要所涉及的矢量的尺寸相同, 它就可以以同样的方式对超过两个维度的矢量起作用。

Before trying this procedure with specific numbers, look at the following pairs of vectors and relative estimates of their dot product:
::在用具体数字尝试这一程序之前,请先看看以下几对矢量及其点产品相对估计值:

Notice how vectors going in generally the same direction have a positive dot product. Think of two forces acting on a single object. A positive dot product implies that these forces are working together at least a little bit. Another way of saying this is the angle between the vectors is less than $\begin{aligned} 90^{\circ} . \end{aligned}$
::注意通向同一方向的矢量通常有一个正点产物。想象一下两个力量在一个对象上行动。积极的点产物意味着这些力量至少合作了一点点。另一种说法是矢量之间的角小于 90\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

There are many important properties related to the dot product that you will prove in the examples and practice problems. The two most important are:
::与点产品相关的许多重要特性,将在实例和实践问题中加以证明。

1) What happens when a vector has a dot product with itself.
:1) 当矢量本身有点产品时会发生什么情况。

2) What is the dot product of two vectors that are perpendicular to each other.
:2) 两个相互垂直的矢量的点产物是什么?

$\begin{aligned} v \cdot v = | v |^{2} \end{aligned}$
::{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F2F2F2F2F2F}{\fn方正黑体简体\fs18\b1\bord1\shad1\3cH2F2F2F2F2F2

$\begin{aligned} v \end{aligned}$ and $\begin{aligned} u \end{aligned}$ are perpendicular if and only if $\begin{aligned} v \cdot u = 0. \end{aligned}$
::v 和 u 具有直视性,如果且仅在v u=0时才具有直视性。

The dot product can help you determine the angle between two vectors using the formula below. Notice that in the numerator, the dot product is required because each term is a vector. In the denominator, only regular multiplication is required because the magnitude of a vector is just a regular number indicating length.
::点产品可以帮助您使用下面的公式确定两个矢量之间的角。请注意, 在分子中, 需要点产品, 因为每个词都是矢量。在分母中, 只需要常规的乘法, 因为矢量的大小只是一个指示长度的普通数字。

$\begin{aligned} \cos θ = \frac{u \cdot v}{| u | | v |} \end{aligned}$

::{\fn华文楷体\fs16\1cHE0E0E0}

The following video discusses the meaning of the dot product, and provides several examples of how to determine the dot product of vectors in two dimensions:
::以下视频讨论点产品的含义,并举例说明如何从两个层面确定矢量的点产品:

Play, Learn, and Explore the Dot Product: .
::玩耍、学习和探索点产品:

Examples
::实例

Example 1
::例1

Show the commutative property holds for the dot product between two vectors. In other words, show that $\begin{aligned} u \cdot v = v \cdot u \end{aligned}$ .
::显示两个矢量之间的点产品对点产品的占位属性。换句话说, 显示 uv=vu 。

Solution:
::解决方案 :

This proof is for , although it holds for any- dimensional vectors. Start with the vectors in component form .
::此证明是用于 , 虽然它支持任何维矢量。以组件形式从矢量开始。

$\begin{aligned} u & =< u_{1}, u_{2} > \\ v & =< v_{1}, v_{2} > \end{aligned}$

::uu1,u2>vv1,v2>

A pply the definition of dot product and rearrange the terms. Then a pply the commutative property .
$\begin{aligned} u \cdot v & =< u_{1}, u_{2} > \cdot < v_{1}, v_{2} > \\ = u_{1} v_{1} + u_{2} v_{2} \\ = v_{1} u_{1} + v_{2} u_{2} \\ =< v_{1}, v_{2} > \cdot < u_{1}, u_{2} > \\ = v \cdot u \end{aligned}$

::应用点产品定义并重新排列术语。然后应用通俗属性 .uvu1,u221,v1,v2u1v1+u2v2=v1u1+v2u2+v2u2_v1,v221,u2>=vu

Example 2
::例2

Find the dot product between the following vectors: $\begin{aligned} < 3, 1 > \cdot < 5, - 4 > . \end{aligned}$
::在以下矢量之间找到点产品: < 3,15,-4>。

Solution:
$\begin{aligned} < 3, 1 > \cdot < 5, - 4 >= 3 \cdot 5 + 1 \cdot (- 4) \\ = 15 - 4 \\ = 11 \end{aligned}$

::解答: < 3,1,1,5,-4,3,5,5+1,1,(-4)=15,4=11

Example 3
::例3

Prove the angle between two vectors formula:
::证明两个矢量公式之间的角 :

$\begin{aligned} \cos θ = \frac{u \cdot v}{| u | | v |} . \end{aligned}$

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

Solution:
::解决方案 :

Use the Law of Cosines.
::使用科辛斯定律。

$\begin{aligned} | u - v |^{2} & = | v |^{2} + | u |^{2} - 2 | v | | u | \cos θ \\ (u - v) \cdot (u - v) & = \\ u \cdot u - 2 u \cdot v + v \cdot v & = \\ | u |^{2} - 2 u \cdot v + | v |^{2} & = \\ - 2 u \cdot v & = - 2 | v | | u | \cos θ \\ \frac{u \cdot v}{| u | | v |} & = \cos θ \end{aligned}$

::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ( ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Example 4
::例4

Return to the question from the Introduction: W hich is the best option for moving Will's heavy dresser if he wants to choose the option which will result in using the least amount of force ?
::回到引言中的问题:如果威尔想选择哪种选择会导致使用最少的武力,那么哪种最佳选择是移动威尔的重衣着?

Solution:
::解决方案 :

Option 1 : For option 1 Will already knows that it will require 450 N of force to lift the dresser onto the truck. To determine how much force option 2 and option 3 will require, we will need to know how much work is done when lifting the dresser onto the truck. For this move, $\begin{aligned} θ \end{aligned}$ is 0 because there is no difference in angle between the force vector and the movement vector.
::备选方案1:对于备选方案1, Will已经知道需要450牛顿的力量才能将装饰器抬上卡车。为了确定需要多少武力来决定选项2和选项3, 我们需要知道在将装饰器抬上卡车时做了多少工作。对于这一移动, 是0, 因为在引力矢量和向量之间没有角度差异。

$\begin{aligned} Work & = \vec{f} \cdot \vec{d} = ‖ \vec{f} ‖ ‖ \vec{d} ‖ \cos θ \\ = 450 \cdot 1.2 \cdot \cos (0) \\ = 540 J \end{aligned}$

::工作=540日工作=540日

If Will uses the 1st option, he will have to do 540 Joules of work. One of the characteristics of work is that finding different ways to perform the task will result in different di stances s and forces but the amount of work will remain the same. We can use this to find the forces required to move the dresser in option 2 and option 3.
::如果威尔使用第一种选择,他必须完成540个工作主体。工作的一个特征是,找到不同的方式来完成这项任务将导致不同的距离和力量,但工作量将保持不变。我们可以用这种方法找到在备选方案2和备选方案3中移动装饰器所需的力量。

Option 2 : The angle at which the dresser is being pushed is 30 °.
::备选办法2:将梳妆台推向的角为30°。

$\begin{aligned} Work & = ‖ \vec{f} ‖ ‖ \vec{d} ‖ \cos θ \\ 540 & = ‖ \vec{f} ‖ \cdot 5.7 \cdot \cos (30) \\ \frac{540}{5.7 \cdot \cos (30)} & = ‖ \vec{f} ‖ \cdot 5.7 \cdot \cos (30) \\ 109.39 N & = ‖ \vec{f} ‖ \end{aligned}$

::工作========================================================================================================================================工作===================================================================================================================================================================================================================================================================================================================================================================================

I f he moves the dresser up the ramp from the ground to the bed of the moving truck, it will take a force of 109.39 N.
::如果他把梳妆台从地面移上斜坡,移到卡车的床上,则需要109.39 N的兵力。

Option 3: The angle between the force base of the front door to the bed of the truck is 27 ° to the horizontal .
::备选办法3:卡车床前门的支架从27度到水平之间的角。

$\begin{aligned} Work & = ‖ \vec{f} ‖ ‖ \vec{d} ‖ \cos θ \\ 540 & = ‖ \vec{f} ‖ \cdot 5.7 \cdot \cos (17) \\ \frac{540 N}{5.7 \cdot \cos (17)} & = ‖ \vec{f} ‖ \cdot 5.7 \cdot \cos (17) \\ 99.07 N & = ‖ \vec{f} ‖ \end{aligned}$

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If Will wants to use as little force as possible, he should choose option 3. He should place the ramp onto the base of his front door .
::如果Will想尽量少使用武力,他应该选择选择3,他应该把斜坡放在前门的底部。

Example 5
::例5

Show the distributive property holds under the dot product .
::显示点产品下的分配属性。

$\begin{aligned} u \cdot (v + w) = u v + u w \end{aligned}$

::u(v+w) = uv+uw

Solution:
::解决方案 :

This proof will work with two-dimensional vectors, although the property does hold in general.
::这一证明对二维矢量有效,尽管该属性在总体上有效。

$\begin{aligned} u =< u_{1}, u_{2} >, v =< v_{1}, v_{2} >, w =< w_{1}, w_{2} > \end{aligned}$

::uu1,u2>,vv1,v2>,ww1,w2>

$\begin{aligned} u \cdot (v + w) & = u \cdot (< v_{1}, v_{2} > + < w_{1}, w_{2} >) \\ = u \cdot < v_{1} + w_{1}, v_{2} + w_{2} > \\ =< u_{1}, u_{2} > \cdot < v_{1} + w_{1}, v_{2} + w_{2} > \\ = u_{1} (v_{1} + w_{1}) + u_{2} (v_{2} + w_{2}) \\ = u_{1} v_{1} + u_{1} w_{1} + u_{2} v_{2} + u_{2} w_{2} \\ = u_{1} v_{1} + u_{2} v_{2} + u_{1} w_{1} + u_{2} w_{2} \\ = u \cdot v + v \cdot w \end{aligned}$

::u(v+w) =u( <v1,v2%w1,w2>) =uv1+w1,v2+w2_u1,u21,v2+w2+w1+w1+u2(v1+w1)+u2(v2+w2) =u1+u1+w1+u2+u2+w2+w2=u1+v1+u2+w2+w2=uv1+v2+w2+w2=uv2+v2+v2+wwww

Example 6
::例6

Find the dot product between the following vectors:
::在以下矢量之间查找点产品 :

$\begin{aligned} (4 i - 2 j) \cdot (3 i - 8 j) . \end{aligned}$

:4i-2j) (3i-8j).

Solution:
::解决方案 :

The standard unit vectors can be written as component vectors.
::标准单位矢量可以作为组成矢量书写。

$\begin{aligned} < 4, - 2 > \cdot < 3, - 8 >= 12 + (- 2) (- 8) = 12 + 16 = 28 \end{aligned}$

Example 7
::例7

What is the angle between $\begin{aligned} v =< 3, 5 > \end{aligned}$ and $\begin{aligned} u =< 2, 8 > \end{aligned}$ ?
::v3-5> 和 u2,8> 之间的角是什么?

Solution:
::解决方案 :

$\begin{aligned} \frac{u \cdot v}{| u | | v |} & = \cos θ \\ \frac{< 3, 5 > \cdot < 2, 8 >}{\sqrt{34} \cdot \sqrt{68}} & = \cos θ \\ \frac{6 + 40}{\sqrt{34} \cdot \sqrt{68}} & = \cos θ \\ \cos^{- 1} (\frac{46}{\sqrt{34} \cdot \sqrt{68}}) & = θ \\ 16.93 ° & \approx θ \end{aligned}$

::2,8>34 68= 6+4034 68= 1 (46346816.93

Summary
::摘要

The dot product is also known as the inner product and scalar product . It produces a scalar number that can be interpreted to tell how much one vector goes in the direction of the other.
::圆点产品也称为内产物和卡路里产品,它产生一个卡路里数,可以被解释为能说明一个矢量向另一个矢量的方向移动多少。

Dot product formula: $\begin{aligned} u \cdot v =< u_{1}, u_{2} > \cdot < v_{1}, v_{2} >= u_{1} v_{1} + u_{2} v_{2} \end{aligned}$
::多特产品公式: uu1, u2v1, v2u1+u2v2

Review
::回顾

Find the dot product for each of the following pairs of vectors:
::为下列每种矢量寻找点产品:

1. $\begin{aligned} < 2, 6 > \cdot < - 3, 5 > \end{aligned}$

2. $\begin{aligned} < 5, - 1 > \cdot < 4, 4 > \end{aligned}$

3. $\begin{aligned} < - 3, - 4 > \cdot < 2, 2 > \end{aligned}$

4. $\begin{aligned} < 3, 1 > \cdot < 6, 3 > \end{aligned}$

5. $\begin{aligned} < - 1, 4 > \cdot < 2, 9 > \end{aligned}$

Find the angle between each pair of vectors below.
::查找下方每对矢量之间的角。

6. $\begin{aligned} < 2, 6 > \cdot < - 3, 5 > \end{aligned}$

7. $\begin{aligned} < 5, - 1 > \cdot < 4, 4 > \end{aligned}$

8. $\begin{aligned} < - 3, - 4 > \cdot < 2, 2 > \end{aligned}$

9. $\begin{aligned} < 3, 1 > \cdot < 6, 3 > \end{aligned}$

10. $\begin{aligned} < - 1, 4 > \cdot < 2, 9 > \end{aligned}$

11. What is $\begin{aligned} v \cdot v \end{aligned}$ ?
::11. 什么是Vv?

12. How can you use the dot product to find the magnitude of a vector?
::12. 您如何使用点产品找到矢量的大小?

13. What is $\begin{aligned} 0 \cdot v \end{aligned}$ ?
::13. 什么是0v?

14. Show that $\begin{aligned} (c u) \cdot v = u \cdot (c v), \end{aligned}$ where $\begin{aligned} c \end{aligned}$ is a constant.
::14. 显示 c 是常数的 (cu)v=u(cv) 。

15. Show that $\begin{aligned} < 2, 3 > \end{aligned}$ is perpendicular to $\begin{aligned} < 1.5, - 1 > \end{aligned}$ .
::15. 显示 <2,3>与 <1.5,-1> 垂直。

16. If Julie applies a force of 5 N at an angle of 30 ° to the horizontal, and moves an object 12 m along a 3° slope, how much work does she do?
::16. 如果Julie在30°角向水平倾角处施以5牛顿的力,并在3°斜坡上移动物体12米,她要工作多少?

17. If Michelle applies a force of 8 N at an angle of 20 ° to the horizontal, and moves an object 18 m along a 12 ° slope, how much work does she do?
::17. 如果米歇尔在20°角向水平倾角处施以8N的力,并在12°斜坡上移动物体18米,她能工作多少?

18. If John applies a force of 10 N at an angle of 10 ° to the horizontal, and moves an object 25 m along a flat surface, how much work does he do?
::18. 如果约翰在10°角向水平倾角处施以10牛顿的力,并在平面上移动物体25米,他要做多少工作?

Review (Answers )
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

Introduction ::导言

Dot Products ::点产品

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers ) ::回顾(答复)

Introduction
::导言

Dot Products
::点产品

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers )
::回顾(答复)