# Is there a rule regarding how vectors and scalars combine in formulae? [closed]

\begin{align} \frac{\vec{d}}{t} &= \vec{v} & \frac{\text{vector}}{\text{scalar}} &= \text{vector} \\ \vec{d}\times \vec{d} &= \vec{A} & \text{vector}\times\text{vector} &= \text{vector} \\ \vec{d}\times \vec{d}\times \vec{d} &= V & \text{vector}\times\text{vector}\times\text{vector} &= \text{scalar} \\ \frac{\vec{F}}{\vec{A}} &= P & \frac{\text{vector}}{\text{vector}} &= \text{scalar} \end{align}

Above are some examples of how various mathematical combinations of vectors and scalars produce a vector or a scalar.

Is there some general rule independent of the quantities involved or must each situation be treated individually ?

• I've deleted a number of obsolete comments and/or responses to them. Commented Jul 20, 2020 at 8:30
• Why would a double cross-product become a scalar? A cross-product operation produces a new vector, regardless of how many times it it performed in a row. Commented Jul 20, 2020 at 10:26
• What is meant by $\vec{F}/\vec{A}$? I suspect that equation should actually be written $P\vec{A} =\vec{F}$. (Multiplying a vector by a scalar gives a vector). If they were numbers you'd be able to divide both sides by $\vec{A}$, but division by a vector is not something that's usually defined. Commented Jul 20, 2020 at 11:58
• Equation in physics are just that : equations and thus do follow general algebra. However notation can vary on how you denote things, distinguish between tensor, vector and scalar - and it can also be sloppy and not always distinguish clearly Commented Jul 20, 2020 at 16:48
• See here, here and here on division by vectors and cross products, which both seem to be confusing you. Also the answers to this question are largely duplicates to the answers on these. Commented Jul 20, 2020 at 22:42

It is very important to know and be careful about which operations you are using.

1. Vectors can be multiplied by scalars so this operation is valid.

2. The cross product of two vectors produces another vector so this is ok.

3. The cross product is not associative so you have to be careful about which order you're performing these cross products in. The cross product of two vectors produces another vector so an operation such as $$\vec v\times \vec v\times \vec v$$ as you've written is valid but the result is a vector, not a scalar.

4. Division of one vector by another is not a defined operation.

Given a 3-dimensional vector space $$V$$ defined over a scalar field $$\Bbb F$$ (either $$\Bbb R$$ or $$\Bbb C$$) the following important operations are defined (not an exhaustive list):

Vector Addition: $$+:V \times V\rightarrow V$$

Scalar Multiplication of Vectors: $$\bullet:V\times \Bbb F\rightarrow V$$

The Dot Product: $$\cdot:V\times V\rightarrow \Bbb F$$

The Cross Product: $$\times: V\times V\rightarrow V$$

• It’s possible that the OP was referring to the triple product of three vectors, $\vec{a} \cdot (\vec{b} \times \vec {c})$, which is actually a scalar. But your broader point about being careful about operations is still a good one. Commented Jul 20, 2020 at 12:32
• In the fourth one, if the two vectors are in the same direction there is sense to it. It becomes a rewrite of the first. Some will argue it is an abuse of notation. Commented Jul 20, 2020 at 13:51
• @RossMillikan I mean in principle you can define any operations you want. However in order for this answer to be useful to someone new to linear algebra in physics it's probably not helpful to go too far off script just for the sake of it :P Commented Jul 20, 2020 at 14:00
• @RossMillikan i'm not sure what you mean, but in all 4 the × refers to the Cartesian product of the two sets. Commented Jul 20, 2020 at 14:35
• The cross product really creates a pseudo-vector. Adding that or setting that equal to a vector is usually wrong. The OP seems to know these operations that you've listed - I'm not clear exactly what's being asked, but "don't set a pseudo-vector equal to a vector" seems like the type of rule being requested. Commented Jul 20, 2020 at 18:07

If in general by division we mean the 'inverse operations' of the product, the only possible 'division between vectors' can be defined starting from the vector product definition. Given two vectors $$\vec a$$ and $$\vec b$$ we define the vector product as:

$$\vec a \times \vec b= \hat n \ ab \sin θ \tag 1$$ where $$\hat n$$ is a normal unit vector both $$\vec a$$ and $$\vec b$$ and $$θ$$ is the angle formed by two vectors. The definition of division between vectors as 'inverse operation' of the vector product necessarily implies two 'limitations'...

(a) the two vectors must be normal to each other;

(b) the resulting 'quotient vector' is a function of the choice of $$θ$$ in $$(1)$$, in the sense that it is different for different values of $$θ$$ and is therefore not uniquely determined.

• Another way of expressing point (b) is that knowing $\hat{n}$ and $\vec{b}$ only allows you to determine the components of $\vec{a}$ that are orthogonal to $\vec{b}$. The parallel components remain undetermined. Commented Jul 20, 2020 at 12:38
• @MichaelSeifert Very kind, thank you very much. I'm very glady to edit my answer to have your support. Thank you again. Commented Jul 20, 2020 at 16:03

Like we never find $$\mathrm{d}x$$ in a denominator of an integral

$$\int \frac{f(x)}{\mathrm{d}x}$$

we never find a vector-valued quantity in denominator

$$\frac{\text{whatever}}{\vec{A}}$$

But don't confuse it with the following legal operations

$$\frac{\text{whatever}}{\vec{A}\cdot \vec{B}}$$ or $$\frac{\text{whatever}}{|\vec{A}|}$$

where the denominators are no longer vector valued quantities.

There is an extension of vectors and scalars called a 'Geometric Algebra' or 'Clifford Algebra' that allows vectors and scalars to be added and multiplied together freely, where multiplication is generally invertible (so division of vectors can be defined), and which includes the usual dot and cross products of vector algebra as special cases. So you can use Geometric Algebra to provide general rules for combining vectors and scalars, and use them to derive the rules for dot and cross products.

A Geometric Algebra for 3D space contains 8 dimensions, one scalar, three for vectors (directed lengths), three for bivectors (a new sort of object representing oriented surface areas), and one for a trivector (representing an oriented volume). You can construct objects, called multivectors, by adding any linear combination of these. So a scalar can be added to a vector, to produce a 'paravector'. A scalar can be added to a bivector to produce a quaternion (or in 2D, a complex number).

The Geometric Algebra multiplies multivectors using the geometric product. If you multiply two vectors together, you get a linear combination of a scalar (equal to the dot product of the vectors) and a bivector (which is dual to the cross product of the vectors). Similarly, you can define division of vectors, which gives a combination of a scalar and a bivector. If the vectors are parallel, only the scalar part is non-zero. If the vectors are perpendicular, only the bivector part is non-zero. Otherwise you get a combination of scalar and bivector.

Geometric Algebra can be defined in spaces of any dimension. In the Geometric Algebra for 4D space, we have one dimension for scalars, four for vectors, six for bivectors, four for trivectors, and one for quadvectors. This is the reason why cross products only work in 3D space. With three dimensions, vectors and bivectors have the same dimension, and you can flip the bivector into a (mostly) equivalent vector, using the dual operation. This is sometimes called an 'axial vector', because it has different transformation properties to the normal 'polar vector'. In 4D, bivectors have six dimensions, and so you can't make a 4D vector out of it by taking the dual. This example shows how using a more general unifying rule can be used to derive and explain features of how the rules for dot and cross products work, and why they are the way they are.

There are plenty of basic introductions to Geometric Algebra on the web.