1️⃣ Basic Definitions & Notation

• Vector: A quantity with both magnitude and direction.
• Notation: $\vec{v}$, $\overrightarrow{AB}$, or $⟨x,y⟩$.
• Position Vector: $\overrightarrow{OP}=⟨x_P,y_P⟩$(from origin $O$ to $P$).
• Zero Vector: $\vec{0}=⟨0,0⟩$ magnitude $0$, direction undefined.
• Unit Vector: $\widehat{u}=\frac{\vec{v}}{\mathrm{\mid }\vec{v}\mathrm{\mid }}$(same direction as $\vec{v}$, length $1$).

🔹 Practice

1. Given $A(2,-1)$and $B(5,3)$, find $\overrightarrow{AB}$ and its magnitude.
2. Find the unit vector in the direction of $\vec{v}=⟨-3,4⟩$.
3. If $\overrightarrow{PQ}=⟨7,-2⟩$ and $P=(-1,4)$, find coordinates of $Q$.

---

2️⃣ Vector Operations

• Addition: $\vec{u}+\vec{v}=⟨u_1+v_1,u_2+v_2⟩$
• Subtraction: $\vec{u}-\vec{v}=⟨u_1-v_1,u_2-v_2⟩$
• Scalar Multiplication: $k\vec{v}=⟨k{v}_1,k{v}_2⟩$
• Properties: Commutative, associative, distributive over addition/scalars.

🔹 Practice

1. If $\vec{a}=⟨2,-5⟩$ and $\vec{b}=⟨-1,3⟩$, compute $3\vec{a}-2\vec{b}$.
2. Find scalars $x,y$x,y such that $x⟨1,2⟩+y⟨3,-1⟩=⟨7,1⟩$.
3. Given $\overrightarrow{OA}=\vec{a}$, $\overrightarrow{OB}=\vec{b}$, express $\overrightarrow{AB}$ in terms of $\vec{a},\vec{b}$.

---

3️⃣ Dot Product & Angles

• Definition: $\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2=\mathrm{\mid }\vec{u}\mathrm{\mid }\mathrm{\mid }\vec{v}\mathrm{\mid }\cos \theta$
• Orthogonality: $\vec{u}\perp \vec{v}⟺\vec{u}\cdot \vec{v}=0$
• Angle Formula: $\cos \theta ={\displaystyle \frac{\vec{u}\cdot \vec{v}}{\mathrm{\mid }\vec{u}\mathrm{\mid }\mathrm{\mid }\vec{v}\mathrm{\mid }}}$​
• Projection of u⃗ onto v⃗: ${\text{proj}}_{\vec{v}}\vec{u}=\left({\displaystyle \frac{\vec{u}\cdot \vec{v}}{\mathrm{\mid }\vec{v}{\mathrm{\mid }}^2}}\right)\vec{v}$

🔹 Practice

1. Compute $\vec{u}\cdot \vec{v}$u⋅v for $\vec{u}=⟨4,-2⟩,\vec{v}=⟨1,3⟩$. Are they perpendicular?
2. Find the angle between $\vec{a}=⟨1,1⟩$ and $\vec{b}=⟨1,-1⟩$.
3. Project $\vec{u}=⟨5,2⟩$onto $\vec{v}=⟨3,4⟩$.

---

4️⃣ Key Geometric Formulas & Theorems

• Midpoint: $\overrightarrow{OM}={\displaystyle \frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}}$​
• Section Formula ($P$P divides $AB$AB in ratio $m:n$m:n): $\overrightarrow{OP}={\displaystyle \frac{n\overrightarrow{OA}+m\overrightarrow{OB}}{m+n}}$​
• Centroid of △ABC△ABC: $\overrightarrow{OG}={\displaystyle \frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}}$​
• Collinearity: $A,B,C$A,B,C collinear $⟺\overrightarrow{AB}=k\overrightarrow{AC}$for some $k\in \mathbb{R}$

🔹 Practice

1. Find the position vector of the midpoint of $A(1,4)$ and $B(7,-2)$.
2. Point $P$P divides $AB$AB in ratio $2:3$. If $A=(0,0)$ and $B=(10,5)$, find $P$.
3. Show that points $(1,2),(4,8),(7,14)$ are collinear using vectors.

---

5️⃣ Magnitude & Useful Identities

• $\mathrm{\mid }\vec{v}\mathrm{\mid }=\sqrt{v_1^2+v_2^2}$
• $\mathrm{\mid }\vec{u}+\vec{v}{\mathrm{\mid }}^2=\mathrm{\mid }\vec{u}{\mathrm{\mid }}^2+\mathrm{\mid }\vec{v}{\mathrm{\mid }}^2+2\vec{u}\cdot \vec{v}$
• $\mathrm{\mid }\vec{u}-\vec{v}{\mathrm{\mid }}^2=\mathrm{\mid }\vec{u}{\mathrm{\mid }}^2+\mathrm{\mid }\vec{v}{\mathrm{\mid }}^2-2\vec{u}\cdot \vec{v}$
• Triangle Inequality: $\mid \mathrm{\mid }\vec{u}\mathrm{\mid }-\mathrm{\mid }\vec{v}\mathrm{\mid }\mid \le \mathrm{\mid }\vec{u}\pm \vec{v}\mathrm{\mid }\le \mathrm{\mid }\vec{u}\mathrm{\mid }+\mathrm{\mid }\vec{v}\mathrm{\mid }$​
• Symmetry Trick: If points $P_1,\dots ,P_n$ are symmetric about $M$, then $\sum \overrightarrow{O{P}_i}=n\overrightarrow{OM}$

🔹 Practice

1. If $\mathrm{\mid }\vec{a}\mathrm{\mid }=3$, $\mathrm{\mid }\vec{b}\mathrm{\mid }=4$, and $\vec{a}\cdot \vec{b}=6$, find $\mathrm{\mid }\vec{a}+\vec{b}\mathrm{\mid }$.
2. Given $\mathrm{\mid }\vec{u}\mathrm{\mid }=5$, $\mathrm{\mid }\vec{v}\mathrm{\mid }=2$, and $\mathrm{\mid }\vec{u}-\vec{v}\mathrm{\mid }=6$, find $\vec{u}\cdot \vec{v}$.
3. Five points are equally spaced on a line segment with midpoint $M$. If each point is at distance $2$ from $M$ alternately left/right, what is ${\sum }_{i=1}^5\overrightarrow{O{P}_i}$ in terms of $\overrightarrow{OM}$?

---

6️⃣ Competition Problem-Solving Strategies

Strategy	When to Use	Example Application
Coordinate Placement	Geometry problems with right angles or symmetry	Place right angle at origin, legs on axes
Vector Decomposition	Unknown points or complex paths	Express $\overrightarrow{XY}$ as combination of basis vectors
Symmetry Summation	Evenly spaced points on a segment/circle	$\sum \overrightarrow{B{P}_k}=n\overrightarrow{BM}$
Dot Product = 0	Prove perpendicularity	Show slopes product = $-1$ via $\vec{u}\cdot \vec{v}=0$
Magnitude Squaring	Find length of sum/difference without coordinates	Use $

🔹 Practice

1. In $\mathrm{△}ABC$, $D$is the midpoint of $BC$. Express $\overrightarrow{AD}$ in terms of $\overrightarrow{AB}$and $\overrightarrow{AC}$.
2. Points $P_1,\dots ,P_6$divide segment $AC$ into $7$ equal parts. If $B$ is not on $AC$, find ${\sum }_{k=1}^6\overrightarrow{B{P}_k}$in terms of $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
3. Given $\mathrm{\mid }\vec{a}\mathrm{\mid }=\mathrm{\mid }\vec{b}\mathrm{\mid }=1$ and $\mathrm{\mid }\vec{a}+\vec{b}\mathrm{\mid }=\sqrt{3}$​, find $\mathrm{\mid }\vec{a}-\vec{b}\mathrm{\mid }$.

---

✅ Answer Key

Section 1

1. $\overrightarrow{AB}=⟨3,4⟩,\mathrm{\mid }\overrightarrow{AB}\mathrm{\mid }=5$AB=⟨3,4⟩,∣AB∣=5
2. $\widehat{u}=⟨-\frac{3}{5},\frac{4}{5}⟩$u^=⟨−53​,54​⟩
3. $Q=(6,2)$Q=(6,2)

Section 2

1. $⟨8,-21⟩$⟨8,−21⟩
2. $x=2,y=\frac{5}{3}$x=2,y=35​
3. $\overrightarrow{AB}=\vec{b}-\vec{a}$AB=b−a

Section 3

1. $\vec{u}\cdot \vec{v}=-2$u⋅v=−2; not perpendicular
2. $\theta ={90}^{\circ }$θ=90∘
3. ${\text{proj}}_{\vec{v}}\vec{u}=\left(\frac{15+8}{25}\right)⟨3,4⟩=\frac{23}{25}⟨3,4⟩=⟨\frac{69}{25},\frac{92}{25}⟩$projv​u=(2515+8​)⟨3,4⟩=2523​⟨3,4⟩=⟨2569​,2592​⟩

Section 4

1. $M=(4,1)⟹\overrightarrow{OM}=⟨4,1⟩$M=(4,1)⟹OM=⟨4,1⟩
2. $P=(4,2)$P=(4,2)
3. $\overrightarrow{AB}=⟨3,6⟩,\overrightarrow{AC}=⟨6,12⟩=2\overrightarrow{AB}⟹$AB=⟨3,6⟩,AC=⟨6,12⟩=2AB⟹ collinear

Section 5

1. $\mathrm{\mid }\vec{a}+\vec{b}\mathrm{\mid }=\sqrt{9+16+12}=\sqrt{37}$∣a+b∣=9+16+12​=37​
2. $36=25+4-2\vec{u}\cdot \vec{v}⟹\vec{u}\cdot \vec{v}=-\frac{7}{2}$36=25+4−2u⋅v⟹u⋅v=−27​
3. Symmetry cancels 4 outer points, middle point is $M$M. Sum $=5\overrightarrow{OM}$=5OM

Section 6

1. $\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$AD=21​(AB+AC)
2. Let $M$M be midpoint of $AC$AC. $\overrightarrow{BM}=\frac{1}{2}(\overrightarrow{BA}+\overrightarrow{BC})$BM=21​(BA+BC). Sum $=6\overrightarrow{BM}=3(\overrightarrow{BA}+\overrightarrow{BC})$=6BM=3(BA+BC)
3. $\mathrm{\mid }\vec{a}+\vec{b}{\mathrm{\mid }}^2=3⟹1+1+2\vec{a}\cdot \vec{b}=3⟹\vec{a}\cdot \vec{b}=\frac{1}{2}$∣a+b∣2=3⟹1+1+2a⋅b=3⟹a⋅b=21​. Then $\mathrm{\mid }\vec{a}-\vec{b}{\mathrm{\mid }}^2=1+1-2(\frac{1}{2})=1⟹\mathrm{\mid }\vec{a}-\vec{b}\mathrm{\mid }=1$∣a−b∣2=1+1−2(21​)=1⟹∣a−b∣=1

---

💡 Tips for Mastery

• Always draw a quick sketch before computing.
• Convert geometric conditions (midpoints, ratios, perpendiculars) into vector equations first.
• In competitions, symmetry + centroid/midpoint formulas often bypass heavy algebra.
• Practice rewriting problems in both coordinate and pure vector forms to build flexibility.

Let me know if you want deeper explanations, more competition-style problems (AMC/AIME level), or a printable PDF version!

向量常用性质和定理总结

一、基本概念

1. 向量的表示

• 几何表示：$\overrightarrow{AB}$AB（从点A指向点B）
• 字母表示：$\vec{a},\vec{b},\mathbf{a},\mathbf{b}$a,b,a,b
• 坐标表示：$\vec{a}=(x,y)$a=(x,y) 或 $\vec{a}=(x,y,z)$a=(x,y,z)

2. 向量的模（长度）

$\mathrm{\mid }\vec{a}\mathrm{\mid }=\sqrt{x^2+y^2}\text{或}\mathrm{\mid }\vec{a}\mathrm{\mid }=\sqrt{x^2+y^2+z^2}$∣a∣=x2+y2​或∣a∣=x2+y2+z2​

---

二、向量运算

1. 加法与减法

• 三角形法则：$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$AB+BC=AC
• 平行四边形法则：$\vec{a}+\vec{b}$a+b 为对角线
• 坐标运算：$(x_1,y_1)\pm (x_2,y_2)=(x_1\pm x_2,y_1\pm y_2)$(x1​,y1​)±(x2​,y2​)=(x1​±x2​,y1​±y2​)

2. 数乘

$k\vec{a}=(kx,ky)$ka=(kx,ky)

• $k>0$k>0：方向不变
• $k<0$k<0：方向相反

3. 点积（数量积） ⭐重要

$\vec{a}\cdot \vec{b}=\mathrm{\mid }\vec{a}\mathrm{\mid }\mathrm{\mid }\vec{b}\mathrm{\mid }\cos \theta =x_1{x}_2+y_1{y}_2$a⋅b=∣a∣∣b∣cosθ=x1​x2​+y1​y2​

性质：

• $\vec{a}\cdot \vec{a}=\mathrm{\mid }\vec{a}{\mathrm{\mid }}^2$a⋅a=∣a∣2
• $\vec{a}\perp \vec{b}⟺\vec{a}\cdot \vec{b}=0$a⊥b⟺a⋅b=0
• 交换律：$\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$a⋅b=b⋅a
• 分配律：$\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}$a⋅(b+c)=a⋅b+a⋅c

---

三、重要定理和公式

1. 共线（平行）条件

$\vec{a}\parallel \vec{b}⟺\vec{a}=k\vec{b}(k\in \mathbb{R})$a∥b⟺a=kb(k∈R) 坐标形式：$\frac{x_1}{x_2}=\frac{y_1}{y_2}$x2​x1​​=y2​y1​​（分母不为0）

2. 三点共线

点A、B、C共线 $⟺\overrightarrow{AB}=k\overrightarrow{AC}$⟺AB=kAC

3. 中点公式

$\overrightarrow{OM}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}$OM=2OA+OB​

4. 定比分点公式

点P分$\overrightarrow{AB}$AB成$\lambda :1$λ:1，则： $\overrightarrow{OP}=\frac{\overrightarrow{OA}+\lambda \overrightarrow{OB}}{1+\lambda }$OP=1+λOA+λOB​

5. 重心公式（三角形）

$\overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$OG=3OA+OB+OC​

---

四、竞赛常用技巧 ⭐

1. 向量分解法

将复杂向量表示为基底向量的线性组合： $\vec{v}=x\vec{a}+y\vec{b}$v=xa+yb

2. 对称点的向量和

若点$P_1,P_2,\dots ,P_n$P1​,P2​,…,Pn​关于点M对称分布，则： ${\sum }_{i=1}^n\overrightarrow{O{P}_i}=n\overrightarrow{OM}$∑i=1n​OPi​​=nOM

证明：$\overrightarrow{O{P}_i}=\overrightarrow{OM}+\overrightarrow{M{P}_i}$OPi​​=OM+MPi​​，对称点$\overrightarrow{M{P}_i}$MPi​​相互抵消

3. 投影公式

$\vec{a}$a在$\vec{b}$b上的投影长度： ${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\mathrm{\mid }\vec{b}\mathrm{\mid }}$projb​a=∣b∣a⋅b​

4. 夹角公式

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{\mathrm{\mid }\vec{a}\mathrm{\mid }\mathrm{\mid }\vec{b}\mathrm{\mid }}$cosθ=∣a∣∣b∣a⋅b​

5. 三角形中的向量关系

• $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}$AB+BC+CA=0
• $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$AB=OB−OA

---

五、常见题型解题策略

类型1：求向量和的模

方法：先求和向量，再求模；或利用平方展开 $\mathrm{\mid }\vec{a}+\vec{b}{\mathrm{\mid }}^2=\mathrm{\mid }\vec{a}{\mathrm{\mid }}^2+\mathrm{\mid }\vec{b}{\mathrm{\mid }}^2+2\vec{a}\cdot \vec{b}$∣a+b∣2=∣a∣2+∣b∣2+2a⋅b

类型2：证明垂直/平行

• 垂直：证$\vec{a}\cdot \vec{b}=0$a⋅b=0
• 平行：证$\vec{a}=k\vec{b}$a=kb

类型3：求最值

利用$\mathrm{\mid }\vec{a}\cdot \vec{b}\mathrm{\mid }\le \mathrm{\mid }\vec{a}\mathrm{\mid }\mathrm{\mid }\vec{b}\mathrm{\mid }$∣a⋅b∣≤∣a∣∣b∣（柯西不等式）

类型4：几何问题

建立坐标系，将几何条件转化为向量运算

---

六、快速记忆口诀

1. 加法：首尾相连，起点到终点
2. 减法：共起点，指向被减数
3. 点积为零：两向量垂直
4. 数乘：伸缩方向看正负
5. 中点：两向量相加除以2
6. 重心：三向量相加除以3

---

七、易错点提醒

⚠️ 向量$\mathrm{\ne }$=数量（向量有方向）
⚠️ $\vec{a}\cdot \vec{b}=0$a⋅b=0不一定$\vec{a}=\vec{0}$a=0或$\vec{b}=\vec{0}$b=0（可能垂直）
⚠️ 向量不能比较大小（只能比较模长）
⚠️ $\vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c}$a⋅b=a⋅c不能推出$\vec{b}=\vec{c}$b=c