# X product refresher

1. Jan 20, 2005

### skiboka33

Kind of a weird question , don't really understand it:

"A student claims to have found a vector A such that:
(2i - 3j + 4k) X A = (4i +3j - k). Do you believe this claim???

2. Jan 20, 2005

### Integral

Staff Emeritus
You need to see if you can find some vector A, such that the given cross product is valid. Simply compute the cross product for a general vector

A = (A1,A2,A3)

Does this vector exist?

3. Jan 20, 2005

### arildno

Alternatively, try to apply the fact that the cross product of two vectors is perpendicular to both of the vectors crossed.

4. Jan 20, 2005

### skiboka33

yeah, just seems like for any two crossed vectors there will always be one perpendicular to both of them, isn't that the case??

5. Jan 20, 2005

### dextercioby

Yes,the three vectors involved in:
$$\vec{A}\times \vec{B}=\vec{C}$$

form a trirectangular trihedron...I hope u know that you problem assumed solving a 3-3 algebraic system...

Daniel.

6. Jan 20, 2005

### arildno

So, since the reputed cross product must be perpendicular to both the other two vectors, the dot product between the (reputed) cross product vector and the other vector you know must be..?

7. Jan 20, 2005

### gerben

The cross-produkt of two vectors is always perpendicular to both the vectors of which you calculate the cross-produkt.
in:
$$v1 \times v2 = v3$$
v3 is perpendicular to both v1 and v2

the dot-produkt of two perpendicular vectors is zero, so in your case
$$(2, -3, 4)\cdot(4, 3, -1)$$
should be zero, but it is -5.

8. Jan 21, 2005

### noelhustler

Calculate the cross product of your first vector and A.
You'll end up with 3 equations, one for each component.
Set these equations to equal the respective components of your final vector.
Then try and solve the equations simultaneously.
If there is a solution, then the vector A exists.

9. Jan 21, 2005

### dextercioby

It's the same advice i gave...It leads to the result,eventually...Unforunately for us,there's a much shorter and "brighter" way.Both Arildno and Gerben pointed it out...

Daniel.

10. Jan 21, 2005

### noelhustler

Of course calculating the dot product is a much easier way to solve the problem.
However, I think it is useful to show that there is more than one way to skin a cat, especially when they don't seem to grasp the first method.