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 $$\vec{v}$$.
Development:
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.
Solution:
$$v_2=\dfrac{3}{\sqrt{5}}$$ and $$v_1=\dfrac{6}{\sqrt{5}} \quad$$ $$\text{proj}_{\vec{v}}(\vec{u})=0$$