Vector Triple product dilemma

1. Nov 5, 2013

Saitama

I am currently going through the book Introduction Of Electrodynamics by Griffiths. I have come across vector triple product which is stated as follows in the book:

$$\textbf{A} \times (\textbf{B} \times \textbf{C})=\textbf{B}(\textbf{A}\cdot \textbf{C})-\textbf{C}(\textbf{A}\cdot \textbf{B})$$
The author then states a few more vector products and says that they can be derived using the vector triple product. One of them I am unable to derive is :
$$(A\times B)\cdot (C\times D)=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C)$$

I don't see how I can derive the above using the vector triple product.

Any help is appreciated. Thanks!

2. Nov 5, 2013

Ray Vickson

First, prove that
$$\mathbf{U} \times \mathbf{V} \cdot \mathbf{W} = \mathbf{U} \cdot \mathbf{V} \times \mathbf{W}$$
for any three vectors U, V, W. (Note: UxV.W means (UxV).W, etc). In other words, in a mixed vector-scalar product you can interchange the 'x' and the '.'. It is pretty easy to prove, just by expanding out both things and comparing the results.

Next, prove that
$$\mathbf{U} \times ( \mathbf{V} \times \mathbf{W}) = (\mathbf{U} \times \mathbf{V}) \times \mathbf{W},$$ just by using $\mathbf{R} \times \mathbf{S} = -\mathbf{S} \times \mathbf{R}$ and applying the vector triple product you are already given.

Now apply these two facts to your problem.

3. Nov 5, 2013

Tanya Sharma

First interchange the signs of '×' and '.'

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

Which gives

$$[(A \cdot C)B - (B \cdot C)A] \cdot D$$

And finally,

$$[(A \cdot C)(B \cdot D) - (B \cdot C)(A \cdot D)]$$

4. Nov 5, 2013

Saitama

Thanks, didn't notice that it was a scalar triple product. :)

I did not require this in proving what I wanted to but still I would like to know how would you prove this. Here's how I tried:

Applying the vector triple product on LHS:
$$V(U\cdot W)-W(U\cdot V)$$

The RHS is:
$$-W \times (U \times V)$$
$$=-(U(W\cdot V)-V(W \cdot U))$$
$$=V(W \cdot U)-U(W\cdot V)$$

They don't turn out to be the same.

5. Nov 5, 2013

D H

Staff Emeritus
That's just wrong, Ray. The cross product is not associative.

6. Nov 5, 2013

D H

Staff Emeritus
You can use both the scalar triple product and vector triple product to prove this identity. Denote $U=C \times D$. With this substitution, $(A \times B) \cdot (C \times D) = (A \times B) \cdot U$. See if you can take it from here, using the scalar triple product and then the vector triple product.

7. Nov 5, 2013

Saitama

I did exactly the same to prove it. I stated it in my previous reply.

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

I have a few more doubts regarding vector proofs but they involve the del operator, should I post them here even though it is a precalculus section?

8. Nov 5, 2013

D H

Staff Emeritus
Sorry, I didn't see that post. I saw your post where you trued (but failed) to prove the associativity of the cross product.

[EDIT]: That should be tried, not trued.

The cross product is not associative in general. Since the cross product is anti-commutative, $(A \times B) \times C = - C \times (A \times B) = C \times (B \times A)$. Now this is in a form where you can use the vector triple product rule: $(A \times B) \times C = C \times (B \times A) = B (A \cdot C) - A (B \cdot C)$. The difference between $A \times (B \times C)$ and $(A \times B) \times C$ is thus $A \times (B \times C) - (A \times B) \times C = C (A \cdot B) - A (B \cdot C)$. This is zero if both $A \cdot B$ and $B \cdot C$ are zero or if C is parallel to A (i.e., $C=\alpha A$ where alpha is a scalar). The difference is non-zero otherwise, and hence the two forms are not equal to one another in general.

Note that the above means that $U \times (V \times U)$ and $(U \times V) \times U$ are equal to one another. This comes up quite often.

It's best to post those questions in the calculus section.

Last edited: Nov 5, 2013
9. Nov 5, 2013

Saitama

Thanks a lot DH for such wonderful tips. I have made a note of these.

10. Nov 5, 2013

D H

Staff Emeritus
A couple more points on this. One is that since $U \times (V \times U) = (U \times V) \times U$ the parentheses can be omitted in this case: It's okay to write $U \times V \times U$. Note that writing $U \times V \times W$ is *not* okay. Parentheses are needed in the general case.

The other is that one reason this comes up quite often is that it is closely related to the component of V that is perpendicular to U, which I'll designate as $V_{\perp_U}$. One way to calculate the component of V that is normal to U is to subtract the projection of V onto U from V. The projection is given by $\frac {U (V \cdot U)} {U^2}$, and thus the normal component is $V_{\perp_U} = V - \frac {U (V \cdot U)} {U^2} = \frac{V (U \cdot U) - U (V \cdot U)} {U^2}$. The numerator is $U\times (V \times U) = U \times V \times U$. Thus $V_{\perp_U} = \frac {U \times V \times U}{U^2}$. If U is a unit vector, this reduces to $U \times V \times U$.

11. Nov 5, 2013

Saitama

Thank you once again D H! :)

12. Nov 5, 2013

Ray Vickson

You are right, sorry. Anyway, you don't need anything like that.