given these vectors: $$ v=\begin{pmatrix} 2 \\ -4 \\ 2 \\ \end{pmatrix} , u = \begin{pmatrix} -4 \\ 1 \\ 3 \\ \end{pmatrix} $$
How can I find vector $w$ that his projection about $span(v)$ is $7v$ and his projection about $span(u)$ is $-8u$ ?
I don't have idea how to do it and I will be happy for help or hints for how to do it.
Hint: the projection of $\,w\,$ onto $\,\operatorname{span}(u)\,$ is $\,\dfrac{w \cdot u}{|u|^2}\,u = -8u\,$, so $\,w \cdot u = -8 |u|^2\,$, and similarly $\,w \cdot v = 7 |v|^2\,$.
Let $\;n=\begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}\,$, then it's easily verified that $n \cdot u = n \cdot v = 0\,$, so $\,\operatorname{span}(u,v,n)=\mathbb{R}^3\,$. Then let $\,w = \lambda u + \mu v + \nu n\,$ and solve the given conditions for $\,\lambda,\mu\,$:
$$\require{cancel} \begin{align} -8 |u|^2 = w \cdot u &= \lambda |u|^2 + \mu \,(u \cdot v) + \bcancel{\nu \, (n \cdot u)} \\ 7 |v|^2 = w \cdot v &= \lambda \,(u \cdot v) + \mu |v|^2 + \bcancel{\nu \, (n \cdot v)} \end{align} $$
Note that $\,\nu\,$ remains undetermined, since the component orthogonal to $\,\operatorname{span}(u,v)\,$ does not affect the projections onto $\operatorname{span}(u)\,$ and $\,\operatorname{span}(v)$.