Unit vectors and direction cosines

Ragnarok7
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Let \vec{A} represent any nonzero vector. Why is \frac{\vec{A}}{A} a unit vector and what is its direction? If θ is the angle that \vec{A} makes with the positive x-axis, explain why \frac{\vec{A}}{A}\cdot\hat{i} 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 \frac{\vec{A}}{A} 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 \vec{A} represent any nonzero vector. Why is \frac{\vec{A}}{A} a unit vector and what is its direction? If θ is the angle that \vec{A} makes with the positive x-axis, explain why \frac{\vec{A}}{A}\cdot\hat{i} is called the direction cosine for that axis.
It isn't made clear, but presumably A = |\vec{A}|. Since that is a scalar, what does dividing a vector by it do to the vector's direction?
If \hat{i} is a unit vector at angle θ to \vec{A}, what is the magnitude of \vec{A}\cdot\hat{i}?
 
Ah, okay. (Yes, A means |\vec{A}|.) Since it's a scalar, then the direction will remain the same.

\vec{A}\cdot \hat{i} is equivalent to |\vec{A}|\cdot|\hat{i}|\cdot\cos\theta, or, since \hat{i} is has a magnitude of 1, just |\vec{A}|\cos\theta. 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 \frac{\vec{A}}{|\vec{A}|} is just 1, and so is the magnitude of \hat{i}. So \frac{\vec{A}}{|\vec{A}|}\cdot\hat{i} is just \cos\theta.

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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