Why isn't projection of $\mathbf{v}$ onto $\mathbf{u}$ defined so that it's perpendicular to the straight line segment connecting $\mathbf{u, v}$? Kuldeep Singh. Linear Algebra: Step by Step
(2013). p 312
Why isn't $\mathbf{proj_u v}$ defined as the red vector below with $\mathbf{v} \perp \mathbf{d}?$ I already know, but can't intuit why, $\mathbf{proj_u v}$ is defined as the green vector below with $\mathbf{p} \perp \mathbf{u}$.



 A: It's so we can reconstruct the original vector easily from its projections. If we have an orthonormal basis $B=\{b_1,\dots, b_n\}$ of a vector space $V$, and any vector $v\in V$, then we have
$$v=\sum_{i=1}^n \operatorname{proj}_{b_i}(v).$$
Essentially, $\operatorname{proj}_uv$ is the "$u$-component" of $v$. For instance, a vector $v=(1,2,3)\in\mathbb R^3$ can be split into its $x$-, $y$-, and $z$-components by projecting it onto the corresponding unit vectors $e_x,e_y,e_z$ and then reconstructed as
$$\operatorname{proj}_{e_x}(v)+\operatorname{proj}_{e_y}(v) +\operatorname{proj}_{e_z}(v)=1e_x+2e_y+3e_z=(1,2,3)=v.$$
This is not as easily done with your way of projecting.
A: By definition, projection of $\vec v$ onto $\vec u$ is vector $\vec w$ parallel to $\vec u$ such that
$$\vec v=\vec w+\vec p$$
therefore, as an intuition, $\vec v$ is the longest vector in the triple.
Indeed, in your alternative sketch, $\vec v$ represents the projection of the red vector onto $\vec v$ direction.
A: The orthogonal parts can be added to give us $u + d = v$ for the canonical version and $v + d = u$ for the version you've drawn in the red box. Note that all you've done is switch the roles of $u$ and $v$ and relabeling the variables doesn't add any additional insight into the problem.
