# College Physics 3.6 – The resultant of two vectors added in reversed order

## Solution:

Reversing the order of the vectors, we come up with the triangle as shown below.

We can actually see that the resultant can be solved by considering the triangle formed by the three vectors. The same with the procedures used in the previous problem, we have $\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}$ $\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=19.4892\:\text{m}$

To solve for the direction, we can solve for the angle included between B and R first. $\displaystyle \frac{12\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }$ $\displaystyle \text{C}=\sin ^{-1}\left(\frac{12\:\text{m}\:\left(\sin 70^{\circ} \right)}{19.4892\:\text{m}}\right)$ $\text{C}=35.35^{\circ}$

Finally to solve for the direction of the resultant, we have $\theta =40^{\circ} -35.35^{\circ} =4.65^{\circ}$

The compass direction is $4.65^{\circ} \:\text{South of West}$.

This result is actually the same as in the previous problem when the order of the vectors are not reversed. No matter what is the order of the vectors we are adding, the result is still the same. This proves that $\vec{A}+\vec{B}=\vec{B}+\vec{A}$.