Axiom Learning·AP Precalculus

Chapter 11: Vectors

向量 — Magnitude, Direction, Operations, and Motion

本章概览

向量（vector）是同时具有大小（magnitude）和方向（direction）的量，与只有大小的标量（scalar）不同。向量在物理（力、速度、位移）和计算机图形学中无处不在。本章将学习向量的表示、运算、点积，以及向量在平面运动建模中的应用。

本章在考纲中的位置 Where this chapter fits

本章对应 CED Unit 4 · Topics 4.8–4.9（Vectors, Vector-Valued Functions）。Unit 4 是 "Additional Topic Available to Schools"：不在 AP Precalculus 考试范围内（Not assessed on the AP Exam），由学校决定是否讲授。向量是物理以及微积分 BC 中平面运动的基础。

Term 术语	Chinese 中文	Meaning
Scalar	标量	Magnitude only (e.g. speed, temperature)
Vector	向量	Magnitude AND direction (e.g. velocity, force)
Component form	分量形式	$\vec{v}=\langle v_1, v_2\rangle$
Magnitude	模长	$\\|\vec{v}\\|=\sqrt{v_1^2+v_2^2}$
Unit vector	单位向量	Vector with magnitude $1$
Dot product	点积/标量积	$\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$ (scalar result)
Resultant force	合力	Vector sum of two or more forces
Position vector	位置向量	$\vec{r}(t)=\langle x(t),y(t)\rangle$
Velocity vector	速度向量	Direction and speed of motion
Speed	速率	Magnitude of velocity vector
Normalizing	单位化	Finding unit vector: $\hat{u}=\vec{v}/\\|\vec{v}\\|$

11.1 Scalars vs Vectors 标量与向量

Scalar 标量

Magnitude only — no direction
Examples: $20\text{ m}$, $50\text{ mph}$, $98.6°F$, mass, time, temperature

Vector 向量

Magnitude and direction
Examples: force $60\text{ lbs NE}$, velocity $50\text{ mph East}$, displacement $\langle -4,3\rangle$

A vector $\vec{v}$ is written in component form（分量形式）: $$\vec{v} = \langle v_1,\, v_2\rangle$$ where $v_1$ is the horizontal component（水平分量）and $v_2$ is the vertical component（竖直分量）.

Equal, Zero, and Negative Vectors

Zero vector（零向量）: $\vec{0}=\langle 0,0\rangle$ — no length, no direction
Equal vectors（相等向量）: same magnitude AND direction (can be at different locations)
Negative vector（相反向量）: $-\vec{v}=\langle -v_1,-v_2\rangle$ — same magnitude, opposite direction

11.2 Component Form and Magnitude 分量形式与模长

Given tail point $P=(x_1,y_1)$ and head point $Q=(x_2,y_2)$, the vector $\overrightarrow{PQ}$ is: $$\overrightarrow{PQ} = \langle x_2-x_1,\; y_2-y_1\rangle$$ Magnitude（模长）: $$\|\vec{v}\| = \sqrt{v_1^2+v_2^2}$$

向量 ≠ 点

向量只有方向和大小，没有固定起点。$\overrightarrow{PQ}$ 和 $\overrightarrow{RS}$ 如果方向和大小相同，就是相等向量，即使它们画在坐标系的不同位置。

🤝 一起做 Together T-1 — Component Form and Magnitude

Point $P$ has coordinates $(3,7)$ and point $Q$ has coordinates $(8,2)$. Find the component form and magnitude of $\overrightarrow{PQ}$.

Component form.
$\overrightarrow{PQ} = \langle 8-3,\; 2-7\rangle = \langle 5,\,-5\rangle$

Magnitude.
$\|\overrightarrow{PQ}\| = \sqrt{5^2+(-5)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$

$\overrightarrow{PQ}=\langle 5,-5\rangle$; $\|\overrightarrow{PQ}\|=5\sqrt{2}$

Direction Angle 方向角

The direction of $\vec v=\langle a,b\rangle$ is the direction of the segment from the origin to the point $(a,b)$. Its direction angle（方向角）$\theta$ is measured counterclockwise from the positive $x$-axis, $0°\le\theta<360°$: $$\cos\theta=\frac{a}{\|\vec v\|},\qquad \sin\theta=\frac{b}{\|\vec v\|},\qquad \tan\theta=\frac{b}{a}$$

$\tan^{-1}\frac{b}{a}$ only gives angles between $-90°$ and $90°$. If $a<0$ (Quadrant II or III), add $180°$; if $a>0$ and $b<0$ (Quadrant IV), add $360°$.
If $a=0$: $\theta=90°$ when $b>0$, and $\theta=270°$ when $b<0$.

Going back from magnitude and angle to components: $\vec v=\langle\|\vec v\|\cos\theta,\;\|\vec v\|\sin\theta\rangle$ (see 11.7).

🤝 一起做 Together T-2 — Magnitude and Direction Angle 模长与方向角

Find the magnitude and the direction angle ($0°\le\theta<360°$) of (a) $\langle-3,4\rangle$, (b) $\langle2,-5\rangle$, (c) $\langle-6,-6\rangle$.

(a) $\langle-3,4\rangle$, Quadrant II. $\|\vec v\|=\sqrt{9+16}=5$. $\tan^{-1}\frac{4}{-3}\approx-53.1°$; since $a<0$, add $180°$: $\theta\approx126.9°$. Check: $\cos\theta=-\frac35$ ✓.

(b) $\langle2,-5\rangle$, Quadrant IV. $\|\vec v\|=\sqrt{4+25}=\sqrt{29}\approx5.385$. $\tan^{-1}\frac{-5}{2}\approx-68.2°$; add $360°$: $\theta\approx291.8°$.

(c) $\langle-6,-6\rangle$, Quadrant III. $\|\vec v\|=\sqrt{72}=6\sqrt2\approx8.485$. $\tan^{-1}\frac{-6}{-6}=\tan^{-1}1=45°$, but $45°$ points into Quadrant I, the opposite direction. Since $a<0$: $\theta=45°+180°=225°$.

Teal arc: $126.9°$ for $\langle-3,4\rangle$; purple arc: $291.8°$ for $\langle2,-5\rangle$.

(a) $5$, $\theta\approx126.9°$ (b) $\sqrt{29}\approx5.385$, $\theta\approx291.8°$ (c) $6\sqrt2\approx8.485$, $\theta=225°$

11.3 Vector Operations 向量运算

Addition 加法

$\vec{u}+\vec{v}=\langle u_1+v_1,\;u_2+v_2\rangle$
Geometric: head-to-tail method（首尾相接法）

Subtraction 减法

$\vec{u}-\vec{v}=\langle u_1-v_1,\;u_2-v_2\rangle$
Same as $\vec{u}+(-\vec{v})$

Scalar Multiplication 数量乘法

$c\vec{v}=\langle cv_1,\;cv_2\rangle$
$c>0$: same dir; $c<0$: opposite dir

Dot Product 点积

$\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$
Result is a scalar（标量）

Head-to-Tail Vector Addition · 向量首尾相接

Translate $\vec{v}$ without changing its length or direction, then connect the first tail to the final head. 平移 $\vec{v}$，再画 resultant。

Step 0 / 2

Choose vector components 选择向量

0 · Compare
Place $\vec{u}$ and $\vec{v}$ at one tail.
1 · Translate
Move $\vec{v}$ to the head of $\vec{u}$.
2 · Resultant
Connect the first tail to the final head.

Vector u Vector v Resultant u + v Original position of v

为什么可以“移动”向量？

向量不固定在某个位置；只要magnitude 和 direction 不变，平移后的向量仍与原向量相等。Head-to-tail method 的核心是：先走 $\vec{u}$，再从 $\vec{u}$ 的终点走 $\vec{v}$；从最初起点直达最终终点的向量，就是 $\vec{u}+\vec{v}$。这与 components 对应相加得到完全相同的结果。

🤝 一起做 Together T-3 — Vector Addition and Scalar Multiplication

Let $\vec{u}=\langle 3,5\rangle$, $\vec{v}=\langle 4,-2\rangle$, and $\vec{w}=\langle -4,5\rangle$. Find $\vec{u}+\vec{v}$. Then find $2\vec{w}$ and $-3\vec{v}$.

Addition.
$\vec{u}+\vec{v}=\langle 3+4,\;5+(-2)\rangle=\langle 7,3\rangle$

Scalar multiplication.
$2\vec{w}=\langle 2(-4),2(5)\rangle=\langle -8,10\rangle$
$-3\vec{v}=\langle -3(4),-3(-2)\rangle=\langle -12,6\rangle$

$\vec{u}+\vec{v}=\langle7,3\rangle$; $2\vec{w}=\langle-8,10\rangle$; $-3\vec{v}=\langle-12,6\rangle$

🤝 一起做 Together T-4 — Vector Subtraction

Let $\vec{u}=\langle 5,6\rangle$, $\vec{v}=\langle -3,2\rangle$. Find $\vec{u}-\vec{v}$.

$\vec{u}-\vec{v}=\langle 5-(-3),\;6-2\rangle=\langle 8,4\rangle$

$\vec{u}-\vec{v}=\langle 8,4\rangle$

📝 练习 E-1 — Vector Arithmetic

Let $\vec{a}=\langle 2,-1\rangle$, $\vec{b}=\langle -3,4\rangle$. Compute: (a) $3\vec{a}-2\vec{b}$, (b) $\|\vec{a}+\vec{b}\|$.

(a) $3\langle2,-1\rangle-2\langle-3,4\rangle=\langle6,-3\rangle-\langle-6,8\rangle=\langle12,-11\rangle$
(b) $\vec{a}+\vec{b}=\langle-1,3\rangle$; $\|\vec{a}+\vec{b}\|=\sqrt{1+9}=\sqrt{10}$

Parallel Vectors 平行向量

A scalar multiple $c\vec v$ is parallel（平行）to $\vec v$: it points in the same direction if $c>0$ and in the opposite direction if $c<0$.
Two nonzero vectors $\vec u=\langle u_1,u_2\rangle$ and $\vec v=\langle v_1,v_2\rangle$ are parallel exactly when $\vec u=c\,\vec v$ for some number $c$; equivalently, $u_1v_2-u_2v_1=0$.

📝 练习 E-2 — Parallel Vectors 平行向量

(a) Are $\langle6,-9\rangle$ and $\langle-4,6\rangle$ parallel? If so, do they point in the same or opposite directions?
(b) Find $k$ so that $\langle k,6\rangle$ is parallel to $\langle2,-3\rangle$. (c) Are $\langle3,5\rangle$ and $\langle5,3\rangle$ parallel?

(a) $\langle-4,6\rangle=-\frac23\langle6,-9\rangle$, so they are parallel and point in opposite directions ($c<0$). Check: $6(6)-(-9)(-4)=36-36=0$ ✓
(b) $\langle k,6\rangle=c\langle2,-3\rangle$: $6=-3c\Rightarrow c=-2$, so $k=2c=-4$.
(c) $3(3)-5(5)=-16\neq0$, so they are not parallel.

11.4 Resultant Force 合力

Forces are vectors. The resultant force（合力）is the vector sum of all forces acting on an object. To find its magnitude when the angle between forces is known, use the Law of Cosines（余弦定理）.

Law of Cosines 余弦定理: $c^2 = a^2+b^2-2ab\cos C$

Law of Sines 正弦定理: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

In vector addition diagrams, use the supplementary angle（补角）$180°-\theta$ between the vectors when forming a triangle.

🤝 一起做 Together T-5 — Resultant Force

Two forces of $25$ lb and $60$ lb act on a body. The angle between them is $30°$. Find the magnitude of the resultant force and the angle it makes with the smaller force.

Step 1 — Set up triangle.
In the parallelogram of forces, the diagonal is the resultant. The two sides are $25$ and $60$. The interior angle of the triangle is $180°-30°=150°$.

The two forces start at the same point; copying the 60 lb force to the head of the 25 lb force makes a triangle whose angle at that corner is $180°-30°=150°$.

Step 2 — Law of Cosines.
$R^2 = 25^2+60^2-2(25)(60)\cos150°$
$= 625+3600-3000\cos150°$
$= 4225-3000\!\left(-\dfrac{\sqrt{3}}{2}\right) = 4225+2598.1 = 6823.1$
$R = \sqrt{6823.1}\approx 82.6\text{ lb}$

Step 3 — Angle with smaller force (Law of Sines).
$\dfrac{\sin\alpha}{60}=\dfrac{\sin150°}{82.6}\Rightarrow \sin\alpha=\dfrac{60\times0.5}{82.6}\approx0.363\Rightarrow\alpha\approx21°$

Resultant $\approx82.6$ lb; angle with $25$ lb force $\approx21°$

🤝 一起做 Together T-6 — Airplane and Wind

A plane flies at $534$ mph. Wind blows at $47$ mph at $38°$ to the plane's direction. Find the resultant speed and the angle between the resultant velocity and the wind vector.

Law of Cosines (interior angle $= 180°-38°=142°$).
$R^2 = 534^2+47^2-2(534)(47)\cos142°$
$= 285156+2209-50196(-0.788)$
$\approx 287365+39554 = 326919$
$R\approx571.8\text{ mph}$

Angle with wind (Law of Sines).
$\dfrac{\sin\beta}{534}=\dfrac{\sin142°}{571.8}\Rightarrow\sin\beta=\dfrac{534\times0.6157}{571.8}\approx0.5752\Rightarrow\beta\approx35.1°$

Resultant speed $\approx571.8$ mph; angle with wind $\approx35.1°$

Resultant by Components 分量法求合力

Component method 分量法 (works for any number of vectors):

Write each vector in components from its magnitude and direction angle: $\langle\|\vec F\|\cos\theta,\;\|\vec F\|\sin\theta\rangle$.
Add the components.
Find the magnitude and the direction angle of the sum (check the quadrant).

🤝 一起做 Together T-7 — Resultant by Components 分量法求合力

(a) Two forces act on an object: $\vec F_1$ has magnitude $40$ N and direction angle $30°$; $\vec F_2$ has magnitude $25$ N and direction angle $135°$. Find the magnitude and direction angle of the resultant.
(b) Check the answer to T-6 with the component method.

(a) Components.
$\vec F_1=40\langle\cos30°,\sin30°\rangle\approx\langle34.641,\;20\rangle$
$\vec F_2=25\langle\cos135°,\sin135°\rangle\approx\langle-17.678,\;17.678\rangle$

Add, then find magnitude and angle.
$\vec R\approx\langle16.963,\;37.678\rangle$, $\|\vec R\|=\sqrt{16.963^2+37.678^2}\approx41.320$ N
Quadrant I: $\theta=\tan^{-1}\dfrac{37.678}{16.963}\approx65.8°$.
Check with the Law of Cosines: the angle between the forces is $135°-30°=105°$, so the triangle angle is $75°$: $\;R^2=40^2+25^2-2(40)(25)\cos75°\approx1707.36$, $R\approx41.32$ ✓

Dashed: the parallelogram. The resultant is its diagonal.

(b) Plane and wind by components. Point the plane's heading along the positive $x$-axis: plane $\langle534,0\rangle$, wind $47\langle\cos38°,\sin38°\rangle\approx\langle37.037,\;28.936\rangle$.
Sum $\approx\langle571.037,\;28.936\rangle$: $\|\vec R\|\approx571.8$ mph ✓. Its direction angle is $\tan^{-1}\frac{28.936}{571.037}\approx2.9°$, so the angle between $\vec R$ and the wind is $38°-2.9°\approx35.1°$ ✓.

(a) $\|\vec R\|\approx41.32$ N, direction angle $\approx65.8°$ (b) $571.8$ mph and $35.1°$, matching the Law of Cosines answer.

🤝 一起做 Together T-8 — Angle from Magnitudes 由模长求夹角

Draw the triangle. Place $\vec v$ at the head of $\vec u$. The sides are $5$, $8$, and $10$; the angle opposite the side $10$ is $\varphi=180°-\theta$.

(a) Law of Cosines. $10^2=5^2+8^2-2(5)(8)\cos\varphi\Rightarrow\cos\varphi=\dfrac{25+64-100}{80}=-\dfrac{11}{80}=-0.1375$
$\varphi\approx97.9°$, so $\theta=180°-\varphi\approx82.1°$.

(b) Law of Sines. $\alpha$ is opposite the side $8$: $\dfrac{\sin\alpha}{8}=\dfrac{\sin\varphi}{10}\Rightarrow\sin\alpha=\dfrac{8\sin97.9°}{10}\approx0.792\Rightarrow\alpha\approx52.4°$.
($\alpha$ is acute: only the angle opposite the longest side can be obtuse.)

Check (uses the dot product from 11.5). $\|\vec u+\vec v\|^2=\|\vec u\|^2+\|\vec v\|^2+2\,\vec u\cdot\vec v$: $\;100=89+2\,\vec u\cdot\vec v\Rightarrow\vec u\cdot\vec v=5.5$, so $\cos\theta=\dfrac{5.5}{5\cdot8}=0.1375$ and $\theta\approx82.1°$ ✓

(a) $\theta\approx82.1°$ (b) $\alpha\approx52.4°$

📝 练习 E-3 — Resultants and Angles 合力与夹角

(a) A boat heads due north at $12$ mph while the current flows due east at $5$ mph. Find the boat's resultant speed and the direction angle of its motion.
(b) Two forces of $30$ N and $50$ N have a resultant of $70$ N. Find the angle between the two forces.

(a) With east as $+x$ and north as $+y$: $\vec R=\langle5,12\rangle$. Speed $=\sqrt{25+144}=13$ mph; direction angle $\tan^{-1}\frac{12}{5}\approx67.4°$ (about $22.6°$ east of north).
(b) Triangle with sides $30$, $50$, $70$: $\cos\varphi=\dfrac{30^2+50^2-70^2}{2(30)(50)}=\dfrac{-1500}{3000}=-0.5\Rightarrow\varphi=120°$. The angle between the forces is $180°-120°=60°$.

11.5 Dot Product 点积

The dot product（点积）of $\vec{u}=\langle u_1,u_2\rangle$ and $\vec{v}=\langle v_1,v_2\rangle$: $$\vec{u}\cdot\vec{v} = u_1v_1+u_2v_2 \qquad \text{(result is a scalar!)}$$ Properties 性质:

Commutative（交换律）: $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$
Distributive（分配律）: $\vec{u}\cdot(\vec{v}+\vec{w})=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$
Magnitude（模长）: $\vec{v}\cdot\vec{v}=\|\vec{v}\|^2$

点积 vs 数量乘法

数量乘法（scalar multiplication）是数 × 向量 = 向量；点积是向量 · 向量 = 数（标量）。这是两个完全不同的运算，不要混淆！

Geometric Meaning: Perpendicularity 几何意义：垂直

For nonzero vectors $\vec{u}$ and $\vec{v}$, if $\theta$ is the angle between them: $$\vec{u}\cdot\vec{v} = \|\vec{u}\|\|\vec{v}\|\cos\theta$$

Dot Product 点积值	Angle Type 角的类型	中文
$> 0$	$0°\le\theta<90°$: acute, or the same direction ($\theta=0°$)	锐角或同向
$< 0$	$90°<\theta\le180°$: obtuse, or opposite directions ($\theta=180°$)	钝角或反向
$= 0$	$\theta=90°$: perpendicular vectors	垂直向量

为什么点积为 0 就垂直？

🤝 一起做 Together T-9 — Computing Dot Products

Let $\vec{u}=\langle 2,3\rangle$, $\vec{v}=\langle 5,-1\rangle$, $\vec{w}=\langle 4,-2\rangle$. Find: (a) $\vec{u}\cdot\vec{v}$, (b) $\vec{u}\cdot(\vec{v}+\vec{w})$.

(a) $\vec{u}\cdot\vec{v}$.
$\vec{u}\cdot\vec{v}=2(5)+3(-1)=10-3=7$

(b) $\vec{u}\cdot(\vec{v}+\vec{w})$.
$\vec{v}+\vec{w}=\langle5+4,-1+(-2)\rangle=\langle9,-3\rangle$
$\vec{u}\cdot\langle9,-3\rangle=2(9)+3(-3)=18-9=9$

(a) $7$ (b) $9$

🤝 一起做 Together T-10 — Finding $k$ for Perpendicular Vectors

Find $k$ such that $\vec{u}=\langle k,-3\rangle$ and $\vec{v}=\langle 7,-4\rangle$ are perpendicular.

Set dot product to zero.
$\vec{u}\cdot\vec{v}=0 \Rightarrow k(7)+(-3)(-4)=0 \Rightarrow 7k+12=0 \Rightarrow k=-\dfrac{12}{7}$

$k=-\dfrac{12}{7}$

11.6 Angle Between Two Vectors 两向量夹角

From the dot product formula: $$\cos\theta = \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\|\vec{v}\|} \qquad \Rightarrow \qquad \theta = \cos^{-1}\!\left(\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\|\vec{v}\|}\right)$$ where $0\le\theta\le\pi$ (or $0°\le\theta\le180°$).

🤝 一起做 Together T-11 — Angle Between Vectors

Find the angle $\theta$ between $\vec{u}=\langle 3,-5\rangle$ and $\vec{v}=\langle -4,-2\rangle$.

Dot product.
$\vec{u}\cdot\vec{v}=3(-4)+(-5)(-2)=-12+10=-2$

Magnitudes.
$\|\vec{u}\|=\sqrt{9+25}=\sqrt{34}$; $\|\vec{v}\|=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$

Angle.
$\cos\theta=\dfrac{-2}{\sqrt{34}\cdot2\sqrt{5}}=\dfrac{-2}{2\sqrt{170}}=\dfrac{-1}{\sqrt{170}}$
$\theta=\cos^{-1}\!\left(\dfrac{-1}{\sqrt{170}}\right)\approx\cos^{-1}(-0.0767)\approx94.4°$

$\theta\approx94.4°$ (obtuse, consistent with $\vec{u}\cdot\vec{v}<0$)

📝 练习 E-4 — Dot Product Applications

(a) Find the angle between $\vec{a}=\langle 1,1\rangle$ and $\vec{b}=\langle 1,0\rangle$.
(b) Are $\vec{p}=\langle 6,4\rangle$ and $\vec{q}=\langle -2,3\rangle$ perpendicular?

(a) $\vec{a}\cdot\vec{b}=1$; $\|\vec{a}\|=\sqrt{2}$, $\|\vec{b}\|=1$; $\cos\theta=\dfrac{1}{\sqrt{2}}\Rightarrow\theta=45°$
(b) $\vec{p}\cdot\vec{q}=6(-2)+4(3)=-12+12=0$ → Yes, perpendicular!

11.7 Unit Vectors 单位向量

A unit vector（单位向量）has magnitude $1$. Standard unit vectors: $$\mathbf{i}=\langle 1,0\rangle, \qquad \mathbf{j}=\langle 0,1\rangle$$ Any vector: $\langle v_1,v_2\rangle = v_1\mathbf{i}+v_2\mathbf{j}$

Normalizing a Vector 单位化: Find the unit vector in the same direction as $\vec{v}$: $$\hat{u} = \frac{\vec{v}}{\|\vec{v}\|}$$ A vector of magnitude $k$ in the same direction: $\;k\,\dfrac{\vec v}{\|\vec v\|}$ (use $-k$ for the opposite direction).

Vector from magnitude and direction 由大小和方向角求向量: $$\vec{v} = \|\vec{v}\|\langle\cos\theta,\sin\theta\rangle = \|\vec{v}\|\cos\theta\;\mathbf{i}+\|\vec{v}\|\sin\theta\;\mathbf{j}$$

🤝 一起做 Together T-12 — Unit Vector and Normalization

(a) Find a unit vector in the same direction as $\vec{v}=2\mathbf{i}-7\mathbf{j}$.
(b) Find $k$ such that $\left\langle\dfrac{2}{3},k\right\rangle$ is a unit vector.

(a) Normalize $\vec{v}=\langle2,-7\rangle$.
$\|\vec{v}\|=\sqrt{4+49}=\sqrt{53}$
$\hat{u}=\dfrac{1}{\sqrt{53}}\langle2,-7\rangle=\left\langle\dfrac{2}{\sqrt{53}},\dfrac{-7}{\sqrt{53}}\right\rangle$

(b) Unit vector condition: $\|\vec{v}\|=1$.
$\left(\dfrac{2}{3}\right)^2+k^2=1\Rightarrow\dfrac{4}{9}+k^2=1\Rightarrow k^2=\dfrac{5}{9}\Rightarrow k=\pm\dfrac{\sqrt{5}}{3}$

(a) $\hat{u}=\left\langle\dfrac{2}{\sqrt{53}},\dfrac{-7}{\sqrt{53}}\right\rangle$ (b) $k=\pm\dfrac{\sqrt{5}}{3}$

🤝 一起做 Together T-13 — Vector from Magnitude and Direction

A ball is thrown at $20$ mph at $45°$ from the positive $x$-axis. Express the velocity vector in terms of $\mathbf{i}$ and $\mathbf{j}$.

$\vec{v}=20\langle\cos45°,\sin45°\rangle=20\left\langle\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right\rangle=\langle10\sqrt{2},10\sqrt{2}\rangle$
In unit vector form: $\vec{v}=10\sqrt{2}\,\mathbf{i}+10\sqrt{2}\,\mathbf{j}$

$\vec{v}=10\sqrt{2}\,\mathbf{i}+10\sqrt{2}\,\mathbf{j}\approx14.1\,\mathbf{i}+14.1\,\mathbf{j}$

📝 练习 E-5 — Unit Vector and $\mathbf{i},\mathbf{j}$ Form

(a) Express $\langle -3,5\rangle$ in $\mathbf{i},\mathbf{j}$ form and find its length.
(b) Find the unit vector in the direction of $\langle -6,8\rangle$.
(c) Find the vector of magnitude $10$ in the same direction as $\langle 3,4\rangle$, and the vector of magnitude $3$ in the opposite direction.

(a) $-3\mathbf{i}+5\mathbf{j}$; $\|\vec{v}\|=\sqrt{9+25}=\sqrt{34}$
(b) $\|\vec{v}\|=\sqrt{36+64}=10$; $\hat{u}=\dfrac{1}{10}\langle-6,8\rangle=\langle-0.6,\,0.8\rangle$
(c) Unit vector $\left\langle\frac35,\frac45\right\rangle$; $\;10\left\langle\frac35,\frac45\right\rangle=\langle6,8\rangle$; $\;-3\left\langle\frac35,\frac45\right\rangle=\langle-1.8,\,-2.4\rangle$

11.8 Vector-Valued Functions 向量值函数

A vector-valued function（向量值函数）gives a vector as output: $$\vec{r}(t) = \langle x(t),\,y(t)\rangle = x(t)\,\mathbf{i}+y(t)\,\mathbf{j}$$ The particle's position at time $t$ is $\bigl(x(t),\,y(t)\bigr)$. Distance from origin 与原点的距离: $$\|\vec{r}(t)\| = \sqrt{[x(t)]^2+[y(t)]^2}$$

🤝 一起做 Together T-14 — Position Vector Function

An object's position is given by the parametric function $x(t)=t^2-t$, $y(t)=-t+2$, $-3\le t\le3$. Write the position as a vector-valued function (in $\langle\;\rangle$ form and in $\mathbf{i},\mathbf{j}$ form) and find the object's distance from the origin at $t=2$.

Vector-valued function.
$\vec{r}(t)=\langle t^2-t,\;-t+2\rangle=(t^2-t)\,\mathbf{i}+(-t+2)\,\mathbf{j}$

Position at $t=2$.
$x(2)=4-2=2$; $y(2)=-2+2=0$. Position: $(2,0)$.

Distance from origin.
$\|\vec{r}(2)\|=\sqrt{2^2+0^2}=2$

$\vec{r}(t)=(t^2-t)\mathbf{i}+(-t+2)\mathbf{j}$; at $t=2$: position $(2,0)$, distance $=2$.

11.9 Velocity and Speed 速度与速率

The velocity vector（速度向量）$\vec{v}(t)=\langle v_x(t),\,v_y(t)\rangle$ gives both speed and direction.

$v_x(t)>0$: moving right; $v_x(t)<0$: moving left
$v_y(t)>0$: moving up; $v_y(t)<0$: moving down

Speed（速率）= magnitude of velocity: $$\text{speed}=\|\vec{v}(t)\|=\sqrt{[v_x(t)]^2+[v_y(t)]^2}$$

速度 vs 速率

速度（velocity）是向量，有方向；速率（speed）是标量，是速度向量的模长，永远非负。类比：速度计显示速率（$60$ km/h），但不显示方向（东/南/西/北）。

Where does $\vec v(t)$ come from? 速度函数从哪来？ In precalculus the velocity function is given: it tells how fast $x$ and $y$ are changing at each instant. Finding it from the position function is a calculus skill.
考纲 4.9.A.2 把速度写成 $v(t)=\langle x(t),y(t)\rangle$，那里的 $x(t)$、$y(t)$ 指水平和竖直方向的速度分量，不是位置。本页记作 $v_x(t)$、$v_y(t)$，以免和位置混淆。

🤝 一起做 Together T-15 — Velocity and Speed

A particle has velocity vector $\vec{v}(t)=\langle 2t,\;2t+1\rangle$. At $t=4$, describe the direction of motion and find its speed.

Velocity at $t=4$.
$\vec{v}(4)=\langle 2(4),\;2(4)+1\rangle=\langle 8,9\rangle$

Direction.
$v_x=8>0$ → moving right; $v_y=9>0$ → moving up. Particle moves up and to the right.

Speed.
$\text{speed}=\|\vec{v}(4)\|=\sqrt{8^2+9^2}=\sqrt{64+81}=\sqrt{145}\approx12.0$

Moving right and up; speed $=\sqrt{145}\approx12.0$ units/time.

🤝 一起做 Together T-16 — Circular Motion Speed

An object has position $\vec{r}(t)=\langle\cos t,\sin t\rangle$ and its velocity is given by $\vec{v}(t)=\langle -\sin t,\cos t\rangle$. Find the distance from origin and speed at any time $t$.

Distance from origin.
$\|\vec{r}(t)\|=\sqrt{\cos^2t+\sin^2t}=\sqrt{1}=1$ (constant — always on the unit circle!)

Speed.
$\|\vec{v}(t)\|=\sqrt{\sin^2t+\cos^2t}=1$ (constant speed 1)

Distance from origin $= 1$ (always); speed $= 1$ (constant).

Motion over Time 按区间分析运动

🤝 一起做 Together T-17 — Reading Motion from $\vec v(t)$ over Time 按时间区间分析速度

A particle's velocity is $\vec v(t)=\langle t^2-4,\;3-t\rangle$ for $0\le t\le5$.
(a) When is the particle moving right? left? up? down? (b) When is it moving left and up at the same time? (c) Find the speed at $t=0$, $t=2$, and $t=3$. (d) Is the particle ever at rest? (e) Describe the motion at $t=4$, including the speed and the direction angle of the velocity.

(a) Signs of the components. $v_x=t^2-4=0$ at $t=2$; $v_y=3-t=0$ at $t=3$.

$t$-interval	$v_x=t^2-4$	$v_y=3-t$	Motion
$0\le t<2$	$-$	$+$	left & up ↖
$2<t<3$	$+$	$+$	right & up ↗
$3<t\le5$	$+$	$-$	right & down ↘

Right for $2<t\le5$, left for $0\le t<2$; up for $0\le t<3$, down for $3<t\le5$.

(b) Left and up together: $0\le t<2$.

(c) Speeds. $\vec v(0)=\langle-4,3\rangle$: speed $5$. $\vec v(2)=\langle0,1\rangle$: speed $1$ (moving straight up). $\vec v(3)=\langle5,0\rangle$: speed $5$ (moving straight right).

(d) At rest? That needs $v_x=0$ and $v_y=0$ at the same time. But $v_x=0$ only at $t=2$ and $v_y=0$ only at $t=3$, so the particle is never at rest.

(e) $t=4$. $\vec v(4)=\langle12,-1\rangle$: moving right and slightly down. Speed $=\sqrt{145}\approx12.042$. Direction angle: $\tan^{-1}\frac{-1}{12}\approx-4.8°$; Quadrant IV, so $\theta\approx355.2°$.

(a) right $2<t\le5$, left $0\le t<2$, up $0\le t<3$, down $3<t\le5$ (b) $0\le t<2$ (c) $5$, $1$, $5$ (d) never (e) right and slightly down, speed $\approx12.042$, direction angle $\approx355.2°$

📝 练习 E-6 — Full Motion Analysis

A particle has position $\vec{r}(t)=\langle t^2-4,\; t\rangle$.
(a) Write the position at $t=3$. (b) Find $\|\vec{r}(3)\|$. (c) If velocity is $\vec{v}(t)=\langle 2t,1\rangle$, find the speed at $t=3$ and describe direction of motion.

(a) $\vec{r}(3)=\langle9-4,3\rangle=\langle5,3\rangle$ → position $(5,3)$
(b) $\|\vec{r}(3)\|=\sqrt{25+9}=\sqrt{34}$
(c) $\vec{v}(3)=\langle6,1\rangle$; speed $=\sqrt{36+1}=\sqrt{37}\approx6.1$; moving right ($v_x=6>0$) and up ($v_y=1>0$)

Chapter 11 Summary 本章总结

Operation / Concept	Formula
Component form	$\overrightarrow{PQ}=\langle x_2-x_1,\;y_2-y_1\rangle$
Magnitude	$\\|\vec{v}\\|=\sqrt{v_1^2+v_2^2}$
Addition	$\vec{u}+\vec{v}=\langle u_1+v_1,\;u_2+v_2\rangle$
Subtraction	$\vec{u}-\vec{v}=\langle u_1-v_1,\;u_2-v_2\rangle$
Scalar multiplication	$c\vec{v}=\langle cv_1,cv_2\rangle$
Dot product	$\vec{u}\cdot\vec{v}=u_1v_1+u_2v_2$ (scalar)
Dot product & angle	$\vec{u}\cdot\vec{v}=\\|\vec{u}\\|\\|\vec{v}\\|\cos\theta$
Perpendicular test	$\vec{u}\perp\vec{v}\iff\vec{u}\cdot\vec{v}=0$
Angle between vectors	$\theta=\cos^{-1}\!\left(\dfrac{\vec{u}\cdot\vec{v}}{\\|\vec{u}\\|\\|\vec{v}\\|}\right)$
Unit vector (normalize)	$\hat{u}=\dfrac{\vec{v}}{\\|\vec{v}\\|}$
Vector from magnitude & direction	$\vec{v}=\\|\vec{v}\\|\langle\cos\theta,\sin\theta\rangle$
Position vector	$\vec{r}(t)=\langle x(t),y(t)\rangle$
Distance from origin	$\\|\vec{r}(t)\\|=\sqrt{x(t)^2+y(t)^2}$
Speed	$\\|\vec{v}(t)\\|=\sqrt{v_x(t)^2+v_y(t)^2}$
Direction angle	$\tan\theta=\frac{b}{a}$; add $180°$ if $a<0$, add $360°$ in Quadrant IV
Parallel test	$\vec u=c\,\vec v$, or $u_1v_2-u_2v_1=0$
Resultant by components	Add $\langle\\|\vec F\\|\cos\theta,\;\\|\vec F\\|\sin\theta\rangle$ for each force, then find magnitude and angle
Angle from magnitudes	Law of Cosines on the triangle with sides $\\|\vec u\\|$, $\\|\vec v\\|$, $\\|\vec u+\vec v\\|$; then $\theta=180°-\varphi$
Vector of magnitude $k$	$k\,\dfrac{\vec v}{\\|\vec v\\|}$
Motion from $\vec v(t)$	Signs of $v_x$, $v_y$ on intervals; at rest only if both are $0$ at the same $t$

考试重点提示

向量加法用分量对应相加，不能混淆水平和竖直分量。
点积结果是标量，不是向量——常见错误！
判断垂直：点积 $=0$；判断角度：用 $\cos\theta=\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\|\vec{v}\|}$。
合力题：画出平行四边形，用补角（$180°-\theta$）代入余弦定理。
速率 $=$ 速度向量的模长（标量，永远 $\ge0$）。
求方向角先看分量所在的象限：计算器的 $\tan^{-1}$ 只给出 $-90°$ 到 $90°$ 之间的角。
只有两个速度分量在同一时刻都为 $0$，粒子才静止。

AP Precalculus · Chapter 11: Vectors 向量 | Axiom Learning 讲义