- #1

Sirius24

- 14

- 0

(A X B)^2 ?

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- Thread starter Sirius24
- Start date

- #1

Sirius24

- 14

- 0

(A X B)^2 ?

- #2

spamiam

- 360

- 0

[tex]

(A \times B) \times (A \times B) ?

[/tex]

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

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

- #3

- 9,568

- 774

(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|

(A x B) dot (A x B)

(A x B) x (A x B)

or something else.

- #4

Sirius24

- 14

- 0

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?

- #5

- 9,568

- 774

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

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

- #6

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

The first two of these are the sameYou 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)

True but probably not what was meant- especially since it is trivial.(A x B) x (A x B)

Assuming that "[itex]A^2[/itex]" for A a vector really means "(length of A) squared"or something else.

then we know that

[tex]|A\times B|= |A||B|sin(\theta)[/tex]

where [itex]\theta[/itex] is the angle between vectors A and B.

So

[tex](A\times B)^2= |A\times B|^2= |A|^2|B|^2 sin^2(\theta)[/tex]

You might also recall that

[tex]cos(\theta)= \frac{A\cdot B}{|A||B|}[/tex]

and, of course, [itex]sin^2(\theta)= 1- cos^2(\theta)[/itex].

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