4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Yes. Use trig to calculate the angle. $\endgroup$
    – nasu
    Sep 1, 2016 at 2:40
  • $\begingroup$ Used inverse tangent 7/3 = 2.3 for an answer of 66.50 degrees from the x axis or is it 7/-3? $\endgroup$
    – Jessica
    Sep 1, 2016 at 2:40

5 Answers 5

4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Amazing answer .. $\endgroup$
    – Cloud JR K
    Jun 15, 2020 at 17:22
3
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $\theta$.

$\endgroup$
-2
$\begingroup$

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
  • $\begingroup$ The angle one gets via this method is 180 degrees away from the correct angle for the vector in the question. $\endgroup$
    – David K
    May 10, 2017 at 13:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .