Projection of v onto orthogonal subspaces are the those with minmum distance to v? This question is similar to my previous question, but adds an additional condition that the subspaces are orthogonal.
Let $V$ be a vector space over field $F$, and $V$ can be decomposed as a direct sum $V = {V_1} \oplus {V_2}$ of two subspaces $V_1$ and $V_2$, then any vector $\bf v$ can be uniquely written as ${\bf v}={\bf v}_1+{\bf v}_2$ for some ${\bf v}_1\in V_1$, ${\bf v}_2\in V_2$. We define ${\bf v}_1$ as the projection of $\bf v$ into subspace $V_1$, denoted by ${\text{pro}}{{\text{j}}_{{V_1}}}{\mathbf{v}} = {{\mathbf{v}}_1}$. 
The projection is unique since the direct sum decomposition of $\bf v$ is unique.
My question is if the following is true:

If $ {V_1} \bot {V_2}$, then ${\text{pro}}{{\text{j}}_{{V_1}}}{\mathbf{v}} = {\arg _{\mathbf{u}}}\mathop {\min }\limits_{{\mathbf{u}} \in {V_1}} \left\| {{\mathbf{u}} - {\mathbf{v}}} \right\|$, where $\|\|$ is any norm. In plain words, ${\text{pro}}{{\text{j}}_{{V_1}}}{\mathbf{v}}$ is the vector in $V_1$ that is nearest to $\bf v$ w.r.t. any norm.

or maybe the following is true?

Let $V$ equip with an inner product $<>$, and $ {V_1} \bot {V_2}$ are orthogonal subspaces w.r.t. the inner product, then ${\text{pro}}{{\text{j}}_{{V_1}}}{\mathbf{v}} = {\arg _{\mathbf{u}}}\mathop {\min }\limits_{{\mathbf{u}} \in {V_1}} \left\| {{\mathbf{u}} - {\mathbf{v}}} \right\|_{<>}$, where $\|\|_{<>}$ is the inner product induced norm. In plain words, ${\text{pro}}{{\text{j}}_{{V_1}}}{\mathbf{v}}$ is the vector in $V_1$ that is nearest to $\bf v$ w.r.t. the inner product induced norm.

Many thanks!

Note the projection is dependent on both $V_1,V_2$. If $V_2$ change, the projection of $\bf v$ onto $V_1$ might change. 

 A: There are two related concepts going on here: orthogonal projection and projection along a vector. 
Orthogonal projection is the one we generally use and think of; it minimizes the distance $\| proj (\vec{v}) - v \|$. It is also the one that comes from $P=A (A^T A)^{-1} A^T$ and all the standard projection formulas. Envision orthogonal projections as the sun casting a shadow from directly overhead. 
For projection along a vector (space), envision casting shadows with the sun off in any old direction.  We need some more terminology. Let's project onto subspace $V$ along the subspace $W$ where $\mathbb{R}^n = V \oplus W$. To do this, we write $\vec{x}$ in terms of  $V$ and $W$:
    $$\vec{x} = \vec{x_v} + \vec{x_w}$$
where $\vec{x_v} \in V$ and $\vec{x_w} \in W$. 
The projection of $\vec{x}$ onto $V$ along $W$ is then found by
 $$proj(\vec{x}) =  \vec{x_v}.$$
There is no reason that this should minimize any sort of metric, unless $W= V^\perp$, in which case we're doing an orthogonal projection.
For example, your picture gives that $\vec{v_1}$ is "the projection of $\vec{v}$ onto $V_1$ along $V_2$" while $\vec{v_1}'$ is "the projection of $\vec{v}$ onto $V_1$ along $V_3$."
As to your first question, not true. This might not even be uniquely defined.  Let $V$ and $W$ be the $x$- and $y$-axes in $\mathbb{R}^2$. The projection of $[ 0, ~3 ]^T$ onto $V$ along $W$ is $[0,~0]^T$, but there are multiple vectors that minimize the taxicab norm. In particular, your definition would have the projection as $[-3,0]^T$ and $[3,0]^T$ and $[0,0]^T$, so it isn't well defined. This is even worse if we take the discrete metric, in which case your definition would give the projection as being the entire $x$-axis. 
As to your second question, true by definition. Distances and orthongality are both defined in terms of the inner product. In some sense your picture is misleading you because you have claimed $V_1 \perp V_2$ or $V_1 \perp V_3$ under some inner products $\langle \bullet | \bullet \rangle_{2,3}$, but have drawn your picture w.r.t. to the standard Euclidean inner product. 
