1. Vector Addition

Definition (Vector). $\boldsymbol{a},\boldsymbol{b}\quad\boldsymbol{a}:=\overrightarrow{AB}\quad\boldsymbol{b}:=\overrightarrow{BC}\quad\boldsymbol{c}:=\overrightarrow{AC}$

$\boldsymbol{c}=\boldsymbol{a}+\boldsymbol{b}$. If $|\boldsymbol{a}|=0$ then $\boldsymbol{a}=\vec{0}$. $-\boldsymbol{a}$ has opposite direction with $\boldsymbol{a}$ and $|-\boldsymbol{a}|=|\boldsymbol{a}|$.

Rules of Vector Addition

1. Associativity $$(\boldsymbol{a}+\boldsymbol{b})+\boldsymbol{c}=\boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c})\quad\forall\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$$

2. Commutativity $$\boldsymbol{a}+\boldsymbol{b}=\boldsymbol{b}+\boldsymbol{a}\quad\forall\boldsymbol{a},\boldsymbol{b}$$

3. Identity $$\forall\boldsymbol{a}\quad\boldsymbol{a}+\vec{0}=\boldsymbol{a}$$

4. Inverse $$\forall\boldsymbol{a}\quad\boldsymbol{a}+(-\boldsymbol{a})=\vec{0}$$

Proofs

Proof of Associativity ($1^\circ$). $$\boldsymbol{a}:=\overrightarrow{OA}\quad\boldsymbol{b}:=\overrightarrow{AB}\quad\boldsymbol{c}:=\overrightarrow{BC}$$ $$\boldsymbol{a}+\boldsymbol{b}=\overrightarrow{OB}\quad(\boldsymbol{a}+\boldsymbol{b})+\boldsymbol{c}=\overrightarrow{OC}$$ $$\boldsymbol{b}+\boldsymbol{c}=\overrightarrow{AC}\quad\boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c})=\overrightarrow{OC}$$ So $(\boldsymbol{a}+\boldsymbol{b})+\boldsymbol{c}=\boldsymbol{a}+(\boldsymbol{b}+\boldsymbol{c})$.

Proof of Commutativity ($2^\circ$). Set $\boldsymbol{a}:=\overrightarrow{OA}\quad\boldsymbol{b}:=\overrightarrow{OB}$ and let $OABC$ be a parallelogram. Then $$\boldsymbol{a}+\boldsymbol{b}=\boldsymbol{b}+\boldsymbol{a}=\overrightarrow{OC}$$

Proof of Identity ($3^\circ$). $$\boldsymbol{a}:=\overrightarrow{OA}\quad\vec{0}:=\overrightarrow{AA}$$ $$\boldsymbol{a}+\vec{0}=\overrightarrow{OA}=\boldsymbol{a}$$

Proof of Inverse ($4^\circ$). $$\boldsymbol{a}:=\overrightarrow{OA}\quad-\boldsymbol{a}:=\overrightarrow{AO}$$ $$\boldsymbol{a}+(-\boldsymbol{a})=\overrightarrow{OO}=\vec{0}$$

2. Vector Subtraction

Definition (Vector Subtraction). $\boldsymbol{a}-\boldsymbol{b}=\boldsymbol{a}+(-\boldsymbol{b})$

Let $\boldsymbol{a}:=\overrightarrow{OA}\quad\boldsymbol{b}:=\overrightarrow{OB}$. Then $$\boldsymbol{a}-\boldsymbol{b}=\overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BO}+\overrightarrow{OA}=\overrightarrow{BA}$$

Since $ABO$ is a triangle, $$|\overrightarrow{BA}|\leq|\overrightarrow{OA}|+|\overrightarrow{OB}|$$ $$\Leftrightarrow|\boldsymbol{a}-\boldsymbol{b}|\leq|\boldsymbol{a}|+|\boldsymbol{b}|$$

Set $\boldsymbol{c}:=-\boldsymbol{b}$. Then $|\boldsymbol{a}+\boldsymbol{c}|\leq|\boldsymbol{a}|+|\boldsymbol{c}|$ is the triangle inequality.

3. Scalar Multiplication

Definition (Scalar Multiplication). $|\lambda\boldsymbol{a}|:=|\lambda|\cdot|\boldsymbol{a}|$

  • When $\lambda>0$, the direction of $\lambda\boldsymbol{a}$ is the same as $\boldsymbol{a}$.
  • When $\lambda<0$, the direction of $\lambda\boldsymbol{a}$ is opposite to $\boldsymbol{a}$.

Special cases: $$\forall\boldsymbol{a}\quad 0\boldsymbol{a}=0\cdot|\boldsymbol{a}|=0 \;\Longrightarrow\; 0\boldsymbol{a}=\vec{0}$$ $$\forall\lambda\quad \lambda\vec{0}=|\lambda||\vec{0}|=0 \;\Longrightarrow\; \lambda\vec{0}=\vec{0}$$

Unit vector. For $\boldsymbol{a}\neq\vec{0}$, $$\frac{\boldsymbol{a}}{|\boldsymbol{a}|}$$ has the same direction as $\boldsymbol{a}$, and $$\left|\frac{\boldsymbol{a}}{|\boldsymbol{a}|}\right|=\frac{1}{|\boldsymbol{a}|}\cdot|\boldsymbol{a}|=1$$ So $\boldsymbol{a}^\circ=\frac{\boldsymbol{a}}{|\boldsymbol{a}|}$ is the unit vector parallel to $\boldsymbol{a}$.

Rules of Scalar Multiplication

1. $1\boldsymbol{a}=\boldsymbol{a},\quad(-1)\boldsymbol{a}=-\boldsymbol{a}$

2. $\lambda(\mu\boldsymbol{a})=(\lambda\mu)\boldsymbol{a}$

3. $(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$

4. $\lambda(\boldsymbol{a}+\boldsymbol{b})=\lambda\boldsymbol{a}+\lambda\boldsymbol{b}$

Proofs

Proof of Rule 1 ($1^\circ$). $1>0$, so $1\boldsymbol{a}$ and $\boldsymbol{a}$ have the same direction. $|1\boldsymbol{a}|=1\cdot|\boldsymbol{a}|=|\boldsymbol{a}|$, so $1\boldsymbol{a}=\boldsymbol{a}$.

$-1<0$, so $(-1)\boldsymbol{a}$ and $-\boldsymbol{a}$ have the same direction. $|(-1)\boldsymbol{a}|=|-1||\boldsymbol{a}|=|\boldsymbol{a}|=|-\boldsymbol{a}|$, so $(-1)\boldsymbol{a}=-\boldsymbol{a}$.

Proof of Rule 2 ($2^\circ$).

If $\lambda\cdot\mu>0$, then $(\lambda\mu)\boldsymbol{a}$ has the same direction as $\boldsymbol{a}$, i.e. $\lambda,\mu>0$ or $\lambda,\mu<0$ implies $\lambda(\mu\boldsymbol{a})$ has the same direction as $\boldsymbol{a}$.

If $\lambda\cdot\mu<0$, then $(\lambda\mu)\boldsymbol{a}$ has the opposite direction to $\boldsymbol{a}$, i.e. $\lambda>0,\mu<0$ or $\lambda<0,\mu>0$ implies $\lambda(\mu\boldsymbol{a})$ has the opposite direction.

If $\lambda\cdot\mu=0$, then $\lambda=\vec{0}$ or $\mu=0$ or $\lambda=\mu=0$, and $\lambda(\mu\boldsymbol{a})=(\lambda\mu)\boldsymbol{a}=\vec{0}$.

In all cases, $\lambda(\mu\boldsymbol{a})$ and $(\lambda\mu)\boldsymbol{a}$ have the same direction, and $$|\lambda(\mu\boldsymbol{a})|=|\lambda|\cdot|\mu\boldsymbol{a}|=|\lambda|\cdot|\mu|\cdot|\boldsymbol{a}|=|\lambda\mu|\cdot|\boldsymbol{a}|=|(\lambda\mu)\cdot\boldsymbol{a}|$$ So $\lambda(\mu\boldsymbol{a})=(\lambda\mu)\boldsymbol{a}$.

Proof of Rule 3 ($3^\circ$). $(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$

If $\lambda=\vec{0}$ or $\mu=0$ or $\boldsymbol{a}=\vec{0}$, equality holds trivially.

Case 1. $\lambda,\mu$ have the same sign. Then $\lambda+\mu$ has the same sign as $\lambda,\mu$, so $(\lambda+\mu)\boldsymbol{a}$ and $\lambda\boldsymbol{a}+\mu\boldsymbol{a}$ have the same direction. $$|(\lambda+\mu)\boldsymbol{a}|=|\lambda+\mu||\boldsymbol{a}|=(|\lambda|+|\mu|)|\boldsymbol{a}|$$ $$=|\lambda||\boldsymbol{a}|+|\mu||\boldsymbol{a}| =|\lambda\boldsymbol{a}|+|\mu\boldsymbol{a}|$$ Since $\lambda\boldsymbol{a}$ and $\mu\boldsymbol{a}$ have the same direction, $$=|\lambda\boldsymbol{a}+\mu\boldsymbol{a}|$$ So $(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$.

Case 2. $\lambda,\mu$ have different signs. Let $\lambda>0,\mu<0$.

(1) $\lambda+\mu=0$. Then $\mu=-\lambda$ and $$(\lambda+\mu)\boldsymbol{a}=\vec{0}=\lambda\boldsymbol{a}-\lambda\boldsymbol{a}$$ So $(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$.

(2) $\lambda+\mu>0$. Since $\lambda+\mu>0$ and $-\mu>0$, use Case 1: $$(\lambda+\mu-\mu)\boldsymbol{a}=(\lambda+\mu)\boldsymbol{a}-\mu\boldsymbol{a}$$ $$\lambda\boldsymbol{a}=(\lambda+\mu)\boldsymbol{a}-\mu\boldsymbol{a}$$ $$(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$$

(3) $\lambda+\mu<0$, $-\lambda<0$: $$(\lambda+\mu-\lambda)\boldsymbol{a}=(\lambda+\mu)\boldsymbol{a}-\lambda\boldsymbol{a}$$ $$(\lambda+\mu)\boldsymbol{a}=\lambda\boldsymbol{a}+\mu\boldsymbol{a}$$

Proof of Rule 4 ($4^\circ$). $\lambda(\boldsymbol{a}+\boldsymbol{b})=\lambda\boldsymbol{a}+\lambda\boldsymbol{b}$

If $\lambda=\vec{0}$ or $\boldsymbol{a}=0$ or $\boldsymbol{b}=0$, equality holds trivially.

Case 1. $\boldsymbol{a}\parallel\boldsymbol{b}$. Then $\exists\mu\in\mathbb{R},\boldsymbol{a}=\mu\boldsymbol{b}$. $$\lambda(\boldsymbol{a}+\boldsymbol{b})=\lambda(\mu\boldsymbol{b}+\boldsymbol{b})$$ $$=\lambda(1+\mu)\boldsymbol{b} \quad\text{(using Rule 3)}$$ $$=(\lambda+\lambda\mu)\boldsymbol{b}=\lambda\boldsymbol{b}+\lambda\mu\boldsymbol{b}=\lambda\boldsymbol{a}+\lambda\boldsymbol{b}$$

Case 2. $\boldsymbol{a}\nparallel\boldsymbol{b}$. $$\boldsymbol{a}:=\overrightarrow{OA}\quad\boldsymbol{b}:=\overrightarrow{AB}$$ $$\boldsymbol{a}+\boldsymbol{b}=\overrightarrow{OB}$$

Set $\overrightarrow{OC}=\lambda\boldsymbol{a}\quad\overrightarrow{CD}=\lambda\boldsymbol{b}$. Then $$\lambda\boldsymbol{a}+\lambda\boldsymbol{b}=\overrightarrow{OD}$$

$\triangle OAB\sim\triangle OCD$, so $$\lambda\overrightarrow{OB}=\overrightarrow{OD}=\lambda(\boldsymbol{a}+\boldsymbol{b})=\lambda\boldsymbol{a}+\lambda\boldsymbol{b}$$

4. Linear Combinations

Definition (Linear Combination). Let $\boldsymbol{a}_1,\cdots,\boldsymbol{a}_n$ be a set of vectors and $k_1,\cdots,k_n\in\mathbb{R}$. Then the vector $$k_1\boldsymbol{a}_1+\cdots+k_n\boldsymbol{a}_n$$ is called a linear combination of $\boldsymbol{a}_1,\cdots,\boldsymbol{a}_n$, and $k_1,\cdots,k_n$ are the coefficients of the linear combination.

5. Collinearity and Coplanarity

Definition (Collinear & Coplanar). If a set of vectors are represented by directed line segments emanating from the same initial point and all lie on a straight line (respectively, a plane), then the vector set is collinear (respectively, coplanar).

  • If $\boldsymbol{a},\boldsymbol{b}$ are collinear, then $\boldsymbol{a}\parallel\boldsymbol{b}$.
  • $\boldsymbol{a}=\lambda\boldsymbol{b}\iff\boldsymbol{a}\parallel\boldsymbol{b}$.

Propositions

Prop 1.1. If $\boldsymbol{a},\boldsymbol{b}$ are collinear and $\boldsymbol{a}\neq\vec{0}$, then there exists a unique $\lambda\in\mathbb{R}$ such that $\boldsymbol{b}=\lambda\boldsymbol{a}$.

Existence. If $\boldsymbol{a},\boldsymbol{b}$ have the same direction, $$\frac{\boldsymbol{a}}{|\boldsymbol{a}|}=\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\;\Longrightarrow\;\boldsymbol{b}=\frac{|\boldsymbol{b}|}{|\boldsymbol{a}|}\boldsymbol{a}$$ If $\boldsymbol{a},\boldsymbol{b}$ have opposite directions, the argument is similar.

Uniqueness. Assume $\boldsymbol{b}=\lambda\boldsymbol{a}=\mu\boldsymbol{a}$. Then $(\lambda-\mu)\boldsymbol{a}=0$. Since $\boldsymbol{a}\neq\vec{0}$, $\lambda-\mu=0\Rightarrow\lambda=\mu$.

Prop 1.2. $\boldsymbol{a},\boldsymbol{b}$ are collinear iff there exist nonzero $\lambda,\mu\in\mathbb{R}$ such that $\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\vec{0}$.

$\Rightarrow$ If $\boldsymbol{a}=\boldsymbol{b}=\vec{0}$, then $1\cdot\boldsymbol{a}+1\cdot\boldsymbol{b}=\vec{0}$. If $\boldsymbol{a},\boldsymbol{b}$ are nonzero, assume $\boldsymbol{a}\neq\vec{0}$. By Prop 1.1, $\exists\lambda\in\mathbb{R}$ such that $\boldsymbol{b}=\lambda\boldsymbol{a}$, so $\lambda\boldsymbol{a}-\boldsymbol{b}=\vec{0}$.

$\Leftarrow$ Assume $\lambda\neq0$. Then $\lambda\boldsymbol{a}=-\mu\boldsymbol{b}$, so $\boldsymbol{a}=-\frac{\mu}{\lambda}\boldsymbol{b}$. Hence $\boldsymbol{a},\boldsymbol{b}$ are collinear.

Corollary 1.1. $\boldsymbol{a},\boldsymbol{b}$ are noncollinear iff $$\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\vec{0}\Rightarrow\lambda=\mu=0$$

Proof. Negating Prop 1.2: $$\neg(\exists\text{ nonzero }\lambda,\mu\in\mathbb{R},\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\vec{0})$$ $$\Leftrightarrow\forall\text{ nonzero }\lambda,\mu\in\mathbb{R},\lambda\boldsymbol{a}+\mu\boldsymbol{b}\neq\vec{0}$$ $$\Leftrightarrow\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\vec{0}\Rightarrow\lambda=\mu=0$$

Prop 1.3. If $\boldsymbol{c}=\lambda\boldsymbol{a}+\mu\boldsymbol{b}$, then $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are coplanar.

Proof. If $\boldsymbol{a}\parallel\boldsymbol{b}$, then $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are collinear, hence coplanar. If $\boldsymbol{a}\nparallel\boldsymbol{b}$, let $\lambda\boldsymbol{a}:=\overrightarrow{OA}$, $\mu\boldsymbol{b}:=\overrightarrow{AB}$. Then $\boldsymbol{c}=\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\overrightarrow{OB}$, so $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ lie on the plane $OAB$.

Prop 1.4. If $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are coplanar and $\boldsymbol{a},\boldsymbol{b}$ are noncollinear, then there exist unique $\lambda,\mu\in\mathbb{R}$ such that $\boldsymbol{c}=\lambda\boldsymbol{a}+\mu\boldsymbol{b}$.

Existence. Set $\boldsymbol{a}:=\overrightarrow{OA}$, $\boldsymbol{b}:=\overrightarrow{OB}$, $\boldsymbol{c}:=\overrightarrow{OC}$. Draw $CD\parallel OB$ with $D$ on line $OA$. Since $OA\parallel OD$ and $CD\parallel OB$, there exist $\lambda,\mu\in\mathbb{R}$ such that $\overrightarrow{OD}=\lambda\boldsymbol{a}$ and $\overrightarrow{DC}=\mu\boldsymbol{b}$. Hence $\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\boldsymbol{c}$.

Uniqueness. Assume $\boldsymbol{c}=\lambda_1\boldsymbol{a}+\mu_1\boldsymbol{b}=\lambda_2\boldsymbol{a}+\mu_2\boldsymbol{b}$. Then $(\lambda_1-\lambda_2)\boldsymbol{a}+(\mu_1-\mu_2)\boldsymbol{b}=0$. By Cor 1.1, $\lambda_1-\lambda_2=0$ and $\mu_1-\mu_2=0$, so $\lambda_1=\lambda_2$ and $\mu_1=\mu_2$.

Prop 1.5. $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are coplanar iff there exist nonzero $k_1,k_2,k_3\in\mathbb{R}$ such that $k_1\boldsymbol{a}+k_2\boldsymbol{b}+k_3\boldsymbol{c}=\vec{0}$.

$\Rightarrow$ If $\boldsymbol{a}\nparallel\boldsymbol{b}$, by Prop 1.4 there exist $\lambda,\mu\in\mathbb{R}$ with $\boldsymbol{c}=\lambda\boldsymbol{a}+\mu\boldsymbol{b}$. Then $\lambda\boldsymbol{a}+\mu\boldsymbol{b}-\boldsymbol{c}=\vec{0}$ (note $-1\neq0$). If $\boldsymbol{a}\parallel\boldsymbol{b}$, there exist nonzero $\lambda,\mu$ with $\lambda\boldsymbol{a}+\mu\boldsymbol{b}=\vec{0}$ (Prop 1.2). Then $\lambda\boldsymbol{a}+\mu\boldsymbol{b}+0\cdot\boldsymbol{c}=\vec{0}$.

$\Leftarrow$ Assume $k_1\neq0$. Then $\boldsymbol{a}=-\frac{k_2}{k_1}\boldsymbol{b}-\frac{k_3}{k_1}\boldsymbol{c}$, so $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are coplanar.

Corollary 1.2. $\boldsymbol{a},\boldsymbol{b},\boldsymbol{c}$ are noncoplanar iff $$k_1\boldsymbol{a}+k_2\boldsymbol{b}+k_3\boldsymbol{c}=\vec{0}\Rightarrow k_1=k_2=k_3=0$$

Proof. Negating Prop 1.5 (analogous to Cor 1.1).

6. Exercises

Ex 1. Prove that a point $M$ lies on the line segment $AB$ iff there exist nonnegative $\lambda,\mu\in\mathbb{R}$ such that $$\overrightarrow{OM}=\lambda\overrightarrow{OA}+\mu\overrightarrow{OB}\quad\text{and}\quad\lambda+\mu=1$$ where $O$ is an arbitrary point.

Proof ($\Rightarrow$). $M$ on $AB$ implies $\overrightarrow{AM}\parallel\overrightarrow{AB}$ and $0\leq|\overrightarrow{AM}|\leq|\overrightarrow{AB}|$. So $\overrightarrow{AM}=k\overrightarrow{AB}$ with $0\leq k\leq1$. $$\overrightarrow{OM}-\overrightarrow{OA}=k(\overrightarrow{OB}-\overrightarrow{OA})$$ $$\overrightarrow{OM}=(1-k)\overrightarrow{OA}+k\overrightarrow{OB}$$ Take $\lambda=1-k$, $\mu=k$. Then $\lambda+\mu=1$ and $\lambda,\mu\geq0$.

Proof ($\Leftarrow$). $$\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=(\lambda-1)\overrightarrow{OA}+\mu\overrightarrow{OB}=\mu\overrightarrow{OB}-\mu\overrightarrow{OA}=\mu(\overrightarrow{OB}-\overrightarrow{OA})$$ $$=\mu\overrightarrow{AB}\quad\text{with}\quad 0\leq\mu\leq1$$ Hence $M$ lies on the line segment $AB$.

Ex 2. Prove that $A,B,C$ are collinear iff there exist nonzero $\lambda,\mu,\nu\in\mathbb{R}$ such that $$\lambda\overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}=\vec{0}\quad\text{and}\quad\lambda+\mu+\nu=0$$ where $O$ is an arbitrary point.

Proof ($\Rightarrow$). $A,B,C$ collinear $\Rightarrow\overrightarrow{AB}\parallel\overrightarrow{AC}$. So there exist nonzero $k,\ell\in\mathbb{R}$ with $$k\overrightarrow{AB}+\ell\overrightarrow{AC}=\vec{0}$$ $$k(\overrightarrow{OB}-\overrightarrow{OA})+\ell(\overrightarrow{OC}-\overrightarrow{OA})=\vec{0}$$ $$(-k-\ell)\overrightarrow{OA}+k\overrightarrow{OB}+\ell\overrightarrow{OC}=\vec{0}$$ Take $\lambda=-k-\ell$, $\mu=k$, $\nu=\ell$. Then $\lambda+\mu+\nu=0$.

Proof ($\Leftarrow$). $\lambda=-\mu-\nu$, so $$(-\mu-\nu)\overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}=\vec{0}$$ $$\mu(\overrightarrow{OB}-\overrightarrow{OA})+\nu(\overrightarrow{OC}-\overrightarrow{OA})=\vec{0}$$ $$\mu\overrightarrow{AB}+\nu\overrightarrow{AC}=\vec{0}$$ If $\mu=\nu=0$, then $\lambda=\mu=\nu=0$, contradicting nonzeroness. So $\mu,\nu$ are not both zero, hence $\overrightarrow{AB}\parallel\overrightarrow{AC}$ and $A,B,C$ are collinear.

References

Qiu, W. S. Analytic Geometry. Peking University Press.