Verify the rule $A \times (B \times C)=B(A \cdot C) - C(A \cdot B)$ by writing out the vectors in component form

Question

1. I'm not sure if this is allowed but I assigned integer values to each vector in order to make the work neater. Did I have to use variables only? Is it imperative to use variables when writing a proof?

2. I used vectors with 3 components because I was confused about the result with vectors of 2 components only. Does this rule apply to vectors of all sizes?

My attempt:

Let $A=(1,1,1)$, $B = (2,1,1)$, and $C = (3,1,1)$.

The left side:

$$A \times B = i(1-1) - j(1-2)+k(1-2) = (0,1,-1)$$ $$A \times B \times C = i(1+1)-j(0+3)+k(0-3)=(2,-3,-3)$$

The right side:

$$A \cdot C = (1 \cdot 3) + (1 \cdot 1) + (1 \cdot 1) = 5$$ $$A \cdot B = (1 \cdot 2) + (1 \cdot 1) + (1 \cdot 1) = 4$$

$$B(A \cdot C) - C(A \cdot B) = ((10-12),(5-4),(5-4))=(-2,1,1)$$

Clearly there is something wrong here...

Answer 1

Mistake in example:

The reason the left hand side of your example working is not equating to the right hand side is because the rule you are trying to verify is this:

$$A \times (B \times C)=B(A \cdot C) - C(A \cdot B)$$

Note the parentheses on the LHS. They mean you need to evaluate $B \times C$ first and then do $A \times$ that resultant working. In you working you do $A \times B$ first and then do that $\times C$ which is equivalent to $A\times (B \times C)$.

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Approaching Proofs

Good job on first approaching a verification of this statement by using a specific example; this is often a good way to initially understand the intuition behind something. However, as the comments have suggested, a rigorous proof should work for any $A, B, C$; not just the specific ones you have picked.

Note: The cross product of vectors is an operation that is only defined in three dimensional space so it only applies to vectors in $\mathbb R^3$.

Answer 2

The "obvious" way to prove this identity is to write $A=\left(A_1\,A_2\,A_3\right)^T$ etc. and express each side in terms of $A_1,\,\cdots,\,C_3$. Without even doing this, we know each term on both sides of the equation will be of the form $\pm A_i B_j C_k$ for some $i,\,j,\,k\in\left\{ 1,\,2,\,3\right\}$. However, it's an approach that makes the result look like an accident of algebra, so instead I'll use a technique that provides more insight by exploiting the geometric significance of $\cdot,\,\times$.

Since $B\times C$ is perpendicular to $B,\,C$ and hence to the plane they span (i.e. vectors of the form $uB+vC$ for numbers $u,\,v$), any vector in $\mathbb{R}^3$ that is in turn perpendicular to $B\times C$ must be in that plane, so numbers $u,\,v$ exist for which $A\times\left( B\times C\right) = uB+vC$. Now we just need to prove $u=A\cdot C,\,v=-A\cdot B$.

Clearly $u,\,v$ will in general depend on $A,\,B,\,C$, but will not change if we rotate the axes we use to represent vectors in $\mathbb{R}^3$. Thinking about the aforementioned $\pm A_i B_j C_k$ terms, we know $u$ consists of $\pm A_i C_k$ terms while $v$ consists of $\pm A_i B_j$ terms. The only way to combine these with rotational invariance is to choose constants $k,\,l$ with $u=kA\cdot C,\,v=lA\cdot B$. Now we want to prove $k=1,\,l=-1$. In fact, since exchanging $B$ with $C$ multiplies the identity by $-1$, we only need check $k=1$.

Note that $k$ is just the $A_1 B_2 C_1$ coefficient in $\left( A\times\left( B\times C\right)\right)_2$. Let's use rotational invariance in one more slick way. We can orient our axes so $A_2=A_3=B_3=0$. Then $$A\times\left( B\times C\right)=A_1\mathbf{i}\times\left( \left( B_1 \mathbf{i} + B_2 \mathbf{j}\right)\times \left( C_1 \mathbf{i} + C_2 \mathbf{j} + C_3 \mathbf{k}\right)\right),$$ so the $_2$ component is $-A_1\left( B\times C\right)_3=A_1 B_2 C_1 - A_1 B_1 C_2$. Notice $k=1$ as required.

Answer 3

Ad 1. You know very well that this is not a correct procedure for a "general proof", and that you have to do the calculations in terms of variables. It is another matter if you had to show that the formula in question is not true in general. In such a case a single (numerical) counterexample would do.

Ad 2. Since you are told to do the proof by "writing out the vectors in component form" you have to start with $A=(a_1,a_2,a_3)$, etcetera, and to compute all components of both sides "the hard way". Of course there are proofs of the formula using the structural and the geometric properties of the involved products. These proofs are simpler.

Concerning dimensions: The cross product $\times$ is defined only in ${\mathbb R}^3$. The $\times$-product of two vectors $a$, $b\in{\mathbb R}^3$ is a vector $a\times b$ living in the same space. There is also a skew product of vectors $a$, $b\in{\mathbb R}^2$. I'd write it as $$a\wedge b:=a_1b_2-a_2b_1\ .$$ The result of this multiplication is a real number, which is equal to the (signed) area of the parallelogram spanned by $a$ and $b$. Numerically this number coincides with the third component of $(a_1,a_2,0)\times(b_1,b_2,0)$.