The problem is as follows:

The figure from below shows vectors $\vec{A}$ and $\vec{B}$. It is known that $A=B=3$. Find $\vec{E}=(\vec{A}+\vec{B})\times(\vec{A}-\vec{B})$.

The alternatives are:

$\begin{array}{ll}

1.&-18\hat{k}\\

2.&-9\hat{k}\\

3.&-\sqrt{3}\hat{k}\\

4.&3\sqrt{3}\hat{k}\\

5.&9\hat{k}\\

\end{array}$

What I've attempted here was to try to decompose each vectors

A⃗ =⟨3cos53∘,3sin53∘⟩A→=⟨3cos53∘,3sin53∘⟩

B⃗ =A⃗ =⟨3cos(53∘+30∘),3sin(53∘+30∘)⟩B→=A→=⟨3cos(53∘+30∘),3sin(53∘+30∘)⟩

But attempting to use these relationships does seem to extend the algebra too much. Does there exist another way? Or could it be that I am overlooking something?

Can someone help me with this?

The figure from below shows vectors $\vec{A}$ and $\vec{B}$. It is known that $A=B=3$. Find $\vec{E}=(\vec{A}+\vec{B})\times(\vec{A}-\vec{B})$.

The alternatives are:

$\begin{array}{ll}

1.&-18\hat{k}\\

2.&-9\hat{k}\\

3.&-\sqrt{3}\hat{k}\\

4.&3\sqrt{3}\hat{k}\\

5.&9\hat{k}\\

\end{array}$

What I've attempted here was to try to decompose each vectors

A⃗ =⟨3cos53∘,3sin53∘⟩A→=⟨3cos53∘,3sin53∘⟩

B⃗ =A⃗ =⟨3cos(53∘+30∘),3sin(53∘+30∘)⟩B→=A→=⟨3cos(53∘+30∘),3sin(53∘+30∘)⟩

But attempting to use these relationships does seem to extend the algebra too much. Does there exist another way? Or could it be that I am overlooking something?

Can someone help me with this?

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