# Triple Scalar Product

• Jun 8th 2008, 05:10 PM
a.a
Triple Scalar Product
What does the triple scalar product of the normals of three planes say about the type of intersection between the planes?

I know that the triple scalar product is 0 when there is a line of intersection, and is not 0 when there is a unique point, what about when there is no intersection?

What other cases of three planes in space?
• Jun 8th 2008, 05:19 PM
PaulRS
If the scalar triple product is 0, this means that the set consisting of the three normals is not linearly independent (since the determinant is 0), this means that all the normal vectors lie in a same plane through the origin.
It can happen that two of them are collinear, but it's not necessary
• Jun 8th 2008, 05:25 PM
a.a
do at least two normals have to be colinear or can they be parallel?
• Jun 8th 2008, 05:31 PM
PaulRS
Quote:

Originally Posted by a.a
do at least two normals have to be colinear or can they be parallel?

Collinears, the vectors have their initial points in the origin. But they are still parallel
• Jun 8th 2008, 05:39 PM
a.a
when the product is not 0 does that meant that there is a unique solution?
• Jun 9th 2008, 12:07 PM
PaulRS
Let $\displaystyle A,B,C$ be our planes and $\displaystyle \vec a,\vec b,\vec c$ be their respective normals

$\displaystyle \vec a \cdot \left( {\vec b \times \vec c} \right) = 0 \Leftrightarrow$ there's a plane such that contains all the three normal vectors and the origin. Let's call it $\displaystyle \Pi$

$\displaystyle \left. \begin{gathered} \vec a \bot A \hfill \\ {\text{the line containing }}\vec a \subset \Pi \hfill \\ \end{gathered} \right] \Rightarrow \Pi \bot A$

Similarly: $\displaystyle \Pi \bot B,\Pi \bot C$

If at least two of the normals are collinear we know that at least 2 of the planes are collinear and thus the intersection can't be a unique point. So we'll asume that this doesn't happen, and so the interesection is always a line.

That is, we are assuming: $\displaystyle \left\{ \begin{gathered} A \cap B \hfill \\ A \cap C \hfill \\ B \cap C \hfill \\ \end{gathered} \right.{\text{ are all lines}}$

$\displaystyle \left. \begin{gathered} \Pi \bot A \hfill \\ \Pi \bot B \hfill \\ \Pi \bot C \hfill \\ \end{gathered} \right] \Rightarrow \left( {A \cap B} \right)||\left( {A \cap B} \right)||\left( {A \cap B} \right)$ (since $\displaystyle \Pi$ is perpendicular to all the intersections)

Thus: $\displaystyle A \cap B \cap C$ is either a line or $\displaystyle \emptyset$ but not a unique point

This shows that if there's such a plane $\displaystyle \Pi$ then the interesection can't be a unique point.

If the intersection is a unique point, there can't be a plane that's perpendicular to all the three planes at the same time (that'd be absurd by the above argument) and therefore $\displaystyle \vec a \cdot \left( {\vec b \times \vec c} \right) \neq 0$

If $\displaystyle \vec a \cdot \left( {\vec b \times \vec c} \right) \neq 0$ then there's no such plane, and therefore the intersections can't be parallel (or there are no two of the planes that are parallel to each other) , and therefore the solution should be a unique point (suppose the contrary and you'll get one of the mentioned cases ).

Thus the interesection of the three planes is a unique point iff $\displaystyle \vec a \cdot \left( {\vec b \times \vec c} \right) \ne 0$