How To Find Angle Between Vector And X Axis

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

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Since we know that any given vector \(\vecs v\) in the \(xy\)-airplane can exist normalized to notice a unit of measurement vector in the aforementioned direction, information technology becomes adequately piece of cake to determine the angle betwixt this vector \(\vecs five\) 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 tin 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 aforementioned management as the vector \(\vecs v\) could be written as \(\vecs west = \frac{\vecs 5}{\|\vecs v\|}.\)

Let'south say this simplifies downwards to a unit vector with components \(\vecs w = \langle a, b\rangle.\)  Since this is a unit vector, we know the kickoff component is equal to \(\cos θ\) and the 2d component is equal to \(\sin θ,\) where \(θ\) is the angle betwixt this vector and the positive \(10\)-centrality.

That is,

\( \cos θ  = a \) and \(\sin θ = b.\)

If \(θ\) is in the outset quadrant, either term volition 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 information technology lies in), nosotros demand to think more than carefully about the bending nosotros obtain to be sure it is in the correct quadrant, making an adjustment as needed or using the other inverse trig function.

Example \(\PageIndex{ten}\)

a. Given \(\vecs 5 = 3 \,\hat{\mathbf i} + four \,\hat{\mathbf j},\) find the angle \(θ\) that this vector makes with the positive \(ten\)-axis (to the nearest hundredth of a degree).

b. Observe the bending \(θ\) between the positive \(x\)-centrality and the vector \(\vecs u = <-ii, four>\) (to the nearest thousandth of a radian and to the nearest 10th of a degree).

c. Find the angle \(θ\) between the positive \(10\)-axis and the vector \(\vecs due west = 1\,\hat{\mathbf i} - 3 \,\hat{\mathbf j}\) (to the nearest thousandth of a radian and to the nearest tenth of a caste).

Solution

a. Since this vector is in the first quadrant, this i is easiest to piece of work out.  First we normalize this vector (find a unit of measurement vector in the same management).

Since \(\|\vecs v\| = \sqrt{3^2 + 4^2} = \sqrt{9+sixteen} = \sqrt{25} = v,\) we get the unit vector,

\[ \vecs u = \frac{iii}{v} \,\hat{\mathbf i} + \frac{four}{5} \,\hat{\mathbf j}.\nonumber\]

Now we tin notice \(θ\) with either component.  That is,

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

b. The signs of the components of this vector tell usa it is in the second quadrant, and then we know that the arccosine office will yet work for us, as it returns an angle in the first or 2nd quadrants (between \(0\) and \(\pi\) radians).  But note that arcsine would not give usa the right angle (correct away) since it always returns an bending between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2},\) (which effectively returns an angle in the first or fourth quadrants).  We could nonetheless utilise it here, knowing information technology would give us an angle in the showtime quadrant here (since the sign of the second component is positive).  Because of the symmetry involved, we would accept to subtract the angle we get from \(\pi\) radians to ontain the correct angle in the second quadrant.  This is not trivial, as it requires a clear agreement and an ability to visualize the angles, but we'll testify how it works below.

Get-go let'southward normalize this vector.

Since \(\|\vecs u\| = \sqrt{(-ii)^two + 4^2} = \sqrt{4+sixteen} = \sqrt{twenty} = 2\sqrt{v},\) we get the unit vector,

\[\langle \frac{-2}{ii\sqrt{5}}, \frac{4}{2\sqrt{5}} \rangle = \langle -\frac{\sqrt{5}}{5}, \frac{2\sqrt{five}}{5}  \rangle. \nonumber\]

Now using the first component, nosotros find,

\[ θ = \arccos  -\frac{\sqrt{5}}{5} \approx ii.034\text{ radians }\approx 116.6°.\nonumber\]

Annotation that this angle is indeed in the 2nd quadrant, as it needed to be!

At present allow's see what would have happened if we used the second component and the arcsine function.

\[ \arcsin  \frac{two\sqrt{5}}{5} \approx one.1071\text{ radians }\approx 63.4°\nonumber\]

As explained above, since this angle is in the commencement quadrant, we'll need to subtract it from \(\pi\) radians to go to the correct angle.

And so hither we get,

\[ θ = \pi -  \arcsin  \frac{2\sqrt{5}}{v} \approx 3.14159 - i.1071 \approx two.034\text{ radians}\nonumber\]

or in degrees,

\[ θ = 180° -  \arcsin  \frac{two\sqrt{5}}{5} \approx 180° - 63.4 \approx 116.6°.\nonumber\]

We can come across that this is the same bending!

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

Here,

\( θ = \arcsin -\frac{3\sqrt{ten}}{10} \approx -ane.249\) radians \(\approx -71.half-dozen°.\)

This is a right respond, but if we were asked for an angle betwixt \(0\) and \(2\pi\) radians, for instance, nosotros would still need to conform it by adding \(2\pi\) radians to the radian version of the angle.  Then \(θ \approx five.034\) radians.

If the vector is in the 3rd quadrant neither the arccosine nor the arcsine give us the right bending.  We could use either one and adjust the bending as we showed with arcsine in Example\(\PageIndex{10}\), part b in a higher place.  But there is some other option.  Although it also may demand an adjustment, this aligning is somewhat easier to make.

If we run into difficulty with the approach above or just want to use a different method, we tin instead use the arctangent function to find the angle \(θ\) a vector \(\vecs v\) makes with the positive \(x\)-centrality.  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.

Figure \(\PageIndex{20}\): The components of a vector grade the legs of a correct triangle, with the vector as the hypotenuse.

As nosotros observe in Figure \(\PageIndex{20}\), given a non-nothing vector, \(\vecs five = <\|\vecs v\|\cos θ, \|\vecs v\|\sin θ>,\) nosotros know that,

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

Still, like the arcsine function, the arctangent part simply returns angles between \(-\tfrac{\pi}{2}\) and \(\tfrac{\pi}{2},\) and then nosotros will still demand 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 volition demand to practise is to add \(\pi\) radians or \(180°\) to our angle to go 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 \(\PageIndex{eleven}\)

Find the angle \(θ\) betwixt the positive \(ten\)-centrality and the vector \(\vecs d = <-5, -i>\) (to the nearest thousandth of a radian and to the nearest tenth of a caste).

Solution

The vector \(\vecs d = <-5, -1>\) is clearly in the third quadrant, and will non be given directly by either the arccosine role or the arcsine function, as mentioned above.  But nosotros do know that the arctangent function will return an angle in the first quadrant in this state of affairs (the two components are the same sign and thus their ratio is positive), and the correct angle in the tertiary quadrant volition be 180° or \(2\pi\) radians greater than the angle we are given by arctangent in the kickoff quadrant.

Remembering that we don't need to make up one's mind the unit vector this time, nosotros instead write the tangent ratio we will need to work with.

\[\tan θ = \frac{-1}{-five}  = \frac{1}{5}\nonumber\]

And then,

\(θ = \pi + \arctan \frac{ane}{5} \approx \pi + 0.197\) radians \(\approx iii.339\) radians \(\approx 191.3°.\)

Note that in degrees, \(\arctan \frac{1}{5} \approx 11.iii°.\)  Adding \(180°\) besides gives us \(191.3°.\)

Practise \(\PageIndex{10}\)

Use the arctangent function to decide the angle \(θ\) between the given vector and the positive \(x\)-axis.  Give the bending accurate to the nearest tenth of a caste.

a. \(\vecs u = <-4, -eight>\)

b. \(\vecs 5 = <10, -fifteen>\)

c. \(\vecs w = <-7, 1>\)

Respond

a. \(θ \approx 243.four°\)
b. \(θ \approx -56.3° = 303.7°\)
c. \(θ \approx 171.9°\)

Source: https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/1%3A_Vectors_in_Space/Finding_the_Angle_a_Given_Vector_Makes_with_the_Positive_x-axis

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