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You can calculate the volume of a parallelepiped by $|(A \times B) \cdot C|$, where $A$, $B$ and $C$ are vectors. I wonder does the order matter? If it does how, is it determined. I know I can just put it in a matrix and calculate the determinant but I would like to know how it is in this case.

Thanks!

TheGeekGreek
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Ghara
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  • see here i hope this will help you https://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/Determinant/Determinant/node11.html – Dr. Sonnhard Graubner Jan 09 '17 at 16:41
  • It does not say if the order matters, thanks anyway. – Ghara Jan 09 '17 at 16:42
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    The order better not matter since the volume shouldn't change under the rigid transformations that swap axes around – MPW Jan 09 '17 at 16:50
  • The first two posted versions of the formula seemed a big confused: one usually uses $\cdot$ for the inner product in contexts where one would use $\times$ for the cross product; and the result of the inner product is a scalar, which requires only $\lvert x\rvert$ and not $\lVert x \rVert$ to find its magnitude. The third version put the formula into a standard notation. I think the spirit of the question is the same despite the notational change. – David K Jan 09 '17 at 17:00

2 Answers2

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If you know that the scalar triple product is equal to the determinant of a matrix whose rows are the components of the vectors, and if you recall the effects of operations on the rows of a matrix, then you can show that swapping any two of the vectors $A,B,C$ in the scalar triple product $(A \times B) \cdot C$ will swap the corresponding rows of the matrix and therefore will flip the sign of the determinant but will not change the magnitude of the determinant. Hence the interchange of any two vectors (which could be $B$ and $C$ or could be $A$ and $C$, not just $A$ and $B$) will likewise flip the sign of the scalar triple product but will not change its magnitude.

Any reordering of the three vectors $A$, $B$, and $C$ can be accomplished by either one or two interchanges of two vectors. For example, to get from $(A,B,C)$ to $(B,C,A)$, swap the first two vectors, then the last two. Hence of the six possible ways to order the three vectors $A$, $B$, and $C$, three orderings will give you positive scalar triple products and three will give you negative scalar triple products, but all scalar triple products will have the same magnitude.

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David K
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We have $a \times b = -(b \times a)$. So you get also a change of sign since the (real) standard inner product is symmetric, so there the order does not matter. The reason why the cross product changes its sign if you permute the arguments, is simply that it is like the determinant an alternating mapping (this is immediate by its definition as the components of the vectors are just determinants of $2\times 2$ matrices).

TheGeekGreek
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