There is nothing special about $\mathbf{u}=(1,\dots,1)$. I work with an arbitrary vector in $(0,\infty)^n$.

**Claim)** Let $n\geq 3$ and $\mathbf{u}=(u_1,\dots,u_n),\mathbf{v}=(v_1,\dots,v_n)\in
(0,\infty)^n$ be two linearly independent vectors. Suppose
$\mathbf{u},\mathbf{v}$ satisfy the followings:

- $\frac{u_1}{u_2}<\frac{v_1}{v_2}$, $\frac{u_2}{u_3}>\frac{v_2}{v_3}$;
- $\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$ where
$$ \begin{split} & A:=\langle\mathbf{u},\mathbf{v}\rangle
\left(1-\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)>0,\\
&B:=|(u_1,u_2,u_3)\times(v_1,v_2,v_3)|^2>0. \end{split} \tag{$\star$}$$
Then there exists a vector $\mathbf{x}\in\Bbb{R}^n$ with both positive
and negative components for which $\left\langle
p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$.

To see this, consider the quadratic form
$$
\begin{split}
&Q(\mathbf{x}):=\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle=
\left\langle\mathbf{x}-\frac{\langle\mathbf{x},\mathbf{u}\rangle}{|\mathbf{u}|^2}\mathbf{u},
\mathbf{x}-\frac{\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle\\
&=|\mathbf{x}|^2-\frac{|\langle\mathbf{x},\mathbf{u}\rangle|^2}{|\mathbf{u}|^2}-\frac{|\langle\mathbf{x},\mathbf{v}\rangle|^2}{|\mathbf{v}|^2}+\frac{\langle\mathbf{x},\mathbf{u}\rangle\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{u}|^2|\mathbf{v}|^2}\langle\mathbf{u},\mathbf{v}\rangle.
\end{split}
$$
on $\Bbb{R}^n$. If $\mathbf{x}$ is a linear combination of the form $\mathbf{u}+t\mathbf{v}$, then
$$
Q(\mathbf{u}+t\mathbf{v})=
\left\langle t\left(\mathbf{v}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{u}|^2}\mathbf{u}\right),
\mathbf{u}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle
=t\overbrace{\langle\mathbf{u},\mathbf{v}\rangle}^{>0}
\overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}.\tag{$\star\star$}
$$
When $n=2$, the vectors $\mathbf{u},\mathbf{v}$ are in the first quadrant. For $t>0$, $Q$ is negative at $\pm(\mathbf{u}+t\mathbf{v})$ which belong to the first or the third quadrant. On the other hand, in order for $\pm(\mathbf{u}+t\mathbf{v})$ to belong to
either the second or fourth quadrants, $t$ should be negative. But then $Q$ becomes positive by $(\star\star)$. This shows that when $n=2$ we have

$\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle\geq 0$
if $\mathbf{x}\notin [0,\infty)^2\cup (-\infty,0]^2$.

Finally, suppose $n>2$. Then there exist non-zero vectors perpendicular to both $\mathbf{u}$ and $\mathbf{w}$. For any such vector $\mathbf{w}$, the formula for $Q$ implies $Q(\mathbf{x}+\mathbf{w})=Q(\mathbf{x})+|\mathbf{w}|^2$. The idea is to find $t>0$ and $\mathbf{w}\perp\mathbf{u},\mathbf{v}$ such that $\mathbf{u}+t\mathbf{v}+\mathbf{w}$ has both positive and negative components (which should be possible since components of $\mathbf{w}$ cannot be all positive or all negative) and

$$
Q(\mathbf{u}+t\mathbf{v}+\mathbf{w})
=\overbrace{t\langle\mathbf{u},\mathbf{v}\rangle}^{>0}
\overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}+|\mathbf{w}|^2<0.
\tag{$\star\star\star$}
$$
We take $\mathbf{w}$ to be
$$
\mathbf{w}:=s(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1,\overbrace{0,\dots,0}^{n-3})\quad s>0.
$$
Clearly, $\mathbf{w}\perp\mathbf{u},\mathbf{v}$, and by our hypotheses, its third component is negative while the first one is positive. For any $s,t>0$ the first component of $\mathbf{u}+t\mathbf{v}+\mathbf{w}$ is positive, and it suffices to arrange $s,t$ so that the third component is negative, and moreover $(\star\star\star)$ holds. The former amounts to
$$
u_3+tv_3<s|u_1v_2-u_2v_1|
$$
while, with the notation from $(\star)$, the latter means
$$
-tA+s^2B<0.
$$
So the problem boils down to $\frac{u_3+tv_3}{|u_1v_2-u_2v_1|}<\sqrt{\frac{tA}{B}}$, because then any $s$ between them qualifies. The last inequality means that

$$
r\mapsto{\frac{v_3}{|u_1v_2-u_2v_1|}}r^2-\sqrt{\frac{A}{B}}r+\frac{u_3}{|u_1v_2-u_2v_1|}
$$
attains negative values over $(0,\infty)$. Since all the coefficients of this quadratic are positive except the linear one, this is equivalent to having positive discriminant:
$$
\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2};
$$
an inequality which was assumed. $\blacksquare$

**Corollary)** Suppose $n\geq 3$ and $\mathbf{u}\in (0,\infty)^n$. Then there exist vectors $\mathbf{v}\in (0,\infty)^n$ and
$\mathbf{x}\in\Bbb{R}^n\setminus([0,\infty)^n\cup(-\infty,0]^n)$ such
that $\left\langle
p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$.

To show this, one only needs to pick $\mathbf{v}=(v_1,\dots,v_n)$ with positive entries such that $\frac{u_1}{u_2}<\frac{v_1}{v_2}$, $\frac{u_2}{u_3}>\frac{v_2}{v_3}$ and
$\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$. If $v_1>0$ is constant and $v_2,v_3$ get small so that $\frac{v_2}{v_3}$ becomes small too, then $\frac{u_1}{u_2}<\frac{v_1}{v_2}$,
$\frac{u_2}{u_3}>\frac{v_2}{v_3}$ hold; and
$\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}\to 0$. But quantities $A$ and $B$ remain bounded from below. Consequently, when $v_2,v_3\to 0^{+}$ such that $\frac{v_2}{v_3}<\frac{u_2}{u_3}$ as $v_1,v_4,\dots,v_n$ are constant, we obtain vectors $\mathbf{v}$ satisfying the desired properties. $\blacksquare$

This argument also provides a recipe for constructing vectors $\mathbf{x}$ and $\mathbf{v}$ for which the dot product under consideration is negative.