# Question #46110

Apr 17, 2017

I think there might be something wrong with part (i) of your question.

Let's assume (taking a very specific case) in 2-D Cartesian:

$\vec{u} = {\left(0 , 2\right)}^{T}$, $\vec{v} = {\left(0 , 1\right)}^{T}$ and $\vec{a} = {\left(1 , 0\right)}^{T}$ Assuming, in the usual course, that your notation means that:

$P r o {j}_{a} \vec{u} = P r o {j}_{a} \vec{v} \implies \vec{u} \cdot \hat{a} = \vec{v} \cdot \hat{a}$

...where $\hat{a}$ is the unit $\vec{a}$ vector , then we have:

$\vec{u} \cdot \hat{a} = \left(0 , 2\right) \left(\begin{matrix}1 \\ 0\end{matrix}\right) = 0$

$\vec{v} \cdot \hat{a} = \left(0 , 1\right) \left(\begin{matrix}1 \\ 0\end{matrix}\right) = 0$

So $P r o {j}_{a} \vec{u} = P r o {j}_{a} \vec{v} = 0$, which is your requirement in this very specific case.

But:

$\vec{u} \cdot \vec{v} = \left(0 , 2\right) \left(\begin{matrix}0 \\ 1\end{matrix}\right) = 2$

And:

$\vec{v} \cdot \vec{a} = \left(0 , 1\right) \left(\begin{matrix}1 \\ 0\end{matrix}\right) = 0$

So:

$\vec{u} \cdot \vec{v} \ne \vec{v} \cdot \vec{a}$

The second bit (ie: part(ii)) is actually a lot easier to show once you add on the necessary conditions.