$$\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}$$
-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