Find Direction Angle of Vector in Plane

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    Plane Vectors
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Discussion Overview

The discussion revolves around how to find the direction angle of a vector in a two-dimensional plane. Participants explore various methods and considerations involved in determining this angle, including the use of trigonometric functions and the vector's components.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest using the tangent function, where the angle θ can be found using the ratio of the vector's components, specifically θ = arctan(b/a).
  • Others emphasize the importance of considering the quadrant in which the vector lies, as this affects the determination of the angle θ.
  • One participant mentions that the direction angle is not uniquely determined by the tangent function alone and requires adjustments based on the vector's quadrant.
  • There is a suggestion that the cosine of the angle could be found using the dot product, although this approach is questioned by other participants regarding its applicability to a single vector.
  • Some participants express confusion over the dot product's relevance to the original question about a single vector.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to find the direction angle of a vector, with multiple competing views and approaches presented throughout the discussion.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the vector's components and the specific context in which the angle is being calculated. The discussion also reflects varying interpretations of the dot product's role in this context.

htk
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can anyone please help me how to find the direction angle of a vector? Thank you!
 
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Think about the triangle that the vector and its components form.
 
htk said:
can anyone please help me how to find the direction angle of a vector? Thank you!
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:
[tex]\theta= arctan(\frac{a}{b})[/tex].
 
Welcome to PF!

Hi htk! Welcome to PF! :smile:
htk said:
can anyone please help me how to find the direction angle of a vector? Thank you!

You find the cosine of the angle …

which you do by finding the dot product. :wink:​
 


tiny-tim said:
Hi htk! Welcome to PF! :smile:


You find the cosine of the angle …

which you do by finding the dot product. :wink:​
His question was about a single vector. What do you want him to take the dot product with?
 
… just answering the question as asked! …

HallsofIvy said:
His question was about a single vector. What do you want him to take the dot product with?

with whatever mysterious entity he had in mind when he specified :wink: …
htk said:
can anyone please help me how to find the direction angle of a vector?
 
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
 


HallsofIvy said:
His question was about a single vector. What do you want him to take the dot product with?

I suppose he could dot it with (1,0).

EDIT: Modulo sign.
 
Last edited:

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