# Unit vectors and direction cosines

1. Nov 14, 2012

### Ragnarok7

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?

2. Nov 14, 2012

### Ragnarok7

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.

3. Nov 14, 2012

### haruspex

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 $\vec{A}\cdot\hat{i}$?

4. Nov 15, 2012

### Ragnarok7

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!!

5. Nov 15, 2012

### Ragnarok7

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?