Is the Triple Scalar Product Always Zero?

member 392791
Hello,

I am confused how vectors that are coplanar will give a triple product of zero? Or is it the case that all 3 vectors must be coplanar for a triple product of zero, or is 2 sufficient? I.e. the vector being dotted with one of the vectors being crossed in the same plane, will this produce a zero scalar product?
 
Mathematics news on Phys.org
Take any two of those coplanar vectors. The cross product is either zero or is normal to both of those vectors -- and every other vector that is coplanar with those first two vectors. What's the dot product of two vectors that are normal to one another?
 
ok, so I see that the textbook specified that if all 3 vectors are coplanar, then its triple scalar product is zero, which makes sense to me because the projection is going to be zero. It's just that the accompanying figure 3.28 doesn't make me think that the vectors are coplanar.
EDIT: here is the figure that I am referring to
 

Attachments

Last edited by a moderator:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
4
Views
3K
Replies
12
Views
2K
Replies
4
Views
3K
Replies
6
Views
1K
Replies
1
Views
1K
Replies
5
Views
1K
Back
Top