$$\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}$$
-Scalar resolute of $\vec{a}$ on $\vec{b} = |\vec{u}| = \vec{a} \cdot \hat{\vec{b}}$
+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 ($\implies \vec{OC} = \lambda \vec{OA} + \mu \vec{OB}$ where $\lambda + \mu = 1$. If $C$ is between $\vec{AB}$, then $0 \lt \mu \lt 1$)
+**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: