inspecting u. v = u.w If the components of u, v and w isn't given is there anything we can learn about the relationship between u and v (not w) from the equation u.v = u.w?
 A: If you know nothing about any of the three vectors, then no. You do not know anything about $v$, and its relation to $u$, because for example, you could have $w=v$ in which case the equation would be trivially true.
All you really know is that, since $u(v-w)=0$, is that $v-w$ must be orthogonal to $u$.

If you know that $\pi_v(w)=\pi_u(w)$ ($\pi_v$ is the projection onto the subspace generated by $v$), then you do know a couple of things:


*

*If $w=0$, then the projection inequality tells you nothing, but in that case you know that $v-w=v$ is orthogonal to $u$, so you know $u,v$ are orthogonal.

*If $w\neq 0$, then you know that $v$ and $u$ are colinear (because $\pi_v(w)=\lambda v$ for some $\lambda$ and $\pi_u(w)=\mu u$ for some $u$).

A: Since scalar product is distributive, we get
$$
\mathbf{u\cdot v = u\cdot w}\\
\mathbf{u\cdot v-u\cdot w = 0}\\
\mathbf{u\cdot(v-w) = 0}
$$
which means that $\mathbf{v-w}$ is orthogonal to $\mathbf u$. We can't get any more information, because the first equation is fulfilled by any two $\mathbf v, \mathbf w$ such that $\mathbf{v-w}$ is orthogonal to $\mathbf u$.
Geometrically this means that the normal line / plane (depending on dimension) of $\mathbf u$ that contains $\mathbf v$ also contains $\mathbf w$.
