节: Scalar 和矢量预测 | 9.6 斯卡路和矢量预测

章节大纲

Introduction
::导言

For an engineer, a pilot, a race car driver, and even a chef, it can be useful to determine the result of different amounts and directions of force applied to a particular action. This information can help decide on the best use of added strength, altitude, speed, or heat to achieve the optimum result.
::对于工程师、飞行员、赛车驾驶员、甚至厨师来说,确定适用于特定行动的不同数量和武力指示的结果是有用的。这些信息有助于决定如何最佳地利用加固的强度、高度、速度或热量,以取得最佳效果。

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 section, you will produce an actual vector, not just a scalar.
::将一个矢量投射到另一个矢量明确回答一个问题, “ 一个矢量有多少向另一个矢量的方向前进? ” 点值产品是有用的, 因为它能产生一个数量级数来帮助解答这个问题。在本节中, 您将产生一个实际的矢量, 而不是一个数量级数。

Why is useful when considering pulling a box in the direction of $\begin{align*}v,\end{align*}$ instead of horizontally in the direction of $\begin{align*}u\end{align*}$ ?
::为什么在考虑将一个盒子划向 v 方向而不是横向划向 u 方向时有用?

Scalar and Vector Projections
::Scalar 和矢量预测

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

The definition of vector projection for the indicated red vector is the called $\begin{align*}proj_u v.\end{align*}$ When you read $\begin{align*}proj_u v,\end{align*}$ you should say "the vector projection of $\begin{align*}v\end{align*}$ onto $\begin{align*}u\end{align*}$ ." This implies that the new vector is going in the direction of $\begin{align*}u\end{align*}$ . The vector projection is the vector produced when one vector is resolved into two component vectors, one that is parallel to the 2nd vector and one that is perpendicular to the 2nd vector. The parallel vector is the vector projection. Conceptually, this means that if someone is pulling the box at an angle and strength of vector $\begin{align*}v\end{align*}$ , then some of their energy would be wasted pulling the box up, and some of their energy would actually contribute to pulling the box horizontally.
::指定的红向量的矢量投影定义为 projuv 。当阅读 projuv 时, 您应该说“ 矢量投影与 u 。 ” 这意味着新的矢量正朝着 u 的方向前进。矢量投影是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和强度 V 上拉动盒子, 那么他们的一些能量就会被浪费, 并且他们的一些能量实际上会帮助水平拉动盒子。

The vector projection formula can be written two ways, as shown below. The version on the left is most simplified, but the version on the right makes the most sense conceptually:
::矢量预测公式可以写成两种方式,如下文所示。左边的版本最简化,但右边的版本在概念上最有意义:

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

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

Given two vectors $\begin{align*}u, v\end{align*}$ , what is $\begin{align*}proj_u v\end{align*}$ ?
::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 $\begin{align*}u\end{align*}$ . This means it will be a product of the unit vector $\begin{align*}\frac{u}{|u|}\end{align*}$ and the length of the red vector (the scalar projection). To find the scalar projection, note the right triangle, the unknown angle $\begin{align*}\theta\end{align*}$ between the two vectors, and the cosine ratio.
::第一个注意点, 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( 斜弧投影) 的产物。要找到斜弧投影, 请注意右三角、两个矢量之间的未知角 , 以及余弦比。

$\begin{aligned} \cos θ = \frac{scalar projection}{| v |} \end{aligned}$
$\begin{align*}\cos \theta=\frac{\text{scalar projection}}{|v|}\end{align*}$
::ocs scalar 投影 v

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

$\begin{aligned} \cos θ & = \frac{scalar projection}{| v |} \\ \frac{u \cdot v}{| u | | v |} & = \frac{scalar projection}{| v |} \\ \frac{u \cdot v}{| u |} & = scalar projection \end{aligned}$
$\begin{align*}\cos \theta &= \frac{\text{scalar projection}}{|v|}\\ \frac{u \cdot v}{|u||v|} &=\frac{\text{scalar projection}}{|v|}\\ \frac{u \cdot v}{|u|} &= \text{scalar projection}\end{align*}$
::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 :

$\begin{aligned} p r o j_{u} v = (\frac{u \cdot v}{| u |}) \frac{u}{| u |} . \end{aligned}$
$\begin{align*}proj_u v=\left(\frac{u \cdot v}{|u|}\right) \frac{u}{|u|}.\end{align*}$
::

The definition of scalar projection is the length of the vector projection. Recall that the dot product of a vector is a scalar quantity describing only the magnitude of a particular vector. A scalar projection is given by the dot product of a vector with a unit vector for that direction.
::calar 投影的定义是矢量投影的长度。提醒注意,矢量的圆点产物是一个标量数量,仅描述特定矢量的大小。一个矢量的圆点产物和该方向的单位矢量提供了标量投影。

When the scalar projection is positive, it means that the angle between the two vectors is less than $\begin{align*}90^\circ\end{align*}$ . When the scalar projection is negative, it means that the two vectors are heading in opposite directions.
::当卡路里投影为正值时,它意味着两个矢量之间的角小于 90 。当卡路里投影为负值时,它意味着两个矢量向相反方向飞去。

For example, the component forms for the vectors shown below are $\begin{align*}AB =\left \langle {4,3} \right \rangle\end{align*}$ and $\begin{align*}D =\left \langle {3,-1.25} \right \rangle\end{align*}$ .
::例如,下列矢量的构成表是AB4,3和D3,2-1.25。

The scalar projection of vector AB onto $\begin{align*}\hat{i}\end{align*}$ is given by
::i 上矢量 AB 矢量 AB 的星级投射 :

$\begin{aligned} \vec{A B} \cdot \hat{i} = (4 \cdot 1) + (3 \cdot 0) = 4. \end{aligned}$
$\begin{align*}\overrightarrow{AB} \cdot \hat{i}=(4 \cdot 1)+(3 \cdot 0)=4.\end{align*}$
::ABi(41)+(30)=4。

The scalar projection of vector AB onto $\begin{align*}\hat{j}\end{align*}$ is given by
::对 j 上矢量 AB 的矢量 AB 的 ACAL 投射

$\begin{aligned} \vec{A B} \cdot \hat{j} = (4 \cdot 0) + (3 \cdot 1) = 3. \end{aligned}$
$\begin{align*}\overrightarrow{AB} \cdot \hat{j}=(4 \cdot 0)+(3 \cdot 1)=3.\end{align*}$
::ABj(40)+(31)=3。

The of AB onto the $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ directions are nonzero numbers because the vector is located in the $\begin{align*}xy\end{align*}$ - plane.
::x 和 Y 方向上的 AB 键为非零数字,因为矢量位于 xy 平面上。

The following video explains how to determine the projection of one vector onto another vector:
::以下视频解释如何确定一个矢量投射到另一个矢量:

Play, Learn, and Explore Vector Projections: .
::播放、学习和探索矢量预测:.

Examples
::实例

Example 1
::例1

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

Solution:
::解决方案 :

As noted earlier , the scalar projection is the magnitude of the vector projection. This was shown to be $\begin{align*}\left(\frac{u \cdot v}{|u|}\right),\end{align*}$ where $\begin{align*}u\end{align*}$ is the vector being projected onto.
::如前所述,卡路里投影是矢量投影的大小,显示是(uvu),其中u是投向的矢量。

$\begin{aligned} \frac{u \cdot v}{| u |} = \frac{< 5, - 12 > \cdot < 3, 4 >}{13} = \frac{15 - 48}{13} = - \frac{33}{13} \end{aligned}$
$\begin{align*}\frac{u \cdot v}{|u|}=\frac{<5, -12> \cdot <3, 4>}{13}=\frac{15-48}{13}=-\frac{33}{13}\end{align*}$
::5 - 12 - 3,4 > 13=15 - 4813 - 33

Example 2
::例2

Find the vector projection of vector $\begin{align*}v=<3, 4>\end{align*}$ onto vector $\begin{align*}u=<5, -12>.\end{align*}$
::将矢量 v3-4 > 的矢量投射到矢量 u5, - 12> 上。

Solution:
::解决方案 :

Since the scalar projection has already been found in Example 1 , you should multiply the scalar by the "onto" unit vector.
::由于在例1中已经发现了星标预测, 您应该将星标乘以“ 上” 单位矢量。

$\begin{aligned} - \frac{33}{13} ⟨ \frac{5}{13}, - \frac{12}{13} ⟩ = ⟨ - \frac{165}{169}, \frac{396}{169} ⟩ \end{aligned}$
$\begin{align*}-\frac{33}{13} \left< \frac{5}{13}, -\frac{12}{13}\right> = \left<-\frac{165}{169}, \frac{396}{169}\right>\end{align*}$

Example 3
::例3

Sketch the vector $\begin{align*}<-2, -2>\end{align*}$ and $\begin{align*}<4, -2>\end{align*}$ . Explain using a sketch why a negative scalar projection of $\begin{align*}<-2, -2>\end{align*}$ onto $\begin{align*}<4, -2>\end{align*}$ 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。

Solution:
::解决方案 :

First, plot the two vectors and extend the "onto" vector. When the vector projection occurs, the vector $\begin{align*}<-2, 2>\end{align*}$ goes in the opposite direction of the vector $\begin{align*}<4, -2>\end{align*}$ . This will create a vector projection going in the opposite direction of $\begin{align*}<4, -2>\end{align*}$ .
::首先,绘制两个矢量并扩展“ 向内” 矢量。当矢量投影发生时, 矢量%2, 2 > 向向向向向向 < 4, - 2 > 。这将产生向相反方向 < 4, - 2 > 的矢量投影。

Example 4
::例4

Recall the problem from the Introduction: W hy is vector projection useful when considering pulling a box in the direction of $\begin{align*}v,\end{align*}$ instead of horizontally in the direction of $\begin{align*}u\end{align*}$ ?
::回顾导言中的问题:为什么矢量投射在考虑将一个盒子划向 v 方向而不是横向划向 u 方向时有用?

Solution:
::解决方案 :

Vector projection is useful in physics applications involving force and work.
::矢量预测在涉及力量和工作的物理应用中非常有用。

When the box is pulled by vector $\begin{align*}v,\end{align*}$ 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.
::当容器被矢量 v 所拉动时, 部分力力在重力下被浪费了。在现实生活中, 这也许有用, 因为摩擦, 但是现在, 这种能量在盒子的横向移动中被低效浪费了。

Example 5
::例5

Determine the vector projection of vector $\begin{align*}\overrightarrow{MN}\end{align*}$ onto the vector $\begin{align*}\overrightarrow{KL}.\end{align*}$
::确定矢量 KL 上矢量 MN 的矢量投影。

Solution:
::解决方案 :

The vector projection of one vector onto a second vector is the dot product of the two vectors and the unit vector defining the direction of the second vector. In this case, $\begin{align*}proj_\overrightarrow{KL} \overrightarrow{MN} &= \left( \frac{\overrightarrow{MN} \cdot \overrightarrow{KL}}{| {\overrightarrow{KL}|}^{2}}\right) \overrightarrow{KL}. \end{align*}$
::将一个矢量投射到第二个矢量上,是两个矢量和确定第二个矢量方向的单位矢量的圆点产物。在这种情况下,projKKLNN(MNQKLKL2)KL。

First, identify the components of the two vectors by using the information given on the graph. In this case, $\begin{align*}\overrightarrow{MN}=\left \langle -2.25,0\right \rangle\end{align*}$ and $\begin{align*}\overrightarrow{KL}=\left \langle 1.5,2 \right \rangle.\end{align*}$
::首先,使用图中提供的信息来标明两个矢量的成分。在这种情况下,MN2.25,0和KL1.5,2。

Next, determine the dot product of the two vectors.
::下一步,确定两个矢量的点产物。

$\begin{aligned} \vec{M N} \cdot \vec{K L} & = (M N)_{x} (K L)_{x} + (M N)_{y} (K L)_{y} \\ = (- 2.25) (1.5) + (0) (2) \\ = - 3.375 \end{aligned}$
$\begin{align*}\overrightarrow{MN} \cdot \overrightarrow{KL} &=(MN)_x \ (KL)_x \ + \ (MN)_y \ (KL)_y \\ &= (-2.25)(1.5) + (0)(2) \\ &= -3.375\end{align*}$
::MNKL(MN)x(KL)x+(MN)yy(KL)y=(-2.25)(1.5)+(0)(2)3.375

Calculate the magnitude of vector $\begin{align*}\overrightarrow{KL}.\end{align*}$
$\begin{aligned} | \vec{K L} | & = \sqrt{(K L)_{x}^{2} + (K L)_{y}^{2}} \\ = \sqrt{(1.5)^{2} + (2)^{2}} \\ = \sqrt{2.25 + 4} \\ = \sqrt{6.25} \\ = 2.5 \end{aligned}$
$\begin{align*}|\overrightarrow{KL}| &=\sqrt{(KL)_x^2+(KL)_y^2} \\ &=\sqrt{(1.5)^2+(2)^2} \\ &=\sqrt{2.25+4} \\ &= \sqrt{6.25} \\ &=2.5\end{align*}$
::计算矢量 KL. KKL(KL)x2+(KL)y2=(1.5) 2+(2)2=2. 25+4=6. 25=2.5

P ut this information into the vector projection formula to obtain
::将这一信息放入矢量预测公式,以获得

$\begin{align*}proj_\overrightarrow{KL} \overrightarrow{MN} &= \left( \frac{-3.375}{{2.5}^{2}}\right) \overrightarrow{KL} \\ &= \left( \frac{-3.375}{6.25}\right) \overrightarrow{KL} \\ &= -0.54\overrightarrow{KL} \end{align*}$
:-3.3752.52)KL(-3375.6.25)KL0.54KL

To multiply vector $\begin{align*}\overrightarrow{KL}\end{align*}$ by the scalar -0.54 , multiply each of the vector components by -0.54.
::要将矢量KL乘以弧度 - 0.54, 将每个矢量组件乘以 - 0.54 。

$\begin{align*}proj_\overrightarrow{KL} \overrightarrow{MN} &= -0.54\left \langle 1.5,2 \right \rangle \\ &= \left \langle -0.54 \cdot 1.5 , -0.54 \cdot 2 \right \rangle \\ &= \left \langle -0.81,-1.08 \right \rangle\end{align*}$
::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:

Example 6
::例6

The diagram below shows both vectors AB and D together on the same grid. Determine the scalar projection of vector AB onto the direction of vector D .
::下图显示同一网格上的矢量 AB 和 D 并列。确定矢量 AB 向矢量 D 方向的星标投影。

Solution:
::解决方案 :

To find the scalar projection onto the direction of another vector, we need to know the unit vector in the direction of vector D .
::要找到向另一个矢量方向的斜线投影, 我们需要知道向矢量D方向的单位矢量。

First, the components of $\begin{align*}\overrightarrow{AB}\end{align*}$ are $\begin{align*}\left \langle 4, 3 \right \rangle\end{align*}$ and $\begin{align*}\overrightarrow{D}\end{align*}$ are $\begin{align*}\left \langle 3,-1.25 \right \rangle\end{align*}$ .
::首先,AB的组成部分为4,3,D组成部分为3,1-1.25。

Now we can use the dot product to calculate the scalar projection of AB onto the direction of vector D .
::现在我们可以用点产品来计算 AB 向矢量 D 方向的斜弧投影。

$\begin{aligned} \vec{A B} \cdot \vec{D} = (4 \cdot 3) + (3 \cdot - 1.25) = 8.25. \end{aligned}$
$\begin{align*}\overrightarrow{AB}\ \cdot \overrightarrow{D} =(4 \cdot 3)+(3 \cdot -1.25)= 8.25.\end{align*}$
::AB__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

The magnitude of $\begin{align*}\overrightarrow{D}\end{align*}$ is
$\begin{aligned} | D | & = \sqrt{(D_{x})^{2} + (D_{y})^{2}} \\ = \sqrt{3^{2} + (- 1.25)^{2}} \\ = \sqrt{9 + 1.5625} \\ = \sqrt{10.5625} \\ = 3.25. \end{aligned}$
$\begin{align*}|D| &=\sqrt{(D_x)^2+(D_y)^2} \\ &=\sqrt{3^2+(-1.25)^2} \\ &=\sqrt{9+1.5625} \\ &=\sqrt{10.5625} \\ &=3.25.\end{align*}$
::D的大小是D(Dx)2+(Dy)2=32+(-1.25)2=9+1.5625=10.5625=3.25。

Thus, t he scalar projection of AB onto the direction of vector D is
::因此,ABA向矢量D方向的星标投射是:

$\begin{aligned} \frac{\vec{A B} \cdot \vec{D}}{| \vec{D} |} = \frac{8.25}{3.25} \approx 2.54. \end{aligned}$
$\begin{align*}\frac{\overrightarrow{AB}\ \cdot \overrightarrow{D}}{|\overrightarrow{D}|} = \frac{8.25}{3.25} \approx 2.54.\end{align*}$ Example 7
::AB___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

A father is pulling his daughter on a sled that has a rope attached to it. They are going up a hill with a $\begin{align*}20^\circ\end{align*}$ incline. The sled sits on the ground, and the father pulls the rope with a force of 100 lbs as he walks. The rope makes a $\begin{align*}39^\circ\end{align*}$ angle with the slope. What is the effective force the father exerts as he pulls his daughter and the sled up the hill?
::一个父亲拉着他女儿的雪橇,上面绑着一条绳子,他们用20英寸的绳子爬上一座山坡,雪橇坐在地上,父亲在走路时用100磅的力拉绳子,绳子用斜坡的39磅角拉绳子。父亲拉着他的女儿和雪橇上山时,有什么有效的力?

Solution:
::解决方案 :

The box below 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.
::下面的方框代表着女孩和雪橇。蓝箭代表着父亲的100磅力。请注意, 问题要求的只是力的大小, 也就是说, 星标投射。由于这不取决于这个山坡, 我们可以旋转我们的角度, 并且仍然得到同样的星标投射。

The components of the father's force vector is $\begin{align*}100 < \cos 39^\circ, \sin 39^\circ >,\end{align*}$ and the "onto" vector is any vector horizontally to the right. Since we are looking only 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 <cos39,sin39,“ 向上” 矢量是向右水平的任何矢量。由于我们只寻找水平的矢量长度, 而您已经拥有两个矢量之间的角, 星际投影是

$\begin{aligned} 100 \cdot \cos 39^{\circ} \approx 77.1 lb . \end{aligned}$
$\begin{align*}100 \cdot \cos 39^\circ \approx 77.1 \text{ lb}.\end{align*}$
::1003977.1磅

Summary
::摘要

The vector projection of a vector onto a given direction has a magnitude equal to the scalar projection.
::向某一方向向矢量投射的矢量的大小与标量投射的数值相等。

The formula for the projection vector is given by $\begin{align*}proj_u v=\left(\frac{u \cdot v}{|u|}\right) \frac{u}{|u|}\end{align*}$ .
::投影矢量的公式由 projuv = (uvu) u u 给出。

A vector $\begin{align*}\overrightarrow{v}\end{align*}$ is multiplied by a scalar s. Its components are given by $\begin{align*}\overrightarrow{sv}=\left \langle sv_x,sv_y \right \rangle\end{align*}$ .
::矢量 v 乘以 scalar s。其组件由 svsvx, svy提供。

A scalar projection is the length of the vector projection. It is given by the dot product of a vector with a unit vector for that direction.
::标量投影是矢量投影的长度,由带有该方向单位矢量的矢量的圆点产物提供。

Review
::回顾

Calculate the result of a scalar multiple of 6 on the vector $\begin{align*}\left \langle 1, 19\right \rangle. \end{align*}$
::计算矢量1,19,1,19,2,2,2,3,2,3,2,3,2,3,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,6,4,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,4,6,6,6,6,6,6,4,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,4,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,

Calculate the result of a scalar multiple of (-8) on the vector $\begin{align*}\left \langle 7, 8 \right \rangle. \end{align*}$
::计算矢量 7,8 上的弧度乘数 (8) 的结果。

Calculate the result of a scalar multiple of 15 on the vector $\begin{align*}\left \langle -7, 3\right \rangle. \end{align*}$
::计算矢量 ++7 3 + 的 15 的弧度乘数结果。

Calculate the result of a scalar multiple of (-11) on the vector $\begin{align*}\left \langle -9, 20\right \rangle. \end{align*}$
::计算矢量 {9,20 } 上的(- 11) 弧数乘以(- 11) 的结果。

Calculate the result of a scalar multiple of 16 on the vector $\begin{align*}\left \langle 9, 10\right \rangle. \end{align*}$
::计算矢量 {9,10} 上的 16 的弧度乘数结果。

Given vector A = $\begin{align*}\left \langle -4, 6 \right \rangle\end{align*}$ and vector B = $\begin{align*}\left \langle 9, 15 \right \rangle\end{align*}$ , what is the projection of A onto B?
::如果矢量A=4,6和矢量B=9,15,A到B的预测值是多少?

What is the projection of $\begin{align*}\left \langle 2, 10 \right \rangle\end{align*}$ onto $\begin{align*}\left \langle 1, 3 \right \rangle\end{align*}$ ?
::2,10对1,3的预测值是多少?

Given vector C = $\begin{align*}\left \langle 1, 2 \right \rangle\end{align*}$ and vector D = $\begin{align*}\left \langle 1, 11 \right \rangle\end{align*}$ , what is the projection of C onto D?
::鉴于矢量C = 1,2 和矢量D = 1,11, C 投射到 D 是什么?

Given vector E = $\begin{align*}\left \langle -5, 3 \right \rangle\end{align*}$ and vector F = $\begin{align*}\left \langle -1, 18 \right \rangle\end{align*}$ , what is the projection of E onto F?
::如果矢量E = 5,3 和矢量F = 1,18,那么E 到 F 的预测值是多少?

What is the projection of $\begin{align*}\left \langle -2, 1 \right \rangle\end{align*}$ onto $\begin{align*}\left \langle -5, 6 \right \rangle\end{align*}$ ?
::2,1对5,6的预测是多少?

Given vector H = $\begin{align*}\left \langle 8i + 11j \right \rangle\end{align*}$ and vector I = $\begin{align*}\left \langle -2i + 15j \right \rangle,\end{align*}$ what is the projection of H onto I?
::鉴于病媒H = 8i+11j和病媒I = 2i+15j,H对I的预测值是多少?

What is the projection of $\begin{align*}\left \langle -1, 6 \right \rangle\end{align*}$ onto $\begin{align*}\left \langle 8, 16 \right \rangle\end{align*}$ ?
::在8,16之间,1,6的预测值是多少?

What is the projection of $\begin{align*}\left \langle -1, 8 \right \rangle\end{align*}$ onto $\begin{align*}\left \langle 7, 8 \right \rangle\end{align*}$ ?
::7,8的预测值是多少?

Given vector J = $\begin{align*}\left \langle -4i + 8j \right \rangle\end{align*}$ and vector K = $\begin{align*}\left \langle -8i + 10j \right \rangle\end{align*}$ , what is the projection of J onto K?
::鉴于矢量 J = 4i+8j和矢量 K = 8i+10j, J 对 K的预测值是多少 ?

Given vector L = $\begin{align*}\left \langle -4,3 \right \rangle\end{align*}$ and vector M = $\begin{align*}\left \langle 7, 17 \right \rangle\end{align*}$ , what is the projection of L onto M?
::鉴于病媒L=4,3和病媒M=7,17,L对M的预测值是多少?

A box is on the side of a hill inclined at $\begin{align*}30^\circ\end{align*}$ . The weight of the box is 40 lbs. What is the magnitude of force required to keep the box from sliding down the hill in the absence of friction ?
::一个盒子在30°C的山坡倾角上。盒子的重量是40磅。要阻止盒子在没有摩擦的情况下滑下山坡, 需要多大的力量?

Sarah is on a sled on the side of a hill inclined at $\begin{align*}60^\circ\end{align*}$ . The weight of Sarah and the sled is 125 lbs. What is the magnitude of the force required for Sam to pull Sarah up the hill?
::莎拉和雪橇的重量是125磅。

A 30-lb force that makes an angle of $\begin{align*}32^\circ\end{align*}$ with an inclined plane is pulling a box up the plane. The inclined plane makes a $\begin{align*}20^\circ\end{align*}$ angle with the horizontal. What is the magnitude of the effective force pulling the box up the plane?
::以倾角平面为角的30磅力将一个箱子拉上飞机。倾角平面以水平为角,取角为20磅。将盒子拉上飞机的有效力有多大?

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

章节大纲

Introduction ::导言

Scalar and Vector Projections ::Scalar 和矢量预测

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Scalar and Vector Projections
::Scalar 和矢量预测

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)