The product of two non zero vectors is equal to the magnitude of one of them times the projection of the other onto it.
In the picture, $$OA'$$ is the projection of the vector $$\vec{u}$$ on $$\vec{v}$$. If we observe the $$OAA'$$ triangle and apply the cosinus definition, we have:
$$$\cos(\alpha)=\dfrac{OA'}{|\vec{u}|} \Rightarrow |OA'| = |\vec{u}|\cdot \cos(\alpha) $$$
Finally, applying to the scalar product formula what we just have found: $$$ \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\alpha)=|\vec{u}|\text{proj}_{\vec{v}}(\vec{u})$$$
Find the projection of the vector $$\vec{u}=(2,3)$$ on $$\vec{v}=(-1,4)$$. $$$ \text{proj}_{\vec{v}}(\vec{u})=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|}= \dfrac{2\cdot (-1)+3\cdot 4}{\sqrt{(-1)^2+4^2}}=\dfrac{10}{\sqrt{17}}$$$