Calculating vector cross product through unit vectors

[greg_rack]

Homework Statement

Derive the formula for calculating: $\vec{U}\times \vec{B}$

Relevant Equations

none

Writing both $\vec{U}$ and $\vec{B}$ with magnitude in all the three spatial coordinates:
$$
\vec{U}\times \vec{B}=
(U_{x}\cdot \widehat{i}+U_{y}\cdot \widehat{j}+U_{z}\cdot \widehat{k})\times
(B_{x}\cdot \widehat{i}+B_{y}\cdot \widehat{j}+B_{z}\cdot \widehat{k})$$
From this point on, I cannot understand the calculations needed to obtain the final formula:
$$
\vec{U}\times \vec{B}=
\widehat{i}(U_{y}B_{z}-U_{z}B_{y})... $$

 

[PeroK]

What are the cross products of the unit vectors?

Have you seen the determinant form of the cross product? I almost always use that.

 

[Delta2]

You have to use the distributive property (i.e $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times\vec{c})$ and also the following basic equations which follow from the definition of the cross product and its application on the unit vectors of a cartesian coordinate system:$$\hat i\times\hat j=\hat k,\hat j\times \hat k=\hat i, \hat k \times \hat i=\hat j$$ and also use the anticommutative property e.g $\hat{j}\times\hat{i}=-(\hat i \times \hat j)=-\hat k$
Also the property $$\lambda \vec{i}\times\mu \vec{j}=\lambda\mu (\vec{i}\times\vec{j})$$.

 

[greg_rack]

What are the cross products of the unit vectors?

$\hat{i}\times \hat{j}=\hat{k}$

Have you seen the determinant form of the cross product? I almost always use that.

Unfortunately not... I know that's what everybody use, but we shouldn't

 

[PeroK]

$\hat{i}\times \hat{j}=\hat{k}$

You use that plus the other rules that @Delta2 posted above.

 

[greg_rack]

Got it guys, thank you! @Delta2 @PeroK
Through the determinant it's way faster and easier, deriving it in this way is a pain