Geometric interpretation of the scalar product
Find a vector $\vec{v}$ which is orthogonal (perpendicular) to the vector $\vec{u}=(2,-4)$which is orthogonal (perpendicular) to the vector $3$. Find the orthogonal projection of $\vec{u}$ on $\vec{v}$.
We want to find a vector $\vec{v}=(v_1,v_2)$ such that its norm is $3$, that is $$ |\vec{v}|=\sqrt{v_1^2+v_2^2}=3 \Rightarrow |\vec{v}|=v_1^2+v_2^2=9$$
and that $\vec{u}\cdot\vec{v}=0$ (we impose perpendicularity): $$ u_1 v_1+u_ 2 v_2=0 \Rightarrow 2v_1+(-4)v_2=0 \Rightarrow v_1=2v_2$$
By substituting $v_1=2v_2$ in the first equality, we obtain: $$ 4v_2^2+v_2^2=5v_2^2=9 \Rightarrow v_2^2=\dfrac{9}{5} \Rightarrow v_2=\dfrac{3}{\sqrt{5}}, \ v_1=\dfrac{6}{\sqrt{5}}$$
To obtain the desired orthogonal projection we use the formula: $\vec{u}\cdot\vec{v}=|\vec{v}|\text{proj}_{\vec{v}}(\vec{u})$. In our case we have:
$$\text{proj}_{\vec{v}}(\vec{u})=\dfrac{\vec{u}\cdot\vec{v}}{|\vec{v}|}= \dfrac{2\cdot\dfrac{6}{\sqrt{5}}+(-4)\cdot\dfrac{3}{\sqrt{5}}}{3}= \dfrac{\dfrac{12}{\sqrt{5}}-\dfrac{12}{\sqrt{5}}}{3}=0$$
We could also have thought that since the vectors are perpendicular, the projection of one onto the other must be zero.
$v_2=\dfrac{3}{\sqrt{5}}$ and $v_1=\dfrac{6}{\sqrt{5}} \quad$ $\text{proj}_{\vec{v}}(\vec{u})=0$