Unit vectors and direction cosines

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Homework Help Overview

The discussion revolves around the properties of unit vectors and direction cosines, specifically focusing on the vector \(\vec{A}\) and its relationship with the positive x-axis. Participants explore the definition of a unit vector and the implications of dividing a vector by its magnitude.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of a unit vector and attempt to demonstrate that the magnitude of \(\frac{\vec{A}}{A}\) is 1. Questions arise regarding the direction of the unit vector and the concept of direction cosines, particularly in relation to the angle \(\theta\) and the unit vector \(\hat{i}\).

Discussion Status

The discussion is active, with participants providing insights and clarifications about the properties of unit vectors and direction cosines. Some have made progress in understanding the relationship between the vector's magnitude and direction, while others are still questioning the implications of the dot product and its interpretation.

Contextual Notes

There is some ambiguity regarding the notation and definitions, particularly the interpretation of \(A\) as the magnitude of \(\vec{A}\). Participants are navigating these assumptions as they explore the concepts.

Ragnarok7
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Let [itex]\vec{A}[/itex] represent any nonzero vector. Why is [itex]\frac{\vec{A}}{A}[/itex] a unit vector and what is its direction? If θ is the angle that [itex]\vec{A}[/itex] makes with the positive x-axis, explain why [itex]\frac{\vec{A}}{A}\cdot\hat{i}[/itex] is called the direction cosine for that axis.


I am self-studying and this question has me stumped. I am familiar with the formula for a unit vector but I don't know why it's true and I have never really heard of a direction cosine. Could anyone give me some hints, perhaps?
 
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Okay, I've thought about the first part a little more. A unit vector is a vector whose magnitude is 1. So we must show that the magnitude of [itex]\frac{\vec{A}}{A}[/itex] is 1. Through algebraic manipulation I can show that pretty easily. I still don't know about the direction or the direction cosines, though.
 
Ragnarok7 said:
Let [itex]\vec{A}[/itex] represent any nonzero vector. Why is [itex]\frac{\vec{A}}{A}[/itex] a unit vector and what is its direction? If θ is the angle that [itex]\vec{A}[/itex] makes with the positive x-axis, explain why [itex]\frac{\vec{A}}{A}\cdot\hat{i}[/itex] is called the direction cosine for that axis.
It isn't made clear, but presumably [itex]A = |\vec{A}|[/itex]. Since that is a scalar, what does dividing a vector by it do to the vector's direction?
If [itex]\hat{i}[/itex] is a unit vector at angle θ to [itex]\vec{A}[/itex], what is the magnitude of [itex]\vec{A}\cdot\hat{i}[/itex]?
 
Ah, okay. (Yes, [itex]A[/itex] means [itex]|\vec{A}|[/itex].) Since it's a scalar, then the direction will remain the same.

[itex]\vec{A}\cdot \hat{i}[/itex] is equivalent to [itex]|\vec{A}|\cdot|\hat{i}|\cdot\cos\theta[/itex], or, since [itex]\hat{i}[/itex] is has a magnitude of 1, just [itex]|\vec{A}|\cos\theta[/itex]. I don't understand what is meant by the magnitude of this dot product, though. It's just a scalar, right?

Thanks so much for your help!
 
I think I've got it. The magnitude of [itex]\frac{\vec{A}}{|\vec{A}|}[/itex] is just 1, and so is the magnitude of [itex]\hat{i}[/itex]. So [itex]\frac{\vec{A}}{|\vec{A}|}\cdot\hat{i}[/itex] is just [itex]\cos\theta[/itex].

So all that is just a long way of saying that the direction cosine is just the cosine of the angle between the x-axis and the vector?
 

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