# Homework Help: Norm of a vector

1. Mar 24, 2009

### physguy09

Ok, so I have no idea how to take the norm of a vector composed of two vectors. I have
$$\vec{q}$$=$$\vec{pi}$$ - $$\vec{pf}$$

we are given:
|$$\vec{pi}$$|=|$$\vec{pf}$$|=|$$\vec{p}$$|

so i know that
|$$\vec{q}$$| $$\neq$$ 0, that would be too easy, and it doesn't make sense.

now, is the following right? it just doesn't seem to be

|$$\vec{q}$$|=$$\sqrt{2}$$*|$$\vec{p}$$|

2. Mar 24, 2009

### CompuChip

You can try working it out explicitly through the inner product.
Note that
$$|\vec q|^2 = \vec q \cdot \vec q$$
so
$$(\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = \ldots?$$

3. Mar 24, 2009

### physguy09

ok so that gets me |$$\vec{q}$$|=$$\sqrt{|\vec{p_{i}^2}|+|\vec{p_{f}^2}|}$$,
would that then be $$\sqrt{2}$$p?

4. Mar 24, 2009

### Tom Mattson

Staff Emeritus
No, it doesn't get you that. Not unless $p_i$ is orthogonal to $p_f$, and the problem statement (as you gave it) doesn't say that.

5. Mar 24, 2009

### CompuChip

Please stop guessing and work it out properly.

$$(\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = (\vec p_i - \vec p_f) \cdot \vec p_i - (\vec p_i - \vec p_f) \cdot \vec p_f = \vec p_i \cdot \vec p_i + \cdots$$

6. Mar 24, 2009

### physguy09

not guessing, just not typing it out entirely. i wrongly assumed pi and pf were orthogonal, but i was trying to make use of the fact that
|p|=|pi|= |pf|.

i know |q|=$$\sqrt{\vec{p_{i}^2} + \vec{p_{f}^2 }-2*\vec{p_i}*\vec{p_f}}$$

but i know |p|=|pi|= |pf|, so |p|$$^2$$ must equal |p|2=|pi|2 = |pf|2, hence, possibly with faulty reasoning,
|q|=$$\sqrt{2*p^2-2*\vec{p_i}*\vec{p_f}}$$

7. Mar 24, 2009

### CompuChip

You are getting closer, it's just not entirely clear to me what
$2 * \vec p_i * \vec p_f$
means. In particular, what do the asterisks mean?

After clearing that up, can you write that expression in terms of p as well?

8. Mar 24, 2009

### physguy09

apologies, I began to get sloppy, as I'm still trying to get used to Latex, it should be
|q|=$$\sqrt{2 p^2-2 \vec{p_i}\cdot\vec{p_f}}$$. I would not know how to write it out in terms of p, and it might be useless in this case, as we are supposed to plug in a known value for p. $$\vec{q}$$ was mainly momentum transfer, and we were given that for this case $$\vec{q} = \vec{p_i} - \vec{p_f}$$. i already turned in the assignment, just wanted to reach the answer for the sake of learning.

9. Mar 25, 2009

### CompuChip

Yes, at first you will work more slowly when you have to do the math and learn how to write it out in $\LaTeX$ - compliments for doing it anyway :)

What I was hinting at was to define $\theta$ as the angle between the ingoing and outgoing momentum, and writing
$$\vec p_i \cdot \vec p_f = |\vec p_i| |\vec p_f| \cos\theta = 2 |\vec p| \cos\theta$$.

Then you can take $\sqrt{2}p$ outside the square root, if you want to write it more beautifully.
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