# Kleppner/Kolenkow Vector Problem

1. Dec 17, 2015

### Cosmophile

Let $\vec {A}$ be an arbitrary vector and let $\hat {n}$ be a unit vector in some fixed direction. Show that $$\vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}.$$

Here is my attempt, which I do believe is fine; I just want to know if there is a better way to approach the problem:

The first term, $(\vec{A} \cdot \hat {n})\hat{n}$ is a new vector $\vec {C}$ with magnitude $A \cos \theta$ in the $\hat {n}$ direction. For the second term, I simply did the operations from left-to-right, first getting $\vec {C} = ( \hat {n} \times \vec {A})$, which if looking down on the plane made by $\vec {A}$ and $\hat {n}$, would point upwards out of the page (I drew $\vec {A} to be above \hat {n}$). Then, I took $\vec {C} \times \hat {n}$ and called this new vector $\vec {D}$ which has a magnitude $A \sin \theta$ and points in a direction $\hat d$, which is perpendicular to $\hat {n}$. So, I end up with $$\vec {A} = A(\cos \theta \hat {n} + \sin \theta \hat{d}).$$

So, I've shown that the statement $\vec {A} = (\vec{A} \cdot \hat {n})\hat{n} + (\hat {n} \times \vec {A}) \times \hat {n}$ just breaks $\vec {A}$ up into its horizontal and vertical components. At least, that's what I make of my work. What do you guys think?

2. Dec 17, 2015

### Cosmophile

I should say that, while I've heard of the triple product, I intentionally didn't use it because (1) it hasn't been taught yet in my book and (2) I don't truly know what it means. I figured there had to be another way to do the problem, so I sought it out.

3. Dec 18, 2015

### Fredrik

Staff Emeritus
Looks good.

The main alternative is to use the definitions of the cross product and the dot product to show that $(\hat n\times\vec A)\times\hat n =\vec A-(\vec A\cdot\hat n)\hat n$.

4. Dec 19, 2015

### Cosmophile

Sorry, I'm not sure I see the difference between what we are saying.

5. Dec 19, 2015

### Fredrik

Staff Emeritus
The definitions I had in mind are these formulas:
\begin{align*}
&\vec x\times\vec y =(x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1)\\
&\vec x\cdot\vec y=\sum_{i=1}^3 x_i y_i.
\end{align*} You didn't use either of these directly. Instead you used two results that can be considered equivalent definitions: If $\theta$ denotes the angle between the vectors, then $\vec x\cdot\vec y$ is equal to $|\vec x||\vec y|\cos\theta$, and $\vec x\times\vec y$ is the vector of magnitude $|\vec x||\vec y|\sin\theta$ in the direction perpendicular to both $\vec x$ and $\vec y$ identified by the right-hand rule.

6. Dec 19, 2015

### Cosmophile

Oh, I see! Sorry, I was thinking of the $AB \cos \theta$ and $AB \sin \theta$ formule.