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I'm trying to figure this out... The mixed product gives us the volume of the parallelepiped which the three vectors form. However, I don't see the connection between a point (let's call it $T_0(x_0, y_0, z_0)$) and two non-colinear vectors (let's name them $\vec a$ and $\vec b$.

Can someone guide me/give me some hints?

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    $\begingroup$ Your question is rather unclear. Before anyone can make sense of your question of how to "derive" an equation from a point and two vectors, you must state what is the desired relation between that equation, that point, and those vectors. For example, if you desire that the solution set of the equation be some geometric object related to that point and those two vectors, then you must say so explicitly, else the problem makes no sense. (Also, $\vec a$ and $\vec b$ are bad names given that two of the coefficients are $a$, $b$). $\endgroup$ – Lee Mosher Apr 1 '17 at 21:07
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I assume that we are talking about a plane in $\mathbb{R}^3$. Observe that each point $(x,y,z)$ on the plane can be characterized by the following property: the vector that points from $T_0$ to $(x,y,z)$ and the two non-colinear vectors than span the plane form a parallelepiped of volume $0$. If you put this in your formula for the mixed product, you will get something that looks like $ax+by+cz+d=0$.

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  • $\begingroup$ Could you clarify your answer a bit? Does $T_0$ refer to the origin? If so, how do I get that two non-colinear vectors that span the plane and that vector for a parallelepiped of volume 0? $\endgroup$ – NumberSymphony Apr 2 '17 at 9:32
  • $\begingroup$ @NumberSymphony $T_0$ is a point on the plane. The two vectors are the ones from the title of this post, i.e. the ones that you call $\vec{a}$ and $\vec{b}$. Each point in the plane can be represented as $T_0+\lambda\vec{a}+\mu\vec{b}$ with $\lambda , \mu \in\mathbb{R}$. If $T=(x,y,z)$ is a point in the plane, then $T-T_0$, $\vec{a}$ and $\vec{b}$ obviuosly form a degenerated parallelepiped of volume 0. $\endgroup$ – Reinhard Meier Apr 2 '17 at 11:50
  • $\begingroup$ Please note that there are a lot of assumption about the problem in my answer, because (as Lee Mosher already correctly mentioned) your description is incomplete, and one can only guess what you are actually talking about. $\endgroup$ – Reinhard Meier Apr 2 '17 at 11:54

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