# Squaring a Vector?

So this one is basic. And should be pretty quick. Lets say that I have a vector $\vec{r}$:

$\vec{r} = \vec{x} + \vec{y} + \vec{z}$

Is this true:

$\vec{r}^{2} = \vec{x}^{2} + \vec{y}^{2} + \vec{z}^{2}$

I know that you can't really multiply a vector by a vector in the normal sense. However you can take the dot product. In which case I think that this would hold. If I saw $\vec{r}^{2}$ would it be safe to read this as $\vec{r} \cdot \vec{r}$?

Then would $\vec{x}^{2} = \vec{x} \cdot \vec{x} = \left|\vec{x} \right|^{2}$

Making $\vec{r}^{2} = \left|\vec{x} \right|^{2} + \left|\vec{y} \right|^{2} + \left|\vec{z} \right|^{2}$

I have encountered strange notation where the square of a vector is written and I was wondering how to interpert it.

• physics.info/vector-multiplication dot product is a scalar, not a vector, I know you know, but just in case – CountTo10 Sep 2 '15 at 20:14
• sTr8_Struggin: Regarding notation: note that the norm of a vector $\vec x$ is often written as $$\| \vec x \|,$$ expressing the conceptual distinction to the absolute value of (real, or complex) numbers. – user12262 Sep 2 '15 at 20:36
• in my experience $\vec{r}^2$ almost always means $\|\vec{r}\|^2$, but occasionally things like $\vec{x}\vec{y}$ can mean tensor product $X\otimes Y$ (in Eric Angle's answer he talks about the inertia tensor, and with vectors written in column matrix form $X\otimes Y=X\,Y^T$) or geometric product (should be pretty obvious from the context), which is a particular way of dealing with the product in a Clifford Algebra – Selene Routley Sep 3 '15 at 2:52
• @KyleKanos At first I thought so too, but there are some notation idiosyncracies peculiar to physicists, so I think it should probably stay here. – Selene Routley Sep 3 '15 at 2:56
• @WetSavannaAnimalakaRodVance I don't think the notation is particularly unique to physicists, so I agree with Kyle that it's suitable for migration. (I'll leave it for a brief comment period and then perform the migration.) And even if the question did use physics-specific notation, I don't think that would automatically make it on topic here. – David Z Sep 3 '15 at 7:57

$${\bf r} \cdot {\bf r} = \left|{\bf r}\right|^2 = \sum_{i,j} r_i r_j {\bf e}_i \cdot {\bf e}_j = \sum_{i,j} r_i r_j \delta_{ij} = \sum_i r_i r_i$$ is often written as ${\bf r}^2$, when it's clear that you don't mean $${\bf r}{\bf r} = \sum_{i,j} r_i r_j {\bf e}_i {\bf e}_j$$ which is a perfectly valid quantity [1].

As for ${\bf r} = {\bf x} + {\bf y} + {\bf z}$, $$\begin{eqnarray} {\bf r} \cdot {\bf r} &=& {\bf x} \cdot {\bf x} + {\bf x} \cdot {\bf y} + {\bf x} \cdot {\bf z} + {\bf y} \cdot {\bf x} + {\bf y} \cdot {\bf y} + {\bf y} \cdot {\bf z} + {\bf z} \cdot {\bf x} + {\bf z} \cdot {\bf y} + {\bf z} \cdot {\bf z} \\ &=& {\bf x} \cdot {\bf x} + {\bf y} \cdot {\bf y} + {\bf z} \cdot {\bf z} + 2\left( {\bf x} \cdot {\bf y} + {\bf x} \cdot {\bf z} + {\bf y} \cdot {\bf z}\right) \end{eqnarray}$$ equals ${\bf x} \cdot {\bf x} + {\bf y} \cdot {\bf y} + {\bf z} \cdot {\bf z}$ for all $\bf x$, $\bf y$, $\bf z$ iff ${\bf x} \cdot {\bf y} + {\bf x} \cdot {\bf z} + {\bf y} \cdot {\bf z} = 0$ -- a special case is when ${\bf x} \cdot {\bf y} = {\bf x} \cdot {\bf z} = {\bf y} \cdot {\bf z} = 0$.

1. For example, both the dot product and regular product are used in the expression for the inertia tensor: $$\int d^3{\bf r} \ \rho\left({\bf r}\right) \left[\left({\bf r} \cdot {\bf r}\right) {\bf I} - {\bf r}{\bf r}\right]$$ where $${\bf I} = \sum_{i,j} \delta_{ij} {\bf e}_i {\bf e}_j$$
• Small suggestion: "in general [2] if" should be "in general [2] iff" (iff = 'if and only if'). – Floris Sep 2 '15 at 22:20
• @Floris No, it was better the first way. The equality can hold even if the dot products are not individually 0. The "in general" doesn't really mean anything here, because Eric is effectively saying, "It's true in general in this special case," which is self-contradictory. You might as well have the special case where the sum is zero as the special case where $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ are all mutually perpendicular. – pwf Sep 3 '15 at 1:06
• I updated my answer. Let me know if you have any comments. – Eric Angle Sep 3 '15 at 2:59
• To be safe I would add a dot after the second factor (and possibly save the unnecessary identity $\mathbf I$) as in $\mathbf T=\mathbf r\cdot\mathbf r-\mathbf r\mathbf r\cdot$ which implies (without ambiguities to be interpreted) that $\mathbf T\mathbf v=|\mathbf r|^2-(\mathbf r\cdot\mathbf v)\mathbf r$ (and the brackets really unnecessary in the last term). – Oskar Limka Jun 12 '20 at 6:40

Why don't you work out the expansion yourself.

$$\vec{r}\cdot \vec{r} = \left( \vec{x}+\vec{y}+\vec{z} \right) \cdot \left( \vec{x}+\vec{y}+\vec{z} \right) =\\= (\vec{x} \cdot \vec{x})+(\vec{y}\cdot\vec{y}) + (\vec{z}\cdot\vec{z}) + 2(\vec{x} \cdot\vec{z}) +2(\vec{y}\cdot\vec{z}) + 2(\vec{x} \cdot\vec{y}) = \\ = \|\vec{x}\|^2 + \|\vec{y}\|^2+\|\vec{z}\|^2 + 2 \left((\vec{x} \cdot\vec{z}) +(\vec{y}\cdot\vec{z}) + (\vec{x} \cdot\vec{y}) \right)$$

IF the basis vectors are orthogonal then all the inner products are zero between vectors and you have $$|\vec{r}|^2 = \|\vec{x}\|^2 + \|\vec{y}\|^2+\|\vec{z}\|^2$$

otherwise no.

• Right, but as the accepted answer says, there could be a random cancellation. – Omry Sep 3 '15 at 4:47

First if nothing is said you can assume that $\vec{x}\vec{y}$ is a dot product.

What you said ${(\vec{x}+\vec{y}+\vec{z})}^{2}=x^2+y^2+z^2$ is only true if the vectors $\vec{x},\vec{y},\vec{z}$ are perpendicular to each other but in the general case:

$${\vec{r}}^{2}={(\vec{x}+\vec{y}+\vec{z})}^{2}\\ =x^2+y^2+z^2+2(xy\cos{\alpha}+yz\cos{\beta}+xz\cos{\gamma})$$

Where : $$x=||\vec{x}||$$

There are two basic ways you can multiply a vector, the dot product, as demonstrated in the link Dot Product, which gives you a scalar, no matter if you are multiplying A.B or squaring it, A.A.

Or you can have the cross product, which is A X B, which gives you another vector, perpendicular to both Cross Product.

The reason there are two different ways to multiply vectors is to deal with different physical situations, so the dot product can be used for work, which is a scalar, F.D, and the cross product is used for torque, say, or angular velocity, which are vectors.

• The cross product only exists in three dimensions and like stated in the answer of Eric Angle, you also have something called the vector direct product. – Kwin van der Veen Sep 3 '15 at 0:04
• There are actually three: you left out the outer product, which produces a 2nd-order tensor (a matrix, essentially) from the two vectors (each a 1st-order tensor). For example, the difference between the Hessian and Laplacian operators is that the former is the outer self-product of the gradient operator $\left(\nabla\nabla\Theta\right)$, and the latter is the inner (dot) self-product $\left(\nabla\cdot\nabla\Theta\right)$. – hBy2Py Sep 3 '15 at 2:57
• @Brian - indeed so. And you can nicely connect them: dot product is the trace of the tensor (a.k.a outer) product, i.e $u_{i}*v_{i}=\delta_{ij}*u_{i}*v_{j}$ and cross product is the dual vector of the tensor product i.e $e_{ijk}*u_{i}*v_{j}=\begin{vmatrix}\underline{e}_{1}&\underline{e}_{2}&\underline{e}_{3}\\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\end{vmatrix}$ Students can then informally see why the cross product only works in 3-space: only then do you get a square determinant. – rdt2 Sep 3 '15 at 10:44
• @rdt2 Yep, unless you generalize the cross product to $n-1$ vectors in $n$-space, or by some other means. For completeness as re my prior comment, note also the curl, which is the cross self-product of the gradient operator: $\nabla\times\nabla\Theta$. – hBy2Py Sep 3 '15 at 10:52

May I suggest a good read on geometric algebra(outer products, r-blades ,...). Check out " An introduction to Geometric Algebra and Calculus " but essentially the answer to your question can be shown using the following definition

$$\vec{a}^2=\vec{a}\vec{a}=\vec{a}\cdot\vec{a}+\vec{a}\bigwedge\vec{a}$$

However

$$\vec{a}\bigwedge\vec{a}=0$$

Hence $$\vec{a}^2=\vec{a}\cdot\vec{a}=||\vec{a}||^2$$