Problem:
For vectors $\mathbf{u}$ and $\mathbf{v}$, we have $ \mathbf{p} = \text{proj}_{\mathbf{v}}(\mathbf{u})$. If $\|\mathbf{u}\| = 11$ and $\|\mathbf{p}\| = 6$, find $\mathbf{p}\cdot\mathbf{u}$.
I don't understand how to do this problem because, to me, it seems like there isn't enough information, but I know that there is. I know that any vector and its projection either form the hypotenuse and a leg of a right triangle, or that they are equal. So I can figure out the cosine of the angle between $\mathbf{u}$ and $\mathbf{p}$. But, I don't know how exactly I would do that? Sorry, I'm just really confused. Any advice? Below is what one of my friends told me to do, but I don't get it.
$$ \mathbf{p} = \text{proj}_{\mathbf{v}}(\mathbf{u}) = \| \mathbf{u} \cos(\theta) \| \, \hat{v}$$
or that
$$\begin{align} \|\mathbf{p}\| &= \|\mathbf{u}\|\cos(\theta) \\ \cos(\theta) &= \frac{\|\mathbf{p}\|}{\|\mathbf{u}\|} \end{align}$$
Remember that the formula for vector projection is $$\mathrm{proj}_{\mathbf{v}}(\mathbf{u}) =\mathbf{p}=\left(\frac{\mathbf{u} \boldsymbol{\cdot} \mathbf{v}}{\| \mathbf{v} \| ^{2}}\right)\mathbf{v}$$ Supposing $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$, this can be written as $$\mathbf{p}=\left(\frac{\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos( \theta )}{\| \mathbf{v} \| ^{2}}\right)\mathbf{v}=\left(\frac{\Vert \mathbf{u}\Vert \cos( \theta )}{\| \mathbf{v} \| }\right)\mathbf{v}$$
Therefore $$\Vert \mathbf{p}\Vert =\left(\frac{\Vert \mathbf{u}\Vert \cos( \theta )}{\| \mathbf{v} \| }\right)\Vert \mathbf{v}\Vert =\Vert \mathbf{u}\Vert \cos( \theta )$$ Now, since $\mathbf{v}$ and $\mathrm{proj}_{\mathbf{v}}(\mathbf{u})$ point in the same direction, the angle between $\mathbf{u}$ and $\mathbf{p}$ is also $\theta$, therefore $$\mathbf{u} \boldsymbol{\cdot} \mathbf{p} =\Vert \mathbf{u}\Vert \Vert \mathbf{p}\Vert \cos( \theta ) =\Vert \mathbf{p}\Vert ^{2} .$$