htk
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can anyone please help me how to find the direction angle of a vector? Thank you!
Find Direction Angle of Vector in Plane
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Dragonfall
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Think about the triangle that the vector and its components form.
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I presume you are talking about vectors in the plane since vectors in three dimensions have three "direction angles". How are you given the vector? If in x,y components, say ai+ bj, then b/a is the tangent of the angle the vector makes with the x-axis:htk said:
[tex]\theta= arctan(\frac{a}{b})[/tex].
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Welcome to PF!
Hi htk! Welcome to PF!
You find the cosine of the angle …
Hi htk! Welcome to PF!
htk said:
You find the cosine of the angle …
which you do by finding the dot product. 
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His question was about a single vector. What do you want him to take the dot product with?tiny-tim said:Hi htk! Welcome to PF!
You find the cosine of the angle …
which you do by finding the dot product.
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… just answering the question as asked! …
with whatever mysterious entity he had in mind when he specified …
HallsofIvy said:
with whatever mysterious entity he had in mind when he specified
…htk said:
Elucidus
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A two-dimensional vector [itex]\langle a,b \rangle[/itex] will have a direction angle [itex]\theta \text{ such that } \tan \theta = b / a[/itex] (not (a/b)) but this does not uniquely determine [itex]\theta[/itex], even if it is restricted to the interval [itex][0, 2 \pi )[/itex].
You also need to consider in which quadrant does the vector lie. You need to adjust the value of [itex]\theta[/itex] so that it falls into the correct quadrant.
For example, the vector [itex]\langle -3, 3 \rangle[/itex] has a direction angle so that [itex]\tan \theta = 3 / -3 = -1 \text{ which implies } \theta = -\pi /4 + n \pi[/itex] for an appropriate choince of integer n. Since the vector is in the second quadrant, we need to select the angle to fall there, so [itex]\theta = 3\pi / 4[/itex] (here n = 1).
I hope this helps.
--Elucidus
You also need to consider in which quadrant does the vector lie. You need to adjust the value of [itex]\theta[/itex] so that it falls into the correct quadrant.
For example, the vector [itex]\langle -3, 3 \rangle[/itex] has a direction angle so that [itex]\tan \theta = 3 / -3 = -1 \text{ which implies } \theta = -\pi /4 + n \pi[/itex] for an appropriate choince of integer n. Since the vector is in the second quadrant, we need to select the angle to fall there, so [itex]\theta = 3\pi / 4[/itex] (here n = 1).
I hope this helps.
--Elucidus
Dragonfall
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HallsofIvy said:
I suppose he could dot it with (1,0).
EDIT: Modulo sign.
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