# Math Help - Please verify.

Vectors $a, b, c$ ( $\vec{i}=i$) make up tetrahedron with volume $V=3$. What's the volume of a tetrahedron made up of vectors $a\times b, b\times c, a\times c$?
Solutions aren't given, so I kindly ask for a recheck.

Setup: $V=\frac{1}{6}(a,b,c)=3\Rightarrow a\cdot(b\times c)=18\Rightarrow b\times c=\frac{18}{a}$

Attempt:
$(a\times b, b\times c, a\times c)=((a\times b)\times(a\times c))\cdot b\times c=(-\langle b, a\times c\rangle\cdot a)\cdot b\times c=(-\langle b, a\times c\rangle\cdot a)\frac{18}{a}=$

$=-18\langle b, a\times c \rangle=-18(b\cdot (a\times c))=-18(b,a,c)=18(a,b,c)=54$

Isn't the volume of the second tetrahedron too large?

2. What does it mean to divide by a vector? What does the expression $\frac{18}{a}$ mean?

3. Let $[\mathbf a,\mathbf b,\mathbf c]$ refer to the scalar triple product $\mathbf a\times\mathbf b\cdot\mathbf c$ so that the volume of the tetrahedron is $V=|[\mathbf a,\mathbf b,\mathbf c]|/6$.

Volume of new tetrahedron is $V'=|[\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]|/6$.

As is well known, $[\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]=[\mathbf a,\mathbf b,\mathbf c]^2$.

Hence $V'=6V^2=54$ as claimed. Although it came close, the method shown by courteous was ultimately incorrect -- that answer is right for the wrong reasons.

4. Originally Posted by halbard
As is well known, $[\mathbf b\times\mathbf c,\mathbf c\times\mathbf a,\mathbf a\times\mathbf b]=[\mathbf a,\mathbf b,\mathbf c]^2$.
Can you expand the "well known" (google is useless)?

BTW, did you intentionally write $c\times a$ instead of $a\times c$ (or it doesn't matter for given problem)?