# How to find the unit vector in a rhombus as a function of a unit vector?

#### Chemist116

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?

Last edited:

#### skeeter

Math Team
$A = 3\hat{u_2} + 3\hat{u_1}$

$B = 6\hat{u_2} - 6\hat{u_1}$

$A+B = 9\hat{u_2} - 3\hat{u_1}$

$|A+B| = 3\sqrt{10}$

#### skipjack

Forum Staff
$|A+B| \neq 3\sqrt{10}$, but all the alternatives given with the problem are incorrect (or there's an error in the diagram).

I think that each side of the rhombus is intended to be 6, but $\left\|\vec{A}\right\| \neq 3$.

#### skeeter

Math Team
You're right, $A+B = 9\hat{u_2} - 3\hat{u_1} \implies |A+B| = 3\sqrt{7}$, making choice (4) almost correct ... if the u's were swapped?

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