Finding the Angle a Given Vector Makes with the Positive x-axis

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Feb 10, 2022
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Since we know that any given vector $\vecs v$ in the $xy$ -plane can be normalized to find a unit vector in the same direction, it becomes fairly easy to determine the angle between this vector $\vecs v$ and the positive $x$ -axis. Once we normalize the vector $\vecs v$ , we know that this unit vector could be placed on the unit circle and can be written in the form:

$<\cos θ, \sin θ> = \cos θ \,\hat{\mathbf i} + \sin θ \,\hat{\mathbf j}\nonumber$

From our work above, we know the unit vector in the same direction as the vector $\vecs v$ could be written as $\vecs w = \frac{\vecs v}{\|\vecs v\|}.$

Let's say this simplifies down to a unit vector with components $\vecs w = \langle a, b\rangle.$ Since this is a unit vector, we know the first component is equal to $\cos θ$ and the second component is equal to $\sin θ,$ where $θ$ is the angle between this vector and the positive $x$ -axis.

That is,

$\cos θ = a$ and $\sin θ = b.$

If $θ$ is in the first quadrant, either term will give us the correct $θ$ using either $θ = \arccos a$ or $θ = \arcsin b.$ If we are in another quadrant (look at the signs of the two components to tell which quadrant it lies in), we need to think more carefully about the angle we obtain to be sure it is in the correct quadrant, making an adjustment as needed or using the other inverse trig function.

Example $\PageIndex{10}$

a. Given $\vecs v = 3 \,\hat{\mathbf i} + 4 \,\hat{\mathbf j},$ find the angle $θ$ that this vector makes with the positive $x$ -axis (to the nearest hundredth of a degree).

b. Find the angle $θ$ between the positive $x$ -axis and the vector $\vecs u = <-2, 4>$ (to the nearest thousandth of a radian and to the nearest tenth of a degree).

c. Find the angle $θ$ between the positive $x$ -axis and the vector $\vecs w = 1\,\hat{\mathbf i} - 3 \,\hat{\mathbf j}$ (to the nearest thousandth of a radian and to the nearest tenth of a degree).

Solution

a. Since this vector is in the first quadrant, this one is easiest to work out. First we normalize this vector (find a unit vector in the same direction).

Since $\|\vecs v\| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5,$ we get the unit vector,

$\vecs u = \frac{3}{5} \,\hat{\mathbf i} + \frac{4}{5} \,\hat{\mathbf j}.\nonumber$

Now we can find $θ$ with either component. That is,

$θ = \arccos \frac{3}{5} = \arcsin \frac{4}{5} \approx 53.13°.\nonumber$

b. The signs of the components of this vector tell us it is in the second quadrant, so we know that the arccosine function will still work for us, as it returns an angle in the first or second quadrants (between $0$ and $\pi$ radians). But note that arcsine would not give us the correct angle (right away) since it always returns an angle between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2},$ (which effectively returns an angle in the first or fourth quadrants). We could still use it here, knowing it would give us an angle in the first quadrant here (since the sign of the second component is positive). Because of the symmetry involved, we would have to subtract the angle we get from $\pi$ radians to obtain the correct angle in the second quadrant. This is not trivial, as it requires a clear understanding and an ability to visualize the angles, but we'll show how it works below.

First let's normalize this vector.

Since $\|\vecs u\| = \sqrt{(-2)^2 + 4^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5},$ we get the unit vector,

$\left\langle \frac{-2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}} \right\rangle = \left\langle -\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5} \right\rangle. \nonumber$

Now using the first component, we find,

$θ = \arccos -\frac{\sqrt{5}}{5} \approx 2.034\text{ radians }\approx 116.6°.\nonumber$

Note that this angle is indeed in the second quadrant, as it needed to be!

Now let's see what would have happened if we used the second component and the arcsine function.

$\arcsin \frac{2\sqrt{5}}{5} \approx 1.1071\text{ radians }\approx 63.4°\nonumber$

As explained above, since this angle is in the first quadrant, we'll need to subtract it from $\pi$ radians to get to the correct angle.

So here we get,

$θ = \pi - \arcsin \frac{2\sqrt{5}}{5} \approx 3.14159 - 1.1071 \approx 2.034\text{ radians}\nonumber$

or in degrees,

$θ = 180° - \arcsin \frac{2\sqrt{5}}{5} \approx 180° - 63.4 \approx 116.6°.\nonumber$

We can see that this is the same angle!

c. Since $\|\vecs w\| = \sqrt{10},$ the unit vector in the same direction as $\vecs w = 1\,\hat{\mathbf i} - 3 \,\hat{\mathbf j}$ is $\frac{\sqrt{10}}{10}\,\hat{\mathbf i} - \frac{3\sqrt{10}}{10} \,\hat{\mathbf j}.$ Because this vector is clearly in the fourth quadrant, the arcsine function is our choice.

Here,

$θ = \arcsin -\frac{3\sqrt{10}}{10} \approx -1.249$ radians $\approx -71.6°.$

This is a correct answer, but if we were asked for an angle between $0$ and $2\pi$ radians, for example, we would still need to adjust it by adding $2\pi$ radians to the radian version of the angle. Then $θ \approx 5.034$ radians.

If the vector is in the third quadrant neither the arccosine nor the arcsine give us the right angle. We could use either one and adjust the angle as we showed with arcsine in Example $\PageIndex{10}$ , part b above. But there is another option. Although it also may need an adjustment, this adjustment is somewhat easier to make.

If we run into difficulty with the approach above or just want to use a different method, we can instead use the arctangent function to find the angle $θ$ a vector $\vecs v$ makes with the positive $x$ -axis. One advantage this approach gives us is that we don't need to normalize the vector first. Since $\tan θ = \dfrac{\sin θ}{\cos θ},$ this ratio of the two vector components automatically removes any scalar that may be present and gives us the correct tangent ratio.

This figure is a right triangle. There is an angle labeled theta. The two sides are labeled “magnitude of v times cosine theta” and “magnitude of v times sine theta.” The hypotenuse is labeled “magnitude of v.” — Figure $\PageIndex{20}$ : The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.

As we observe in Figure $\PageIndex{20}$ , given a non-zero vector, $\vecs v = <\|\vecs v\|\cos θ, \|\vecs v\|\sin θ>,$ we know that,

$\tan θ = \frac{\|\vecs v\|\sin θ}{\|\vecs v\|\cos θ} = \frac{\sin θ}{\cos θ}. \nonumber$

However, like the arcsine function, the arctangent function only returns angles between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2},$ so we will still need to adjust the resulting angle based on which quadrant in which we know it must lie. At least in this case, the worst that we will need to do is to add $\pi$ radians or $180°$ to our angle to get it into the correct quadrant. [Note that this is because the tangent will give the same results when the signs of the sine and cosine are the same, and it will give the same results when the signs of the two ratios are different.]

Example $\PageIndex{11}$

Find the angle $θ$ between the positive $x$ -axis and the vector $\vecs d = <-5, -1>$ (to the nearest thousandth of a radian and to the nearest tenth of a degree).

Solution

The vector $\vecs d = <-5, -1>$ is clearly in the third quadrant, and will not be given directly by either the arccosine function or the arcsine function, as mentioned above. But we do know that the arctangent function will return an angle in the first quadrant in this situation (the two components are the same sign and thus their ratio is positive), and the correct angle in the third quadrant will be 180° or $\pi$ radians greater than the angle we are given by arctangent in the first quadrant.

Remembering that we don't need to determine the unit vector this time, we instead write the tangent ratio we will need to work with.

$\tan θ = \frac{-1}{-5} = \frac{1}{5}\nonumber$

So,

$θ = \pi + \arctan \frac{1}{5} \approx \pi + 0.197$ radians $\approx 3.339$ radians $\approx 191.3°.$

Note that in degrees, $\arctan \frac{1}{5} \approx 11.3°.$ Adding $180°$ also gives us $191.3°.$

Exercise $\PageIndex{10}$

Use the arctangent function to determine the angle $θ$ between the given vector and the positive $x$ -axis. Give the angle accurate to the nearest tenth of a degree.

a. $\vecs u = <-4, -8>$

b. $\vecs v = <10, -15>$

c. $\vecs w = <-7, 1>$

Answer: a. $θ \approx 243.4°$
b. $θ \approx -56.3° = 303.7°$
c. $θ \approx 171.9°$

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