Course: 7.3 矢量成构件的分辨率

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矢量成构件的分辨率

Sometimes working with horizontal and vertical components of a vector can be significantly easier than working with just an angle and a magnitude . This is especially true when combining several forces together.

Consider four siblings fighting over a candy in a four way tug of war. Lanie pulls with 8 lb of force at an angle of $\begin{aligned} 41^{\circ} \end{aligned}$ . Connie pulls with 10 lb of force at an angle of $\begin{aligned} 100^{\circ} \end{aligned}$ . Cynthia pulls with 12 lb of force at an angle of $\begin{aligned} 200^{\circ} \end{aligned}$ . How much force and in what direction does poor little Terry have to pull the candy so it doesn’t move?
::想想四个兄弟姐妹以四种方式为糖果而战斗。兰尼以8磅的武力在41磅的角拉动。康妮以10磅的武力在100磅的角拉动。辛西娅以12磅的武力在200磅的角拉动。可怜的小泰瑞需要多少武力和方向拉动糖果才能不动?

The Unit Vector and Component Form
::单位矢量和组件表

A unit vector is a vector of length one. Sometimes you might wish to scale a vector you already have so that it has a length of one. If the length was five, you would scale the vector by a factor of $\begin{aligned} \frac{1}{5} \end{aligned}$ so that the resulting vector has magnitude of 1. Another way of saying this is that a unit vector in the direction of vector $\begin{aligned} \vec{v} \end{aligned}$ is $\begin{aligned} \frac{\vec{v}}{| \vec{v} |} \end{aligned}$ .
::单位矢量是长度为一的矢量。有时,您可能想要缩放您已有的矢量,使其有一个长度。如果长度为5,那么您会将矢量缩放15倍,从而使由此产生的矢量的大小为1。另一种说法是,朝向矢量 v__ 方向的单位矢量是 vv} 。

There are two standard unit vectors that make up all other vectors in the coordinate plane. They are $\begin{aligned} \vec{i} \end{aligned}$ which is the vector $\begin{aligned} < 1, 0 > \end{aligned}$ and $\begin{aligned} \vec{j} \end{aligned}$ which is the vector $\begin{aligned} < 0, 1 > \end{aligned}$ . These two unit vectors are perpendicular to each other. A linear combination of $\begin{aligned} \vec{i} \end{aligned}$ and $\begin{aligned} \vec{j} \end{aligned}$ will allow you to uniquely describe any other vector in the coordinate plane in component form . For instance the vector $\begin{aligned} < 5, 3 > \end{aligned}$ is the same as $\begin{aligned} 5 \vec{i} + 3 \vec{j} \end{aligned}$ .
::有两种标准单位矢量构成坐标平面上所有其他矢量。它们是 i , 即矢量 < 1,0 > 和 j , 即矢量 < 0,1 > 。这两个单位矢量是垂直的。 i 和 j 的线性组合将允许您以组件形式独有地描述坐标平面上的任何其他矢量。例如, 矢量 < 5,3 > 与 5i3j相同。

Often vectors are initially described as an angle and a magnitude rather than in component form. Working with vectors written as an angle and magnitude requires extremely precise geometric reasoning and excellent pictures. One advantage of rewriting the vectors in component form is that much of this work is simplified. Remember that component form is the form $\begin{aligned} < x, y > \end{aligned}$ and t o translate from magnitude $\begin{aligned} r \end{aligned}$ and direction $\begin{aligned} θ \end{aligned}$ to component form , use the relationship $\begin{aligned} < r \cdot \cos θ, r \cdot \sin θ >=< x, y > \end{aligned}$ . In this situation $\begin{aligned} r \end{aligned}$ is the magnitude and $\begin{aligned} θ \end{aligned}$ is the direction.
::矢量通常最初被描述为一个角度和范围,而不是以组件形式。将矢量作为角度和规模进行工作需要非常精确的几何推理和极佳的图片。将矢量以组件形式重新写入的一个优点是,大部分这项工作都是简化的。请记住,组成部分的形式是 <x,y > ,并将之从数量表和方向表 * 转换为组件形式,使用关系<rcos,rsinx,y>。在这种情况下, r 是数量, 也是方向。

Take a plane that has a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ and is going 350 mph. To find the component form of the velocity of the airplane, note that a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ is the same as a $\begin{aligned} 30^{\circ} \end{aligned}$ on the unit circle . This corresponds to the point $\begin{aligned} (\frac{\sqrt{3}}{2}, \frac{1}{2}) \end{aligned}$ which is the same as $\begin{aligned} (\cos 30, \sin 30) \end{aligned}$ . When written as a vector $\begin{aligned} < \frac{\sqrt{3}}{2}, \frac{1}{2} > \end{aligned}$ is a unit vector because it has magnitude 1. Now you just need to scale by a factor of 350 and you get your answer of $\begin{aligned} < 175 \sqrt{3}, 175 > \end{aligned}$ . Thus, the velocity of the airplane in component form is $\begin{aligned} < 175 \sqrt{3}, 175 > \end{aligned}$ . This is the same as using the relationship $\begin{aligned} < r \cdot \cos θ, r \cdot \sin θ >=< x, y > \end{aligned}$ .
::选择一个具有60°C 和 350 mph 的方位。要找到飞机速度的组件形式, 请注意, 60°C 的方位与单位圆上的 30°C 相同。这相当于点( 32°C, 12°C) , 与( cos= 30, sin= 30°C) 相同。当以矢量 < 32, 12> 写入为矢量时, 是单位矢量, 因为它的大小为 1 。现在您只需要以 350 的乘数进行缩放, 就可以得到 < 1753, 175 > 的回答。因此, 组件形式的飞机速度为 < 1753, 175 > 。这与使用 rcos {, rsin=x,y 相同。

Examples
::实例

Example 1
::例1

Earlier, you were asked to consider four siblings fighting over a candy in a four way tug of war. Lanie pulls with 8 lb of force at an angle of $\begin{aligned} 41^{\circ} \end{aligned}$ . Connie pulls with 10 lb of force at an angle of $\begin{aligned} 100^{\circ} \end{aligned}$ . Cynthia pulls with 12 lb of force at an angle of $\begin{aligned} 200^{\circ} \end{aligned}$ . How much force and in what direction does poor little Terry have to pull the candy so it doesn’t move?
::早些时候,有人要求你考虑四个兄弟姐妹在四条拔河的道路上为糖果而战斗。兰妮在41岁的角度上用8磅的力拉。康妮以10磅的力拉在100岁的角度上。辛西娅以12磅的力拉在200岁的角度上。可怜的小泰瑞需要多少力量和方向来拉糖果才能不动?

To add the three vectors together would require several iterations of the Law of Cosines. Instead, write each vector in component form and set equal to a zero vector indicating that the candy does not move.
::要将三种矢量加在一起,需要多次重复使用《科辛斯定律》。相反,将每种矢量以元件形式写入,设定为等于零矢量,表示糖果不会移动。

\begin{aligned} \vec{L} + \vec{C O N} + \vec{C Y N} + \vec{T} =< 0, 0 > \end{aligned}

::科诺·塞纳·塔罗0,0>

\begin{aligned} < 8 \cdot \cos 41^{\circ}, 8 \cdot \sin 41^{\circ} > + < 10 \cdot \cos 100^{\circ}, 10 \cdot \sin 100^{\circ} > \\ + & < 12 \cdot \cos 200^{\circ}, 12 \cdot \sin 200^{\circ} > + \vec{T} =< 0, 0 > \end{aligned}

::<8418110101010010110112122212222000000000000101010101010101010101010101101212122220000000000000000000

Use a calculator to add all the $\begin{aligned} x \end{aligned}$ components and bring them to the far side and the $\begin{aligned} y \end{aligned}$ components and then subtract from the far side to get:
::使用计算器添加所有 x 组件, 并将其带至远端和 Y 组件, 然后从远端减去以获得 :

\begin{aligned} \vec{T} \approx< 6.98, - 10.99 > \end{aligned}

::T6.98,-10.99>

Turning this component vector into an angle and magnitude yields how hard and in what direction he would have to pull. Terry will have to pull with about 13 lb of force at an angle of $\begin{aligned} {302.4}^{\circ} \end{aligned}$ .
::将此元件矢量转换为角度和大小, 将决定他必须往何方拉动的难度。 Terry 将不得不用大约13磅的力力拉动, 角度为 302.4 。

Example 2
::例2

Consider the plane that has a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ and is going 350 mph. If there is wind blowing with the bearing of $\begin{aligned} 300^{\circ} \end{aligned}$ at 45 mph, what is the component form of the total velocity of the airplane?
::考虑一下具有60°C和350mph的轴承的飞机。如果风吹着300°C和45mph的轴承,飞机总速度的构成形式是什么?

A bearing of $\begin{aligned} 300^{\circ} \end{aligned}$ is the same as $\begin{aligned} 150^{\circ} \end{aligned}$ on the unit circle which corresponds to the point $\begin{aligned} (- \frac{\sqrt{3}}{2}, \frac{1}{2}) \end{aligned}$ . You can now write and then scale the wind vector.
::在与点对应的单位圆(- 32, 12)上,直径为300°Z的大小与150°Z的单位圆相同。您现在可以写作,然后缩放风向矢量。

\begin{aligned} 45 \cdot < - \frac{\sqrt{3}}{2}, \frac{1}{2} >=< - \frac{45 \sqrt{3}}{2}, \frac{45}{2} > \end{aligned}

Since both the wind vector and the velocity vector of the airplane are written in component form, you can simply sum them to find the component vector of the total velocity of the airplane.
::由于飞机的风向矢量和速度矢量都以组件形式写成,所以可以简单地将它们相加,以找到飞机总速度的组件矢量。

\begin{aligned} < 175 \sqrt{3}, 175 > + < - \frac{45 \sqrt{3}}{2}, \frac{45}{2} >=< \frac{305 \sqrt{3}}{2}, \frac{395}{2} > \end{aligned}

Example 3
::例3

Consider the plane from Example 2 with the same wind and velocity. Find the actual ground speed and direction of the plane (as a bearing).
::以相同的风速和速度从例2中考虑飞机。找到飞机的实际地面速度和方向(方位)。

Since you already know the component vector of the total velocity of the airplane, you should remember that these components represent an $\begin{aligned} x \end{aligned}$ distance and a $\begin{aligned} y \end{aligned}$ distance and the question asks for the hypotenuse.
::既然您已经知道飞机总速度的组件矢量, 您应该记住, 这些组件代表 x 距离和 y 距离, 且问题要求下限。

\begin{aligned} {(\frac{305 \sqrt{3}}{2})}^{2} + {(\frac{395}{2})}^{2} & = c^{2} \\ 329.8 & \approx c \end{aligned}

30532)2+(3952)2=c2329.8c

The airplane is traveling at about 329.8 mph.
::飞机在大约329.8英里处行驶。

Since you know the $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ components, you can use tangent to find the angle. Then convert this angle into bearing.
::由于您知道 x 和 y 组件, 您可以使用正切值找到角度。然后将这个角度转换为轴承。

\begin{aligned} \tan θ & = \frac{(\frac{395}{2})}{(\frac{305 \sqrt{3}}{2})} \\ θ & \approx {36.8}^{\circ} \end{aligned}

::{\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}

An angle of $\begin{aligned} {36.8}^{\circ} \end{aligned}$ on the unit circle is equivalent to a bearing of $\begin{aligned} {53.2}^{\circ} \end{aligned}$ .
::单位圆上角为36.8,等于53.2。

Note that you can do the entire problem in bearing by just switching sine and cosine, but it is best to truly understand what you are doing every step of the way and this will probably involve always going back to the unit circle.
::请注意,你只要按正弦和正弦来做就可以解决整个问题, 但最好真正理解你正在做的每一步, 这可能涉及总是回到单位圆。

For Examples 4 and 5, use the following information:
::关于实例4和5,请使用以下信息:

$\begin{aligned} \vec{v} =< 2, - 5 >, \vec{u} =< - 3, 2 >, \vec{t} =< - 4, - 3 >, \vec{r} =< 5, y > \end{aligned}$
::2,-5>,u3,2>,t4,-3>,r5,y>

$\begin{aligned} B = (4, - 5), P = (- 3, 8) \end{aligned}$
::B=(4,-5),P=(-3,8)

Example 4
::例4

Find the unit vectors in the same direction as $\begin{aligned} \vec{u} \end{aligned}$ and $\begin{aligned} \vec{t} \end{aligned}$ .
::查找单位向量与 u 和 t 相同方向的单位向量。

To find a unit vector, divide each vector by its magnitude.
::要找到一个单位矢量, 将每个矢量除以其大小。

\begin{aligned} \frac{\vec{u}}{| \vec{u} |} =< \frac{- 3}{\sqrt{13}}, \frac{2}{\sqrt{13}} >, \frac{\vec{t}}{| \vec{t} |} =< \frac{- 4}{5}, \frac{- 3}{5} > \end{aligned}

::u313,213>,tt45,-35>

Example 5
::例5

Find the point 10 units away from $\begin{aligned} B \end{aligned}$ in the direction of $\begin{aligned} P \end{aligned}$ .
::在P方向找到距离B10点的点

The vector $\begin{aligned} \vec{B P} \end{aligned}$ is $\begin{aligned} < - 7, 13 > \end{aligned}$ . First take the unit vector and then scale it so it has a magnitude of 10.
::矢量 BP为 7,13>。先选取单位矢量, 然后缩放它, 它的大小为 10 。

\begin{aligned} \frac{B P}{| B P |} & =< \frac{- 7}{\sqrt{218}}, \frac{13}{\sqrt{218}} > \\ 10 \cdot \frac{B P}{| B P |} & =< \frac{- 70}{\sqrt{218}}, \frac{130}{\sqrt{218}} > \end{aligned}

::BPBP7218,13218>10BPBP70218,130218>

You end up with a vector that is ten units long in the right direction. The question asked for a point from $\begin{aligned} B \end{aligned}$ which means you need to add this vector to point $\begin{aligned} B \end{aligned}$ .
::问题要求从 B 点得出一个点,这意味着您需要在 B 点中添加此矢量。

\begin{aligned} (4, - 5) + < \frac{- 70}{\sqrt{218}}, \frac{130}{\sqrt{218}} >\approx (- 0.74, 3.8) \end{aligned}

Summary

A unit vector is a vector of length one.
::单位矢量是长度 1 的矢量。

A unit vector in the direction of vector $\begin{aligned} \vec{v} \end{aligned}$ is $\begin{aligned} \frac{\vec{v}}{| \vec{v} |} \end{aligned}$
::向矢量 v 方向的单位矢量是 vv

The two standard unit vectors in the coordinate plane are $\begin{aligned} \vec{i}, \end{aligned}$ which is the vector $\begin{aligned} < 1, 0 >, \end{aligned}$ and $\begin{aligned} \vec{j}, \end{aligned}$ which is the vector $\begin{aligned} < 0, 1 > . \end{aligned}$ These two unit vectors are perpendicular to each other.
::坐标平面上的两个标准单位矢量是 i ,即矢量 < 1,0 > 和 j ,即矢量 < 0,1 > 。这两个单位矢量相互垂直。

A linear combination of $\begin{aligned} \vec{i} \end{aligned}$ and $\begin{aligned} \vec{j} \end{aligned}$ allows you to uniquely describe any other vector in the coordinate plane in component form.
::i 和 j 的线性组合使您能够以组件形式单独描述坐标平面中的任何其他矢量。

To convert from magnitude $\begin{aligned} r \end{aligned}$ and direction $\begin{aligned} θ \end{aligned}$ to component form, use the relationship $\begin{aligned} < r \cos θ, r \sin θ >=< x, y > . \end{aligned}$
::要从数值 r 和方向转换为元件窗体, 请使用 < rcos, rsinx,y > 的关系。

Review
::回顾

Use the following defined vectors and points to answer 1-8.
::使用以下定义的矢量和点对回答 1-8 使用如下矢量和点。

$\begin{aligned} \vec{v} =< 1, - 3 >, \vec{u} =< 2, 5 >, \vec{t} =< 9, - 1 >, \vec{r} =< 2, y > \end{aligned}$
::1,-3>,u2,5>,t9,-1>,r2,y>

$\begin{aligned} A = (- 3, 2), B = (5, - 2) \end{aligned}$
::A=(-3,3,2),B=(-5,2)

1. Solve for $\begin{aligned} y \end{aligned}$ in vector $\begin{aligned} \vec{r} \end{aligned}$ to make $\begin{aligned} \vec{r} \end{aligned}$ perpendicular to $\begin{aligned} \vec{t} \end{aligned}$ .
::1. 在矢量 r 中为 y 解决 y ,使 r 垂直为 t 。

2. Find the unit vector in the same direction as $\begin{aligned} \vec{u} \end{aligned}$ .
::2. 查找单位矢量的方向与u相同。

3. Find the unit vector in the same direction as $\begin{aligned} \vec{t} \end{aligned}$ .
::3. 查找单位矢量的方向与t相同。

4. Find the unit vector in the same direction as $\begin{aligned} \vec{v} \end{aligned}$ .
::4. 在与 v 相同的方向上查找单位矢量。

5. Find the unit vector in the same direction as $\begin{aligned} \vec{r} \end{aligned}$ .
::5. 在与r相同的方向上找到单位矢量。

6. Find the point exactly 3 units away from $\begin{aligned} A \end{aligned}$ in the direction of $\begin{aligned} B \end{aligned}$ .
::6. 在B方向找到距离A3个单位的点。

7. Find the point exactly 6 units away from $\begin{aligned} B \end{aligned}$ in the direction of $\begin{aligned} A \end{aligned}$ .
::7. 在A方向找到距离B6个单位的点。

8. Find the point exactly 5 units away from $\begin{aligned} A \end{aligned}$ in the direction of $\begin{aligned} B \end{aligned}$ .
::8. 在B方向找到距离A5个单位的点。

9. Jack and Jill went up a hill to fetch a pail of water. When they got to the top of the hill, they were very thirsty so they each pulled on the bucket. Jill pulled at $\begin{aligned} 30^{\circ} \end{aligned}$ with 20 lbs of force. Jack pulled at $\begin{aligned} 45^{\circ} \end{aligned}$ with 28 lbs of force. What is the resulting vector for the bucket?
::9. 杰克和吉尔上山去取水桶,他们到山顶时非常渴,所以每人拉上桶,吉尔用20磅的武力拉了30磅,杰克用28磅的武力拉了45磅。

10. A plane is flying on a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ at 400 mph. Find the component form of the velocity of the plane. What does the component form tell you?
::10. 一架飞机在400米时方位60°Z的方位飞行。找到飞机速度的部件形式。部件形式告诉你什么?

11. A baseball is thrown at a $\begin{aligned} 70^{\circ} \end{aligned}$ angle with the horizontal with an initial speed of 30 mph. Find the component form of the initial velocity.
::11. 棒球投向70°Z角,水平,最初速度为30mph。寻找初始速度的构成形式。

12. A plane is flying on a bearing of $\begin{aligned} 200^{\circ} \end{aligned}$ at 450 mph. Find the component form of the velocity of the plane.
::12. 一架飞机在450英里处以200英寸的速度飞行,找到飞机速度的构成形式。

13. A plane is flying on a bearing of $\begin{aligned} 260^{\circ} \end{aligned}$ at 430 mph. At the same time, there is a wind blowing at a bearing of $\begin{aligned} 30^{\circ} \end{aligned}$ at 60 mph. What is the component form of the velocity of the plane?
::13. 一架飞机在430英里处飞行,方位260°C。同时,在60英里处的方位30°C时有风吹。飞机速度的构成形式是什么?

14. Use the information from the previous problem to find the actual ground speed and direction of the plane.
::14. 利用上一个问题的信息寻找飞机的实际地面速度和方向。

15. Wind is blowing at a magnitude of 40 mph with an angle of $\begin{aligned} 25^{\circ} \end{aligned}$ with respect to the east. What is the velocity of the wind blowing to the north? What is the velocity of the wind blowing to the east?
::15. 东风的风速为40米,角度为25英寸,向北风速是多少?东风速是多少?

Review (Answers)
::回顾(答复)

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。

Section outline

General

矢量成构件的分辨率

The Unit Vector and Component Form ::单位矢量和组件表

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)