Calculating vector cross product through unit vectors

  • #1
greg_rack
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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:
$$
\vec{U}\times \vec{B}=
\widehat{i}(U_{y}B_{z}-U_{z}B_{y})... $$
 
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  • #2
What are the cross products of the unit vectors?

Have you seen the determinant form of the cross product? I almost always use that.
 
  • #3
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 $$\lambda \vec{i}\times\mu \vec{j}=\lambda\mu (\vec{i}\times\vec{j})$$.
 
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  • #4
PeroK said:
What are the cross products of the unit vectors?
##\hat{i}\times \hat{j}=\hat{k}##

PeroK said:
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
 
  • #5
greg_rack said:
##\hat{i}\times \hat{j}=\hat{k}##
You use that plus the other rules that @Delta2 posted above.
 
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  • #6
Got it guys, thank you! @Delta2 @PeroK
Through the determinant it's way faster and easier, deriving it in this way is a pain
 
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