# Figure represented by a triple product

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

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.

• 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)? Apr 30, 2018 at 22:26
• 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?
– amd
Apr 30, 2018 at 22:37
• 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.
– user211599
Apr 30, 2018 at 22:49
• @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 Apr 30, 2018 at 23:07
• $|x|=|y| \implies x= \pm y$, 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 $\pi$.
– user211599
Apr 30, 2018 at 23:10