# A shorter way to prove the identity on vectors

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 $\vec{A}$ in terms of $\hat{i}$, $\hat{j}$ and $\hat{k}$ 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.

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There is no general cross product in $n$ dimensions. – Ted Dec 5 '12 at 17:33
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. – Matthew Pressland Dec 5 '12 at 17:36
Yes, that is what I want. – Erwin Dec 5 '12 at 17:36

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.

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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.

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

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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 $(\vec{n} \times \vec{A}) \times \vec{n} = b \vec{v} \times \vec{n} = b \vec{u}$.

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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 $\gamma = 1$.

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