Cross Product Identity: Solving for (A X B)^2

[Sirius24]

I've been trying to find what the square of two cross products is and can't find it. Can anyone tell me the identity for
(A X B)^2 ?

 

[spamiam]

$A \times B$ is a vector, so do you want
$$(A \times B) \times (A \times B) ?$$

If so, it's not too difficult. For any vector $a$, can you simplify $a \times a$?

(For that matter, which definition of the cross product are you using?)

 

[LCKurtz]

I've been trying to find what the square of two cross products is and can't find it. Can anyone tell me the identity for
(A X B)^2 ?

You don't usually talk about squaring a vector. So you have to tell us what that notation means or otherwise correct your question. It might mean any of these:

|A x B|²
(A x B) dot (A x B)
(A x B) x (A x B)

or something else.

 

[Sirius24]

You don't usually talk about squaring a vector. So you have to tell us what that notation means or otherwise correct your question. It might mean any of these:

|A x B|²
(A x B) dot (A x B)
(A x B) x (A x B)

or something else.

The notation is exactly as I posted it. My homework has two vectors, V1 and V2. The part I'm trying to work with says (V1 X V2)^2. I worked through (V1 X V2) dot (V1 X V2) and (V1 X V2) x (V1 X V2). The result is zero in either case for this problem, but will that always be the case?

 

[LCKurtz]

The notation is exactly as I posted it. My homework has two vectors, V1 and V2. The part I'm trying to work with says (V1 X V2)^2. I worked through (V1 X V2) dot (V1 X V2) and (V1 X V2) x (V1 X V2). The result is zero in either case for this problem, but will that always be the case?

That's not the point. You need to know what that notation means before it makes sense to calculate it. Answer this:

If A = 3i -2j + 5k, what is A²?

Somewhere you must have a definition of what you mean by squaring a vector.

 

[HallsofIvy]

You don't usually talk about squaring a vector. So you have to tell us what that notation means or otherwise correct your question. It might mean any of these:

|A x B|²
(A x B) dot (A x B)

The first two of these are the same

(A x B) x (A x B)

True but probably not what was meant- especially since it is trivial.

or something else.

Assuming that "$A^2$" for A a vector really means "(length of A) squared"
then we know that
$$|A\times B|= |A||B|sin(\theta)$$
where $\theta$ is the angle between vectors A and B.

So
$$(A\times B)^2= |A\times B|^2= |A|^2|B|^2 sin^2(\theta)$$

You might also recall that
$$cos(\theta)= \frac{A\cdot B}{|A||B|}$$
and, of course, $sin^2(\theta)= 1- cos^2(\theta)$.