$$\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2$$
+**on CAS:** `dotP([a b c], [d e f])`
+
## Scalar product properties
1. $k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k{b})$
2. $\boldsymbol{a \cdot 0}=0$
3. $\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}$
+4. $\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1$
+5. If $\boldsymbol{a} \cdot \boldsymbol{b} = 0$, $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular
+6. $\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2$
For parallel vectors $\boldsymbol{a}$ and $\boldsymbol{b}$:
$$\boldsymbol{a \cdot b}=\begin{cases}
## Finding angle between vectors
+**positive direction**
+
$$\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}$$
+**on CAS:** `angle([a b c], [a b c])` (Action -> Vector -> Angle)
+
## Vector projections
$$\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}$$
+Scalar resolute of $\vec{a}$ on $\vec{b} = |\vec{u}| = \vec{a} \cdot \hat{\vec{b}}$ (results in a scalar)
+Vector resolute of $\vec{a}$ perpendicular to $b$ is equal to $\vec{a} - \vec{u}$ where $\vec{u}$ is vector projection of $\vec{a}$ on $\vec{b}$
+
## Vector proofs
**Concurrent lines -** $\ge$ 3 lines intersect at a single point
-**Collinear points -** $\ge$ 3 points lie on the same line
+**Collinear points -** $\ge$ 3 points lie on the same line ($\implies \vec{OC} = \lambda \vec{OA} + \mu \vec{OB}$ where $\lambda + \mu = 1$. If $C$ is between $\vec{AB}$, then $0 < \mu < 1$)
Useful vector properties:
For $\vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}$ which makes angles $\alpha, \beta, \gamma$ with positive direction of $x, y, z$ axes:
$$\cos \alpha = {a_1 \over |\vec{a}|}, \quad \cos \beta = {a_2 \over |\vec{a}|}, \quad \cos \gamma = {a_3 \over |\vec{a}|}$$
+**on CAS:** `angle([a b c], [1 0 0])` for angle between $a\vec{i} + b\vec{j} + c\vec{k}$ and $x$-axis
+
+## Collinearity
+
+Points $A, B, C$ are collinear iff $\vec{AC}=m\vec{AB} \text{ where } m \ne 0$