The problem is as follows:

The figure from below shows a set of vectors. Find the unit vector of $\vec{A}+\vec{B}$ as a function of the unit vectors $\vec{u_1}$ and $\vec{u_2}$. It is known that the sides of the rhombus is $6$.

The alternatives given are as follows:

$\begin{array}{ll}

1.&\frac{(3\vec{u_1}+\vec{u_2})}{\sqrt{10}}\\

2.&\frac{(3\vec{u_1}+\vec{u_2})}{\sqrt{13}}\\

3.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{10}}\\

4.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{7}}\\

5.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{13}}\\

\end{array}$

I'm totally lost at this problem exactly how to find the resultant. The information regarding the sides of the rhombus is a little bit ambiguous to me as it is stated. Does it mean that the sum of all sides is $6$ or each side is $6$? How should I made the interpretation? Because if it means $4x=6$ then $x=\frac{3}{2}$. But if it means that each side is $6$. then $\left\|\vec{B}\right\|=6$ and $\left\|\vec{A}\right\|=3$.

Can somebody help me here?

The figure from below shows a set of vectors. Find the unit vector of $\vec{A}+\vec{B}$ as a function of the unit vectors $\vec{u_1}$ and $\vec{u_2}$. It is known that the sides of the rhombus is $6$.

The alternatives given are as follows:

$\begin{array}{ll}

1.&\frac{(3\vec{u_1}+\vec{u_2})}{\sqrt{10}}\\

2.&\frac{(3\vec{u_1}+\vec{u_2})}{\sqrt{13}}\\

3.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{10}}\\

4.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{7}}\\

5.&\frac{(3\vec{u_1}-\vec{u_2})}{\sqrt{13}}\\

\end{array}$

I'm totally lost at this problem exactly how to find the resultant. The information regarding the sides of the rhombus is a little bit ambiguous to me as it is stated. Does it mean that the sum of all sides is $6$ or each side is $6$? How should I made the interpretation? Because if it means $4x=6$ then $x=\frac{3}{2}$. But if it means that each side is $6$. then $\left\|\vec{B}\right\|=6$ and $\left\|\vec{A}\right\|=3$.

Can somebody help me here?

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