Section: 矢量投射 | 7.5 矢量投射

Section outline

Projecting one vector onto another explicitly answers the question: “how much of one vector goes in the direction of the other vector?” The dot product is useful because it produces a scalar quantity that helps to answer this question. In this concept, you will produce an actual vector not just a scalar.
::将一个矢量投射到另一个矢量中,明确回答了这样一个问题:“一个矢量向另一个矢量的方向移动了多少?” 点值产品是有用的,因为它能产生一个可帮助回答这个问题的定量数量。在这个概念中,你将产生一个实际的矢量,而不仅仅是一个标量。

Why is vector projection useful when considering pulling a box in the direction of $\begin{aligned} v \end{aligned}$ instead of horizontally in the direction of $\begin{aligned} u \end{aligned}$ ?
::为什么在考虑将一个盒子划向 v 方向而不是横向划向 u 方向时,矢量投射是有用的?

Projections
::预测预测数

Consider the question from above.
::审议上述问题。

The definition of vector projection for the indicated red vector is the called $\begin{aligned} p r o j_{u} v \end{aligned}$ . When you read $\begin{aligned} p r o j_{u} v \end{aligned}$ you should say “the vector projection of $\begin{aligned} v \end{aligned}$ onto $\begin{aligned} u \end{aligned}$ . ” This implies that the new vector is going in the direction of $\begin{aligned} u \end{aligned}$ . The vector projection is the vector produced when one vector is resolved into two component vectors, one that is parallel to the second vector and one that is perpendicular to the second vector. The parallel vector is the vector projection. Conceptually, this means that if someone was pulling the box at an angle and strength of vector $\begin{aligned} v \end{aligned}$ then some of their energy would be wasted pulling the box up and some of the energy would actually contribute to pulling the box horizontally.
::指定红向量的矢量投影定义为 projuv 。当阅读 projuv 时, 您应该说 : “ 矢量投影或 oto u ” 。这意味着新的矢量正在朝u 方向前进。矢量投影是指当一个矢量溶解成两个组成矢量时产生的矢量, 一个与第二个矢量平行, 一个与第二个矢量垂直。平行矢量是矢量投影。从概念上看, 这意味着如果有人在矢量的角度和强度上拉动盒子, 那么他们的一些能量就会被浪费, 拉动盒子, 一些能量实际上会有助于横向拉动盒子。

The definition of is simply the length of the vector projection. When the scalar projection is positive it means that the angle between the two vectors is less than $\begin{aligned} 90^{\circ} \end{aligned}$ . When the scalar projection is negative it means that the two vectors are heading in opposite directions.
::定义只是矢量投影的长度。当标量投影为正值时, 它意味着两个矢量之间的角小于 90 。当标量投影为负值时, 它意味着两个矢量朝相反的方向向前进。

The vector projection formula can be written two ways. The version on the left is most simplified, but the version on the right makes the most sense conceptually. A.
::矢量投影公式可以写成两种方式。左边的版本最简化, 但右边的版本在概念上最合理。 A 。

$\begin{aligned} p r o j_{u} v = (\frac{v \cdot u}{| u |^{2}}) u = (\frac{v \cdot u}{| u |}) \frac{u}{| u |} \end{aligned}$
::

The proof of the vector projection formula is as follows:
::矢量预测公式的证明如下:

Given two vectors $\begin{aligned} u, v \end{aligned}$ , what is $\begin{aligned} p r o j_{u} v \end{aligned}$ ?
::给两个矢量u,v,什么是projuv?

First note that the projected vector in red will go in the direction of $\begin{aligned} u \end{aligned}$ . This means that it will be a product of the unit vector $\begin{aligned} \frac{u}{| u |} \end{aligned}$ and the length of the red vector (the scalar projection). In order to find the scalar projection, note the right triangle, the unknown angle $\begin{aligned} θ \end{aligned}$ between the two vectors and the cosine ratio.
::第一个注意是红色的预测矢量将朝u的方向前进。这意味着它将是单位矢量 uu 和红色矢量长度的产物( 彩虹投影 ) 。为了找到 scalar 投影, 请注意右边的三角形, 两个矢量之间未知的角和余弦比。

$\begin{aligned} \cos θ = \frac{s c a l a r p r o j e c t i o n}{| v |} \end{aligned}$
::ocs scalar 投影 v

Recall that $\begin{aligned} \cos θ = \frac{u \cdot v}{| u | | v |} \end{aligned}$ . Now just substitute and simplify to find the length of the scalar projection.
::回想一下cosuvuuv。现在只需替换和简化,以找到卡路里投影的长度。

\begin{aligned} \cos θ & = \frac{s c a l a r p r o j e c t i o n}{| v |} \\ \frac{u \cdot v}{| u | | v |} & = \frac{s c a l a r p r o j e c t i o n}{| v |} \\ \frac{u \cdot v}{| u |} & = s c a l a r p r o j e c t i o n \end{aligned}

::或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或或

Now you have the length of the vector projection and the direction you want it to go:
::现在您有了矢量投影的长度和方向, 您想要它去的方向 :

$\begin{aligned} p r o j_{u} v = (\frac{u \cdot v}{| u |}) \frac{u}{| u |} \end{aligned}$
::

Examples
::实例

Example 1
::例1

Earlier, you were asked why vector projection useful when considering pulling a box in the direction of instead of horizontally in the direction of $\begin{aligned} u \end{aligned}$ .Vector projection is useful in physics applications involving force and work.
::早些时候,有人问到,为什么在考虑将矢量投射划向方向而不是横向划向U.Vector投射方向时,矢量投射对涉及力量和工作的物理应用是有用的。

When the box is pulled by vector $\begin{aligned} v \end{aligned}$ some of the force is wasted pulling up against gravity. In real life this may be useful because of friction, but for now this energy is inefficiently wasted in the horizontal movement of the box.
::当盒子被矢量拉扯时, 部分力力被浪费在重力拉动上。在现实生活中, 这也许有用, 因为摩擦, 但现在这种能量在盒子的横向移动中被低效浪费了。

Example 2
::例2

Find the scalar projection of vector $\begin{aligned} v =< 3, 4 > \end{aligned}$ onto vector $\begin{aligned} u =< 5, - 12 > \end{aligned}$ .
::在矢量 u5,-12> 上找到矢量 v3,4> 的星际投影。

As noted earlier , the scalar projection is the magnitude of the vector projection. This was shown to be $\begin{aligned} (\frac{u \cdot v}{| u |}) \end{aligned}$ where $\begin{aligned} u \end{aligned}$ is the vector being projected onto.
::如前所述,卡路里预测是矢量预测的大小,显示是(uvu),其中u是向上投放的矢量。

\begin{aligned} \frac{u \cdot v}{| u |} = \frac{< 5, - 12 > \cdot < 3, 4 >}{13} = \frac{15 - 48}{13} = - \frac{33}{13} \end{aligned}

::5 - 12 - 3,4 > 13=15 - 4813 - 33

Example 3
::例3

Find the vector projection of vector $\begin{aligned} v =< 3, 4 > \end{aligned}$ onto vector $\begin{aligned} u =< 5, - 12 > \end{aligned}$
::查找矢量 v3-4> 的矢量投射到矢量 u5, - 12> 的矢量投影

Since the scalar projection has already been found in Example 2, you should multiply the scalar by the “onto” unit vector.
::由于在例2中已经发现了卡路里预测值,所以您应该将卡路里乘以“上”单位矢量。

\begin{aligned} - \frac{33}{13} < \frac{5}{13}, - \frac{12}{13} >=< - \frac{165}{169}, \frac{396}{169} > \end{aligned}

Example 4
::例4

Sketch the vector $\begin{aligned} < - 2, - 2 > \end{aligned}$ and $\begin{aligned} < 4, - 2 > \end{aligned}$ . Explain using a sketch why a negative scalar projection of $\begin{aligned} < - 2, - 2 > \end{aligned}$ onto $\begin{aligned} < 4, - 2 > \end{aligned}$ makes sense.
::将矢量 #%2, ~2 > 和 < 4, -2 > 拼贴在矢量 #2, ~2 > 和 < 4, 2-2 > 上。使用草图解释为什么在 < 4, 2-2 > 上进行负弧投影是有道理的。

First plot the two vectors and extend the “onto” vector. When the vector projection occurs, the vector $\begin{aligned} < - 2, 2 > \end{aligned}$ goes in the opposite direction of the vector $\begin{aligned} < 4, - 2 > \end{aligned}$ . This will create a vector projection going in the opposite direction of $\begin{aligned} < 4, - 2 > \end{aligned}$ .
::第一个绘图两个矢量并扩展“ 向上” 矢量。当矢量投影发生时, 矢量 #%2, 2> 向向向向向 < 4, - 2 > 。这将产生向相反方向 < 4, - 2 > 的矢量投影。

Example 5
::例5

A father is pulling his daughter up a hill. The hill has a $\begin{aligned} 20^{\circ} \end{aligned}$ incline. The daughter is on a sled that sits on the ground and has a rope that the father pulls with a force of 100 lb as he walks. The rope makes a $\begin{aligned} 39^{\circ} \end{aligned}$ angle with the slope. What is the effective force that the father exerts moving his daughter and the sled up the hill?
::父亲将女儿拉上山丘,山上有20英寸的斜坡。女儿坐在雪橇上,坐在地上,有绳子,父亲在走路时用100磅的力拉着绳子。绳子与斜坡的角是39英寸的角。父亲用什么有效力将女儿和雪橇抬上山?

The box represents the girl and the sled. The blue arrow indicates the father’s 100 lb force. Notice that the question asks for simply the amount of force which means scalar projection. Since this is not dependent on the slope of this hill, we can rotate our perspective and still get the same scalar projection.
::盒子代表的是女孩和雪橇。蓝箭代表的是父亲的100磅力。请注意,问题要求的只是意味着卡路里投射的强度。由于这不取决于这座山的坡坡,我们可以旋转我们的观点,并且仍然得到同样的卡路里投射。

The components of the father’s force vector is $\begin{aligned} 100 < \cos 39^{\circ}, \sin 39^{\circ} > \end{aligned}$ and the “onto” vector is any vector horizontally to the right. Since we are only looking for the length of the horizontal component and you already have the angle between the two vectors, the scalar projection is:
::父矢量的成分为 100 < cosç39,sin39,“ 向上” 矢量是向右水平的任何矢量。由于我们只寻找水平组件的长度,而您已经拥有了两个矢量之间的角,因此,星际预测是:

$\begin{aligned} 100 \cdot \cos 39^{\circ} \approx 77.1 l b \end{aligned}$
::1003977.1磅

Summary

Vector projection is the process of resolving one vector into two component vectors, one parallel and one perpendicular to a second vector. The parallel vector is the vector projection.
::矢量预测是将一个矢量溶解成两个组成矢量的过程,一个平行矢量,另一个垂直矢量。平行矢量是矢量预测。

Scalar projection is the length of the vector projection, indicating the amount of one vector going in the direction of the other vector.
::Scalar 预测是矢量预测的长度,表示向另一矢量方向前进的一个矢量。

Positive scalar projection means the angle between the two vectors is less than 90 degrees, while negative scalar projection means the vectors are heading in opposite directions.
::正弧投影表示两个矢量之间的角小于90度,负弧投影表示矢量朝相反方向飞去。

The vector projection formula can be written as:

$\begin{aligned} p r o j_{u} v = (\frac{v \cdot u}{| u |^{2}}) u = (\frac{v \cdot u}{| u |}) \frac{u}{| u |} . \end{aligned}$

::矢量投影公式的写法为: projuv=(vuuu2)u=(vuu)uuu。

Review
::回顾

1. Sketch vectors $\begin{aligned} < 2, 4 > \end{aligned}$ and $\begin{aligned} < 2, 1 > \end{aligned}$ .
::1. 牵线矢量 < 2,4 > 和 < 2,1 > 。

2. What is the vector projection of $\begin{aligned} < 2, 4 > \end{aligned}$ onto $\begin{aligned} < 2, 1 > \end{aligned}$ ? Sketch the projection.
::2. <2,4>对 <2,1>的矢量投射值是多少?

3. Sketch vectors $\begin{aligned} < - 2, 1 > \end{aligned}$ and $\begin{aligned} < - 1, 3 > \end{aligned}$ .
::3. 箭头矢量2,1 > 和1,3>。

4. What is the vector projection of $\begin{aligned} < - 1, 3 > \end{aligned}$ onto $\begin{aligned} < - 2, 1 > \end{aligned}$ ? Sketch the projection.
::4. 1,3>的矢量投射在2,1>上是多少?

5. Sketch vectors $\begin{aligned} < 6, 2 > \end{aligned}$ and $\begin{aligned} < 8, 1 > \end{aligned}$ .
::5. 牵线矢量 < 6.2 > 和 < 8.1 > 。

6. What is the vector projection of $\begin{aligned} < 6, 2 > \end{aligned}$ onto $\begin{aligned} < 8, 1 > \end{aligned}$ ? Sketch the projection.
::6. < 6,2 > 的矢量投射到 < 8,1 > 上是什么?

7. Sketch vectors $\begin{aligned} < 1, 7 > \end{aligned}$ and $\begin{aligned} < 6, 3 > \end{aligned}$ .
::7. 牵线矢量 < 1,7 > 和 < 6,3>。

8. What is the vector projection of $\begin{aligned} < 1, 7 > \end{aligned}$ onto $\begin{aligned} < 6, 3 > \end{aligned}$ ? Sketch the projection.
::8. <1,7>对<6,3>的矢量投射值是多少?

9. A box is on the side of a hill inclined at $\begin{aligned} 30^{\circ} \end{aligned}$ . The weight of the box is 40 pounds. What is the magnitude of the force required to keep the box from sliding down the hill?
::9. 盒子在山坡倾角的侧面,30磅,重量为40磅,使盒子不滑下山坡所需的力有多大?

10. Sarah is on a sled on the side of a hill inclined at $\begin{aligned} 60^{\circ} \end{aligned}$ . The weight of Sarah and the sled is 125 pounds. What is the magnitude of the force required for Sam to keep Sarah from sliding down the hill.
::10. Sarah身在山坡边的雪橇上,倾角为60磅,Sarah和雪橇的重量为125磅,Sam阻止Sarah滑下山坡所需的力有多大。

11. A 1780 pound car is parked on a street that makes an angle of $\begin{aligned} 15^{\circ} \end{aligned}$ with the horizontal. Find the magnitude of the force required to keep the car from rolling down the hill.
::11. 一辆1 780磅的汽车停在一条街道上,其角度为15英寸,水平为15英寸,找到阻止汽车从山上滚下来所需的力量。

12. A 1900 pound car is parked on a street that makes an angle of $\begin{aligned} 10^{\circ} \end{aligned}$ with the horizontal. Find the magnitude of the force required to keep the car from rolling down the hill.
::12. 一辆1900磅的汽车停在一条街上,以水平为10英寸的角,找到阻止汽车从山上滚下来所需的力量。

13. A 30 pound force that makes an angle of $\begin{aligned} 32^{\circ} \end{aligned}$ with an inclined plane is pulling a box up the plane. The inclined plane makes a $\begin{aligned} 20^{\circ} \end{aligned}$ angle with the horizontal. What is the magnitude of the effective force pulling the box up the plane?
::13. 30磅重的力使角角为32°Q 和倾斜平面的角拉上一个箱子,倾斜平面使角为20°C,水平为20°C。有效力将盒子拉上平面的角有多大?

14. A 22 pound force that makes an angle of $\begin{aligned} 12^{\circ} \end{aligned}$ with an inclined plane is pulling a box up the plane. The inclined plane makes a $\begin{aligned} 25^{\circ} \end{aligned}$ angle with the horizontal. What is the magnitude of the effective force pulling the box up the plane?
::14. 一股22磅的力,使角度为12°Q 与倾斜平面的角拉上一个箱子,倾斜平面使角度为25°C,水平为25°C。将盒子拉上平面的有效力有多大?

15. Anne pulls a wagon on a horizontal surface with a force of 50 pounds. The handle of the wagon makes an angle of $\begin{aligned} 30^{\circ} \end{aligned}$ with the ground. What is the magnitude of the effective force pulling the wagon?
::15. Anne用50磅的威力在水平表面拉着一辆马车。马车的手柄与地面的角角为30英寸。拉着马车的有效力有多大?

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

Projections ::预测预测数

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)