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I noticed something when I was doing a proof of the BAC-CAB rule, and wanted to check if my intuition was correct.

First, when I actually multiplied out $\vec{B} (\vec{A} \cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B}) $ I noticed that I got the same thing. That is, the two quantities are equal. This makes sense since they are dot products. But that further implies that the expression goes to zero.

Well, that means that $\vec{A} \times (\vec{B} \times \vec{C})$ would have to be zero too. And when I thought about it it occurred to me that $\vec{B} \times \vec{C}$ describes a vector perpendicular to both B and C; it's orthogonal. But $\vec{A} \times$ (any vector) will describe a vector perpendicular to (any vector) and in this case, that's $\vec{B} \times \vec{C}$. That is going to be parallel with the vector described by $\vec{B} \times \vec{C}$ itself and that further implies that the cross product of the two is equal to zero. Which is equal to the BAC-CAB quantity above.

So, what is the big honking flaw in my reasoning here? :-)

bubba
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Jesse
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    The big flaw is in the first step. Note that each term is a dot product, resulting in a scalar, multiplied by a vector. So the result is a vector. – Harald Hanche-Olsen Sep 18 '13 at 13:42
  • Those two quantities very much aren't equal. Take $A = \hat{i}$, $B = \hat{j}$, $C = \hat{i}$. Then $$ B(A \cdot C) = \hat{j}(\hat{i} \cdot \hat{i}) = \hat{j} \neq 0 = \hat{i}(\hat{i} \cdot \hat{j}) = C(A \cdot B). $$ – Branimir Ćaćić Sep 18 '13 at 13:44
  • Also, cross products aren't even associative, so you have to be careful to write $A \times (B \times C)$ or $(A \times B) \times C$, depending on what you actually want. – Branimir Ćaćić Sep 18 '13 at 13:46
  • Ah, I see -- I assumed that $B(A \cdot C)$ would end up as a scalar. Thanks for pointing it out :-) – Jesse Sep 18 '13 at 13:50

2 Answers2

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Two possible flaws, neither of them honkers, in my view:

(1) Your computation of $\vec{B} (\vec{A} \cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B})$ must have some errors -- it should not come out to be the zero vector.

(2) You argument that $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{0}$ also has errors. Draw a picture, and I think you'll see why. $\vec{A} \times (\vec{B} \times \vec{C})$ is perpendicular to $\vec{A}$, and it's perpendicular to $\vec{B} \times \vec{C}$. I don't see how you conclude from this that it must be the zero vector.

bubba
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  • Is there a nice geometric, coordinate-free proof of the identity? I can do it in coordinates but there must be an easier way... – user7530 Sep 18 '13 at 13:50
  • i was thinking that if we look at B x C as a vector (call it D) and then look at that vector crossed with A, you would get one perpendicular to both, and since A x (B xC) describes a vector orthogonl to those two... and the cross product of any parallel vector is zero... – Jesse Sep 18 '13 at 13:54
  • @user7530 -- Yes. Let $P$ be the plane containing $\vec{B}$ and $\vec{C}$. Then the vector $\vec{N} = \vec{B} \times \vec{C}$ is normal to this plane. But then $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{N}$ is perpendicular to $\vec{N}$, so it must lie in the plane $P$. So there exist scalars $h$ and $k$ with $\vec{A} \times (\vec{B} \times \vec{C}) = h\vec{B} + k\vec{C}$. Taking dot products of both sides with $\vec{B}$ and $\vec{C}$ in turn gives you two equations for $h$ and $k$. The solutions are $h= \vec{A} \cdot\vec{C}$ and $k = \vec{A} \cdot\vec{B}$. – bubba Sep 18 '13 at 14:11
  • @Jesse, So you think that vector $D$ which is perpendicular to product $B\times C$ is a vector parallel to both? Just think over it. Let's say $B = \hat{i}$ and $C = \hat {j}$. So, $B\times C = \hat{k}$. Now, consider any vector in the $x-y$ plane. What is the angle between that vector and $\hat{k}$? – Parth Thakkar Sep 18 '13 at 14:13
  • @Jesse -- your reasoning is still tangled up. Again, draw a picture (in 3D). Please. – bubba Sep 18 '13 at 14:14
  • @user7530 -- there's a better explanation here: http://math.stackexchange.com/questions/400641/do-the-bac-cab-identity-for-triple-vector-product-have-some-intepretation?rq=1 – bubba Sep 19 '13 at 01:57
  • @bubba Right now I am struggling with the solution of $h$ and $k$. What does it mean by taking the dot product of both side with $\vec{b}$ and $\vec{c}$? How to solve h and k with these equations? Someone said "take dot product with $\vec{a}$", but then I still cannot solve the unknowns... – Jason Ng Sep 22 '18 at 16:12
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Usually I use matrices. Because

$\mathbf{v}\times\,\mathbf{w}=\begin{bmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \\\end{bmatrix}\cdot\begin{bmatrix}w_1\\w_2\\w_3\end{bmatrix} $

Then

enter image description here

wmora2
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