Determine the magnitude of the vector sum

Question

Determine the magnitude of the vector sum $V = V1 + V2$ and the angle $θx$ which $V$ makes with the positive x-axis.

I get $B$ to be 37.0414 but I am not sure what to add it do to get the correct angle.

Answer 1

Here's the easiest way: find the components of $V_1$ and $V_2$ explicitly to find $V$. Once you have the components of $V$, it is easy to get its magnitude and the angle it makes with the $x$-axis.

Let $\theta_1$ be the angle $V_1$ makes with the x axis and $\theta_2$ be the same for $V_2$.

To start, we have $$ \tan\theta_1=3/6=\frac12 $$ From there, do some quick manipulation to find $\sin\theta_1$ and $\cos\theta_1$. We have $$ V_1=(9\cos\theta_1,9\sin\theta_1)=(8.05,4.02) $$ Now, $V_2$ makes an angle of $180˚-63˚=117˚=\theta_2$ with the x-axis. So, we have $$ V_2=(12\cos\theta_2,12\sin\theta_2)=(-5.45,10.69) $$ From there, we have $$ V=V_1+V_2=(2.60,14.71) $$ Now, it's simply a matter of finding $$ |V|=\sqrt{2.60^2+14.71^2}=14.9\\ \theta_x=\arctan(14.71/2.6)=80˚ $$

Answer 2

The angle is between the two vectors. So it is

$(90-{\arctan{{\frac{1}{2}}}})+(90-63)$,

such that the ${\arctan{{\frac{1}{2}}}}$ is due to the $v_{1}$

Answer 3

We have

${\vert}{\vec{v}}{\vert}={\vert}{{\vec{v_{1}}}+{\vec{v_{2}}}}{\vert}$

such that holds

${\vert}{\vec{v}}{\vert}={\sqrt{{v^{2}_{1}}+v^{2}_{2}+2v_{1}v_{2}{\cos{\theta}}}}$,

where ${\theta}$ is the angle between the two vectors. We know that

$v^{2}_{1}=81$ and $v^{2}_{2}=144$, and

$2(9)(12)cos{\theta}=216(-7,591{\times}10^{-3})=-1639,656{\times}10^{-3}$

So, we have

${\sqrt{144+81-1,639656=223,360344}}=14,94524486={\vert}{\vec{v}}{\vert}$

About the angle:

${\theta}=(90-63)+(90-{\alpha})$,

where ${\alpha}$ is the angle between the positive axe-x and the $v_{1}$ such that the angle we compute is the resulting one, an angle ${\beta}$, for instance, where ${\beta}$ is the angle between the positive axe-y and the $v_{1}$.