You are given two vectors $\vec a = -3.00\hat i + 7.00\hat j$ and $\vec b= 4.00\hat i + 2.00\hat j$. Let the counterclockwise angles be positive.
What angle $\theta (\vec a)$ where $0^\circ \le \theta (\vec a) < 360^\circ $, does $\vec a$ make with the $+x$-axis?
I drew a right triangle with a $\vec ax$ component of $-3$ and an $\vec ay$ component of $7$. Do I just use trig to find the angle off the $x$-axis?
$\endgroup$ 25 Answers
$\begingroup$A general rule for finding the counterclockwise angle from the positive $x$ axis to the direction of a two-dimensional vector is explained inHow to convert components into an angle directly (for vectors)?
The basic idea is that you first take the arc tangent of the $y$-component divided by the $x$-component:$$ \theta_1 = \arctan\frac{a_y}{a_x}. $$
One complication is that this procedure gives the same resulting $\theta_1$for the vector $\langle a_x,a_y \rangle$ and the vector $\langle -a_x,-a_y \rangle,$ which are two vectors in directions $180$ degrees apart. So $\theta_1$ may be in the direction you want, or in the direction$180$ degrees opposite.
In fact $\theta_1$ will be in the correct direction whenever $a_x > 0,$since that is how the arc tangent function is designed to work. (It always produces angles in the range $-\frac\pi2$ to $\frac\pi2$ radians, that is, $-90$ to $90$ degrees, which correspond to vectors with positive $x$-components.)
The cases where $\theta_1$ is wrong are precisely the cases where$a_x < 0,$ so if $a_x < 0$ (as it is in your particular question) you have to add $180$ degrees to $\theta_1.$
A second complication (which does not actually come up in your particular question) occurs because the procedures given above produce a result in the range $-90$ degrees to $270$ degrees, but people usually want an answer in the range $-180$ to $180$ or $0$ to $360.$ The solution to this, of course, is to add or subtract $360$ degrees as needed to get an answer in the desired angle range.
$\endgroup$ 1 $\begingroup$You can use trig, but since these are vectors, there's a far easier way. Use dot products!
Suppose you have a vector $\vec{a}$ and you need to find the angle it makes with the $x$-axis. So take a unit vector along the $x$-axis, viz. $\hat{i}$. Using the dot product of these vectors,
$$ \vec{a}\cdot\hat{i} = |\vec{a}||\hat{i}|\cos\theta $$
where $\theta$ is the angle between the two vectors. Since $|\hat{i}|$ = 1, and $\vec{a}\cdot\hat{i} = a_x$, hence
$$ \cos\theta = \frac{a_x}{|\vec{a}|} \implies \theta = \cos^{-1} \frac{a_x}{|\vec{a}|}. $$
$\endgroup$ $\begingroup$Problem: Counterclockwise angle between $ +x-$ axis and vector $\vec{a} = (-3,7)$.
Draw a $ x-,y- $ axes, and vector $\vec{a}$ pointing from the origin $(0,0) $ to $ (-3,7)$. Verify that we are in the second quadrant , I.e. $\angle \theta $ is obtuse.
Normalize vector $\vec{a}$ , call it
$\vec{n}$ : $(1/√58) (-3,7)$.
Dot product: $\vec{n} \cdot \vec{e_x}$ = $cos(\theta)$.
$cos(\theta)$ = $(1/√58) (-3, 7) \cdot (1, 0) $= $-3 (1/√58)$.
Left: Look up $arccos ( \theta)$ to find the obtuse angle $\theta$.
$\endgroup$ $\begingroup$Given the vector $(-3,7),$ the length is $\sqrt{(-3)^2+7^2}=\sqrt{58},$ so the vector becomes $$\sqrt{58}\left(\frac{-3}{\sqrt{58}},\frac{7}{\sqrt{58}}\right).$$ Thus, $$\cos\phi=\frac{-3}{\sqrt{58}}$$ and $$\sin\phi=\frac{7}{\sqrt{58}},$$ where $\phi$ is the angle that the vector makes with the positive $x$-axis.
There are two angles with this cosine between $0$ and $360$ degrees -- one acute and one obtuse, so from the sign of the cosine we get that the angle we seek is the obtuse one. Can you continue now?
$\endgroup$ $\begingroup$Yeah you just use trig. Tan is the easiest. I saw your question about whether it is arctan(7/3) or arctan(7/-3) it is the 7/-3. Now this will give you a negative angle. This is outside your range so add 360 to it to get the answer that is in your range.
$\endgroup$ 1
