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

Jun 2017
399
6
Lima, Peru
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
Jul 2011
3,355
1,848
Texas
$A = 3\hat{u_2} + 3\hat{u_1}$

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

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

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

unit_vectors.jpg
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
$|A+B| \neq 3\sqrt{10}$, 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
Jul 2011
3,355
1,848
Texas
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?