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I have to solve the following exercise. Let $P_0, P_1, P_2$ three not aligned points. What figure does the set of points $P$ such that $$ (P-P_0)\times(P_1-P_0)\cdot(P_2-P_0)=h $$ represent? And what figure does the condition $$ |(P-P_0)\times(P_1-P_0)\cdot(P_2-P_0)|=|h| $$ represent?

My attempt. The height of the parallelepiped with sides $P_1-P_0$, $P_2-P_0$ and $P-P_0$ is fixed and it is equal to $$ (\star)\ \ \ \ \ \ \ \ \ \ \ height=\frac{|h|}{|(P_1-P_0)\times(P_2-P_0)|},\quad h=\text{volume of the parallelepiped}. $$ On the other hand, the height coincides with the distance between $P$ and $\pi$ = the plane containing $P_0, P_1$ and $P_2$. So I conclude that the required figure is a plane parallel to $\pi$. But what about the figure represented by all $P$ such that $$ |(P-P_0)\times(P_1-P_0)\cdot(P_2-P_0)|=|h|? $$ Some hints?

Thank You

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The absolute value corresponds to $h$ or $(P-P_0)\times(P_1-P_0)\cdot(P_2-P_0)$ being either positive or negative, so it represents two planes, one $h$ units above $\pi$ and one $h$ units below.

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  • $\begingroup$ So, do the points $P$ belong above $\pi$ or below $\pi$ according to the sign of $h$ (if $h$ is positive, above - if $h$ is negative, below)? $\endgroup$
    – Redeldio
    Apr 30, 2018 at 22:26
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    $\begingroup$ Above and below relative to what? Take, for example, $P_0=(0,0,0)$, $P_1=(0,1,0)$ and $P_2=(0,0,1)$. What does “above” mean for this plane? $\endgroup$
    – amd
    Apr 30, 2018 at 22:37
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    $\begingroup$ It's arbitrary. The only information that matters is that the planes are reflections across $\pi$. If you want to define an orientation using something like the "right-hand rule," feel free, but that information is irrelevant under the symmetry of the solution. $\endgroup$
    – user211599
    Apr 30, 2018 at 22:49
  • $\begingroup$ @JefferyOpoku-Mensah ok, but you write "the absolute value corresponds to $h$ or $(P-P_0)\times(P_1-P_0)\cdot(P_2-P_0)$ being either positive or negative". What do you mean for "positive" and "negative"? Thank You $\endgroup$
    – Redeldio
    Apr 30, 2018 at 23:07
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    $\begingroup$ $|x|=|y| \implies x= \pm y$, or equivalently $y = \pm x$. And as $h$ is the signed height of the parallelepiped, it corresponds to 2 planes, one with positive height and one with negative height below $\pi$. $\endgroup$
    – user211599
    Apr 30, 2018 at 23:10

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