Course: 9.4 向量单位 | The Way To Learn

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单位矢量
Introduction
::导言

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.
::有时,与矢量的横向和纵向组成部分合作可能比仅仅以一个角度和一个规模开展工作要容易得多。当将数种力量结合在一起时尤其如此。

Consider four siblings fighting over a box of candy in a four-way tug of war. Lanie pulls with 8 lbs of force at an angle of $\begin{aligned} 41^{\circ} \end{aligned}$ . Connie pulls with 10 lbs of force at an angle of $\begin{aligned} 100^{\circ} \end{aligned}$ . Cynthia pulls with 12 lbs 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?
::想想四个兄弟姐妹在四边的战争中为了一盒糖果而战吧。 Lanie在41°C的角度上用8磅的武力拉拉。Connie在100°C的角度上用10磅的武力拉拉拉。Cynthia在200°C的角度上用12磅的武力拉拉拉。可怜的小泰瑞需要多少武力和方向拉动糖果才能不动?

Unit Vectors
::单位矢量

A unit vector is a vector of length 1. Sometimes you might wish to scale a vector you already have so it has a length of 1. If the length was 5, you'd scale the vector by a factor of $\begin{aligned} \frac{1}{5}, \end{aligned}$ so the resulting vector has a magnitude of 1. A unit vector of vector $\begin{aligned} \vec{v} \end{aligned}$ is notated as

$\begin{aligned} \frac{\vec{v}}{| \vec{v} |} . \end{aligned}$

::单位矢量是长度1 的矢量。有时,您可能想要缩放您已经拥有的矢量,其长度为 1。如果长度为 5,您会将矢量缩放15 倍,因此由此产生的矢量的大小为 1 。矢量的单位矢量 v 以 vv 表示。

Two standard unit vectors make up all other vectors in the coordinate plane. They are $\begin{aligned} \hat{i}, \end{aligned}$ which is the vector $\begin{aligned} < 1, 0 >, \end{aligned}$ and $\begin{aligned} \hat{j} \end{aligned}$ which is the vector $\begin{aligned} < 0, 1 > \end{aligned}$ . When using unit vectors, it is common to use hat notation, which refers to the caret or hat on top of the letter representing the vector, rather than the arrow which we have been using thus far. For example, we write the unit vector of i as $\begin{aligned} \hat{i} \end{aligned}$ instead of $\begin{aligned} \vec{i} \end{aligned}$ because the hat identifies it as a unit vector. These two unit vectors, $\begin{aligned} \hat{i} \end{aligned}$ and $\begin{aligned} \hat{j}, \end{aligned}$ are perpendicular to each other. A linear combination of $\begin{aligned} \hat{i} \end{aligned}$ and $\begin{aligned} \hat{j} \end{aligned}$ will allow you to uniquely describe any other vector in the coordinate plane.The use of $\begin{aligned} \hat{i} \end{aligned}$ and $\begin{aligned} \hat{j} \end{aligned}$ when describing vector components is called unit vector notation. The vector $\begin{aligned} < 5, 3 > \end{aligned}$ is the same as $\begin{aligned} 5 \hat{i} + 3 \hat{j} . \end{aligned}$
::两个标准单位矢量组成坐标平面上所有其他矢量。它们是 i 。即矢量 < 1,0 > 和 j 。当使用单位矢量 < 0,1 > 时,通常使用顶部标记, 指代表矢量的字母顶部的护带或顶部帽, 而不是我们迄今一直使用的箭头。例如, 我们将单位矢量 i 写成i i , 而不是 i , 因为顶部识别它为单位矢量。这两个单位矢量, 即 i 和 j , 相互垂直。将i 和 j 的线性组合允许您在坐标平面上独有描述任何其他矢量。当描述矢量组件时, i 和 j 的用途被称为单位矢量记号。矢量 < 5, 3 > 与 5 {3j} 相同。

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.
::与作为角度和规模的矢量一起工作需要非常精确的几何推理和极佳的图片。将矢量重写成组件的一个优点是,这项工作大部分都简化了。

The video below shows how to determine a unit vector given a vector. It also explains how to determine the component form of a vector in standard position that intersects the unit circle .
::下面的视频展示了如何确定单位矢量给定矢量的单位矢量。它也解释了如何在单位圆之间标准位置上确定矢量的构成形式。

Play, Learn, and Explore : .
::玩耍、学习和探索:

Examples
::实例

Example 1
::例1

A plane has a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ and is going 350 mph. Write the vector of the velocity of the airplane in unit vector notation.
::平面的方位为60°C,正向350mph方向移动。将飞机速度的矢量写成单位向量标记。

Solution:
::解决方案 :

A bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ is the same as a $\begin{aligned} 30^{\circ} \end{aligned}$ angle on the unit circle, which corresponds to the point $\begin{aligned} (\frac{\sqrt{3}}{2}, \frac{1}{2}) \end{aligned}$ . When written as a vector, $\begin{aligned} ⟨ \frac{\sqrt{3}}{2}, \frac{1}{2} ⟩ \end{aligned}$ is a unit vector because it has magnitude of 1.
::轴承60°Z与单位圆上30°Z角相同,与点(32,12)对应,当以矢量写入时,32,12°Z是单位矢量,因为它的大小为1。

Since the plane is going 350 mph, scale the vector by a factor of 350.
::由于飞机正在飞行350米, 将矢量乘以350。

$\begin{aligned} 350 \cdot ⟨ \frac{\sqrt{3}}{2}, \frac{1}{2} ⟩ = ⟨ 175 \sqrt{3}, 175 ⟩ \end{aligned}$

Thus, the velocity of the airplane in component form is $\begin{aligned} ⟨ 175 \sqrt{3}, 175 ⟩ \end{aligned}$ .
::因此,飞机部件形式的速度是1753,175。

W rite this in unit vector notation as $\begin{aligned} 175 \sqrt{3} \hat{i} + 175 \hat{j} \end{aligned}$
::将此写为单位矢量符号 1753 i 175 j

Example 2
::例2

Consider the plane flying in Example 1. If there is wind blowing with a bearing of $\begin{aligned} 300^{\circ} \end{aligned}$ at 45 mph, what is the vector of the total velocity of the airplane written in unit vector notation?
::试例1 :如果风吹起,在45毫米时的距离为300°Z,则以单位矢量记号书写的飞机总速度的矢量是什么?

Solution:
::解决方案 :

A bearing of $\begin{aligned} 300^{\circ} \end{aligned}$ is the same as a $\begin{aligned} 150^{\circ} \end{aligned}$ angle on the unit circle, which corresponds to the point $\begin{aligned} (- \frac{\sqrt{3}}{2}, \frac{1}{2}) \end{aligned}$ .
::轴承为300的角与单位圆上与点(-32,12)对应的150角相同。

Since the wind is blowing at 45 mph, scale the vector by a factor of 45.
::由于风在45米时吹,向量乘以45。

$\begin{aligned} 45 \cdot ⟨ - \frac{\sqrt{3}}{2}, \frac{1}{2} ⟩ = ⟨ - \frac{45 \sqrt{3}}{2}, \frac{45}{2} ⟩ \end{aligned}$

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。

$\begin{aligned} ⟨ 175 \sqrt{3}, 175 ⟩ + ⟨ - \frac{45 \sqrt{3}}{2}, \frac{45}{2} ⟩ = ⟨ \frac{305 \sqrt{3}}{2}, \frac{395}{2} ⟩ \end{aligned}$

W rite this in unit vector notation as $\begin{aligned} \frac{305 \sqrt{3}}{2} \hat{i} + \frac{395}{2} \hat{j} \end{aligned}$
::将此写为单位矢量符号 30532 i3952 j

Example 3
::例3

Consider the plane and wind in Examples 1 and 2. Find the actual ground speed and direction of the plane (as a bearing).
::在例1和2中考虑飞机和风,找出飞机的实际地面速度和方向(方位)。

Solution:
::解决方案 :

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.
::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, 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: You can do the entire problem in bearing just by switching sine and cosine, but it is best to truly understand what you are doing every step of the way. This will probably involve always going back to the unit circle.
::注意 : 您可以通过切换正弦和正弦, 来解决整个问题, 但最好真正理解你正在做的每一步。这可能会涉及到总是返回单位圆。

Example 4
::例4

Recall the problem from the Introduction: Four siblings are fighting over a box of candy in a four-way tug of war. Lanie pulls with 8 lbs of force at an angle of $\begin{aligned} 41^{\circ} \end{aligned}$ from the positive $\begin{aligned} x \end{aligned}$ -axis. Connie pulls with 10 lbs of force at an angle of $\begin{aligned} 100^{\circ} \end{aligned}$ from the positive $\begin{aligned} x \end{aligned}$ -axis . Cynthia pulls with 12 lbs of force at an angle of $\begin{aligned} 200^{\circ} \end{aligned}$ from the positive $\begin{aligned} x \end{aligned}$ -axis . How much force and in what direction does poor little Terry have to pull the candy so it doesn't move?
::回顾引言中的问题:四个兄弟姐妹在四条战线上为一盒糖果而战斗。 Lanie用8磅的武力从正x轴的41英寸角拉开。Connie用10磅的武力从正x轴的100英寸的角拉开。Cynthia用12磅的武力从正x轴的200英寸的角拉开。可怜的小泰瑞需要多少力量和方向拉开糖果才能不动?

Solution:
::解决方案 :

The problem can be illustrated by the following diagram:
::以下图表可以说明问题:

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 0 vector, indicating that the candy does not move.
::要将三种矢量加在一起,需要多次重复《科辛斯定律》。相反,将每种矢量以组成形式写入,并设定为等于 0 矢量,表明糖果不会移动。

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

::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 right side of the equation and the $\begin{aligned} y \end{aligned}$ components, and then subtract from the right side to get
::使用计算器来添加所有 x 组件并把它们加到公式和 Y 组件的右侧,然后从右侧减去以获取

$\begin{aligned} \vec{T} \approx ⟨ 6.98, - 10.99 ⟩ . \end{aligned}$

::T6.98, -10.99。

Turning this component vector into an angle and magnitude yields how hard and in what direction Terry will have to pull. He'll have to pull with about 13 lbs 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 Terry 如何艰难和朝何方向拉动。他将不得不以大约13磅的力力拉动到 302.4 的角。

Example 5
::例5

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

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

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 。

Solution:
::解决方案 :

$\begin{aligned} \vec{t} \end{aligned}$ has slope $\begin{aligned} \frac{3}{4}, \end{aligned}$ which means that $\begin{aligned} \vec{r} \end{aligned}$ must have slope $\begin{aligned} - \frac{4}{3} \end{aligned}$ . A vector's slope is found by putting the $\begin{aligned} y \end{aligned}$ component over the $\begin{aligned} x \end{aligned}$ component, just as with $\begin{aligned} \frac{r i s e}{r u n} \end{aligned}$ .
::t有斜坡34,这意味着 r必须有斜坡-43. 矢量的斜坡通过将 y 组件放在 x 组件上方的方式找到,就像上升一样。

$\begin{aligned} \frac{y}{5} & = - \frac{4}{3} \\ y & = - \frac{20}{3} \end{aligned}$

::y5... 43y... 203

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

Solution:
::解决方案 :

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>

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

Solution:
::解决方案 :

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

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

::BPBP7218,1321810BPBP70218,130218

You end up with a vector that is 10 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 $\begin{aligned} B \end{aligned}$ .
::您最终的矢量是方向正确的10个单位。问题要求从 B 点得出一个点, 这意味着您需要将此矢量添加到 B 点。

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

Summary
::摘要

A unit vector is a vector of magnitude 1.
::单位矢量是星级 1 的矢量。

The standard unit vectors are $\begin{aligned} \hat{i}, \end{aligned}$ which is the vector $\begin{aligned} < 1, 0 >, \end{aligned}$ and $\begin{aligned} \hat{j}, \end{aligned}$ which is the vector $\begin{aligned} < 0, 1 > . \end{aligned}$
::标准单位矢量是 i ,即矢量 < 1,0 > 和 j ,即矢量 < 0,1 > 。

A linear combination of vectors $\begin{aligned} \vec{u} \end{aligned}$ and $\begin{aligned} \vec{v} \end{aligned}$ means a multiple of one plus a multiple of the other.
::矢量 u 和 v 的线性组合表示一个乘以一个乘以另一个乘以一个。

Review
::回顾

Use the following defined vectors and points to answer 1-8: $\begin{aligned} \vec{v} =< 1, - 3 >, \vec{u} =< 2, 5 >, \vec{t} =< 9, - 1 >, \vec{r} =< 2, y > \end{aligned}$
::使用以下定义的矢量和点来回答1-8:v1,-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 in unit vector notation?
::9. 杰克和吉尔上山去取水桶,他们到山顶时非常口渴,因此每人拉上水桶,吉尔用20磅的武力拉上30磅,杰克用28磅的武力拉上45磅。

10. A plane is flying on a bearing of $\begin{aligned} 60^{\circ} \end{aligned}$ at 400 mph. Find the vector of the velocity of the plane in unit vector notation. What do the components of the unit vector notation form tell you?
::10. 飞机在400米方位上以60°Z的方位飞行。在单位矢量标记中找到飞机速度的矢量。单位矢量标记表的组件告诉您什么?

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 vector of the initial velocity in unit vector notation.
::11. 棒球投向70°Z角,水平,最初速度为30mph。在单位矢量标记中找到初始速度的矢量。

12. A plane is flying on a bearing of $\begin{aligned} 200^{\circ} \end{aligned}$ at 450 mph. Find the vector of the velocity of the plane in unit vector notation.
::12. 一架飞机在450米时方位200°Z的方位飞行,在单位矢量标记中找到飞机速度的矢量。

13. A plane is flying on a bearing of $\begin{aligned} 260^{\circ} \end{aligned}$ at 430 mph. At the same time, a wind is blowing at a bearing of $\begin{aligned} 30^{\circ} \end{aligned}$ at 60 mph. What is the vector of the velocity of the plane in unit vector notation?
::13. 一架飞机在430英里处飞行,方位260°C,同时,风力吹向30°C,方位60mph,飞机速度的矢量在单位矢量标记中是多少?

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,北风速是多少?东风速是多少?东风速是多少?

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

General

单位矢量

Introduction ::导言

Unit Vectors ::单位矢量

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Unit Vectors
::单位矢量

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)