Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to prove that $\vec{A}=(\vec{A}\cdot \vec{n})\vec{n}+(\vec{n}\times\vec{A})\times\vec{n} $ where $\vec{n}$ is a unit vector and $\times$ indicates the cross product.

I am dealing with vectors in 3-dimensions in Klepner's book on mechanics, and so I assigned A vector in terms of i-cap, j-cap and k-cap and tried to do the same with the unit vector. That made my solution hideous.

I was wondering if someone could show me how to do the problem in $3$ dimensions elegantly.

share|improve this question
1  
There is no general cross product in $n$ dimensions. –  Ted Dec 5 '12 at 17:33
1  
There may still be a coordinate-free proof in 3 dimensions though, which I think is what the OP probably had in mind. But yes, it will be specific to dimension 3 so there actually is a cross product. –  Matt Pressland Dec 5 '12 at 17:36
    
Yes, that is what I want. –  Erwin Dec 5 '12 at 17:36

5 Answers 5

The vector $(A\cdot n)\,n$ is essentially the projection of $A$ onto $n$, and subsequently it remains to be seen that $(n\times A)\times n$ provides us with the projection of $A$ onto the plane with unit normal $n$. Without loss of generality assume that $n$ is on the positive $z$ axis so the plane $n^\perp$ is the $xy$-plane (the idea is just easier to visualize this way). Note that $n\times A$ is located on $n^\perp$ and forms a right angle (in the counterclockwise direction!) to the projection $p$ of $A$ onto $n^\perp$, so by the right-hand rule $(n\times A)\times n$ will not only be another vector on $A$ but will form a right angle clockwise to $n\times A$, hence will point in the same direction as $p$. Now it only remains to be seen that the magnitude is the correct.

Since $n\times A\perp n$, we have $\|(n\times A)\times n\|=\|n\times A\|\cdot\|n\|\sin\frac{\pi}{2}=\|A\|\sin\theta$, where $\theta$ is the angle $A$ makes with $n$. But $\|A\|\sin\theta$ is precisely the length of $A$'s projection $p$ onto the $xy$-axis! Q.E.D.

share|improve this answer

Extracting a little bit from anon's answer: You can assume without loss of generality that $\vec n$ is in the positive $z$ direction, i.e., that it is the vector you called "k-cap". Now an algebraic solution won't be hideous.

The reason you can assume a particular direction for $\vec n$ is that all the ingredients of the problem (dot products, cross products, addition of vectors, and multiplication by scalars) are invariant under rotation.

share|improve this answer

Repeated indicies are summed below, and I used a property of the Levi-Civita symbol: $$ \begin{eqnarray} \left({\bf A} \bullet {\bf n}\right) {\bf n} + \left({\bf n} \times {\bf A}\right) \times {\bf n} &=& \left(A_i n_i n_m + n_i A_j n_r \epsilon_{i j k} \epsilon_{r m k} \right) {\bf e}_m \\ &=& \left[A_i n_i n_m + n_i A_j n_r \left(\delta_{ir} \delta_{jm} - \delta_{im} \delta_{jr}\right) \right] {\bf e}_m \\ &=& \left(A_i n_i n_m + n_i A_m n_i - n_m A_j n_j \right) {\bf e}_m \\ &=& n_i A_m n_i {\bf e}_m \\ &=& \left({\bf n} \bullet {\bf n} \right){\bf A} \\ &=& {\bf A} \\ \end{eqnarray} $$

share|improve this answer

Choose unit vectors $\vec{u}$ and $\vec{v}$ so $\vec{n}, \vec{u}, \vec{v}$ form a right-handed orthogonal basis (thus $\vec{n} \times \vec{u} = \vec{v}$, $\vec{u} \times \vec{v} = \vec{n}$, $\vec{v} \times \vec{n} = \vec{u}$) and $\vec{A}$ is in the span of $\vec{n}$ and $\vec{u}$. Thus $\vec{A} = a \vec{n} + b \vec{u}$ for scalars $a$, $b$. Now $\vec{A} \cdot \vec{n} = a$ and $(\vec{n} \times \vec{A}) \times \vec{n} = b \vec{v} \times \vec{n} = b \vec{u}$.

share|improve this answer

Here's another approach. $\def\VA{{\bf A}} \def\VB{{\bf B}} \def\VC{{\bf C}} \def\n{{\bf n}} \def\a{\alpha} \def\b{\beta} \def\g{\gamma}$

We assume $\VA \ne {\bf 0}$ and that $\VA$ and $\n$ are not parallel (or antiparallel) so the vectors $\{\n, \n\times\VA, (\n\times\VA)\times \n\}$ form an orthogonal basis. Then $$\begin{align*} \VA &= \a \n + \b \n\times\VA + \g(\n\times\VA)\times\n. \tag{1} \end{align*}$$ Using the triple product identity $\VA\times(\VB\times\VC) = \VB(\VA\cdot\VC) - \VC(\VA\cdot\VB)$ we find $$\VA = \a \n + \b \n\times\VA + \g[\VA - (\n\cdot\VA)\n].$$ We must have $\gamma = 1$, $\alpha = (\n\cdot\VA)$, and $\beta = 0$ so $$\VA = (\VA\cdot\n) \n + (\n\times\VA)\times\n$$ as claimed.


We can also find the coefficients of (1) in the usual way,
$$\begin{align*} \a &= \n\cdot\VA \\ \b &= \frac{(\n\times\VA)\cdot \VA}{(\n\times\VA)^2} \\ \g &= \frac{[(\n\times\VA)\times\n]\cdot \VA}{[(\n\times\VA)\times\n]^2}. \end{align*}$$ We immediately have $\beta = 0$. Using the triple product identity we find $$\begin{align*} [(\n\times\VA)\times\n]\cdot \VA &= -[\n(\n\cdot\VA)-\VA(\n\cdot\n)]\cdot \VA \\ &= \VA^2 - (\n\cdot\VA)^2 \\ [(\n\times\VA)\times\n]^2 &= [\n(\n\cdot\VA)-\VA(\n\cdot\n)]\cdot[\n(\n\cdot\VA)-\VA(\n\cdot\n)] \\ &= (\n\cdot\VA)^2 - 2(\n\cdot\VA)^2 +\VA^2 \\ &= \VA^2 - (\n\cdot\VA)^2, \end{align*}$$ so $\gamma = 1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.