How Does Distributive Property Apply in Vector Cross Products?

[razored]

I was reading my physics book, and I stumbled across this : $$A_{x} \hat{i} \times B_{y} \hat{j} = (A_{x}B_{y})\hat{ i} \times \hat{ j}$$.
I am trying to figure out, how can they use the distribute property ( I presume) like that? How did they factor the Ax and Bx out? I would have assumed it would have multiplied out like this : $$(A_{x}B_{y})\hat{i} \times \hat{j} = (A_{x}B_{y})\hat{i\times}(A_{x}B_{y}) \hat{j}$$ I thought those were cross products, not multiplication signs.

Can anyone clear up things please?

Thanks beforehand.

 

[granpa]

if you double A in the first equation what happens to the answer. if you double B what happens?

 

[granpa]

BTW, cross product is not a real vector. its a pseudovector.

 

[razored]

I'm still lost

 

[Defennder]

It's a property of the cross product. To show that it is permissible, ask yourself what is (AxBy) i x j? Then what is Axi X Byj ? How would you get the magnitude of the latter product? What formula should you use?

 

[HallsofIvy]

The standard way to take the cross product of vectors $A_x\vec{i}+ A_y\vec{j}+ A_z\vec{k}$ and $B_x\vec{i}+B_y\vec{j}+ B_z\vec{k}$ is to use the (symbolic) determinant:
$$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z\end{array}\right|$$

Here, $A_y= A_z= B_x= B_z= 0$ so that is
$$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ A_x & 0 & 0 \\ 0 & B_y & 0\end{array}\right|$$
What is that?