# How do I find the angle a vector makes to the $+x$ axis?

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?

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• Yes. Use trig to calculate the angle. – nasu Sep 1 '16 at 2:40
• Used inverse tangent 7/3 = 2.3 for an answer of 66.50 degrees from the x axis or is it 7/-3? – Jessica Sep 1 '16 at 2:40

A general rule for finding the counterclockwise angle from the positive $x$ axis to the direction of a two-dimensional vector is explained in How 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.

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

$$\cos\theta = \frac{a_x}{|\vec{a}|} \implies \theta = \cos^{-1} \frac{a_x}{|\vec{a}|}.$$

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 $\theta$.

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?

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.

• The angle one gets via this method is 180 degrees away from the correct angle for the vector in the question. – David K May 10 '17 at 13:48