How can the vector triple product be used to derive other vector products?

[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!

 

[Ray Vickson]

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!

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.

 

[Tanya Sharma]

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!

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,

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

 

[Saitama]

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.

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

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.

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.

 

[D H]

Next, prove that
$$\mathbf{U} \times ( \mathbf{V} \times \mathbf{W}) = (\mathbf{U} \times \mathbf{V}) \times \mathbf{W},$$

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

 

[D H]

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

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.

 

[Saitama]

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.

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?

 

[D H]

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

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.

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?

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

 

[Saitama]

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

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.

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

 

[D H]

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.

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 $U \times V \times U$.

 

[Saitama]

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 $U \times V \times U$.

Thank you once again D H! :)

 

[Ray Vickson]

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.

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