Why Dot Product is Scalar? why not vector? as cross product is vector?

Question

why Dot Product is Scalar? why not vector. As cross product is vector. Anyone can define this Please?

Answer 1

The simplest answer is: they are defined that way, so that's the way it is. But of course the motivation for having them defined in this way, is that they are useful expressions in many contexts.

Dot product

This can be defined equivalently algebraically or geometrically.

Algebraic definition

Generally in $n$ dimenstions: $$\mathbf{a}\cdot\mathbf{b}=\sum_{i=1}^na_ib_i$$

This seems like the simplest way to define a mapping from two vectors to a scalar that includes all elements of the vectors $\mathbf{a},\mathbf{b}$. Also it has some nice mathematical properties as it is: commutative, distributive, bilinear,...

Geometric definition

In arbitrary n-dimensional Euclidean vector space (but more easily visualized in 2 or 3 dimensions), the dot product is the product of the lengths/magnitudes of the vectors and the cosine of the angle between them: $$\mathbf{a}\cdot\mathbf{b}=||\mathbf{a}|| ||\mathbf{b}|| \cos\theta$$

You could therefore use the scalar product to calculate the angle between two vectors and find if two vectors are perpendicular (scalar product = 0 )but there are other applications...

Geometrically, $||\mathbf{a}|| \cos\theta$ is the (scalar) projection of the vector $\mathbf{a}$ on the direction of $\mathbf{b}$. In a physical context, if you have some vector (e.g. a force or velocity), and you split it in two vectors as in the parallelogram of force, this projection will give you the length of any of the two resulting vector in their chosen directions.

As an example, mechanical work when displacing an object by $\mathbf{s}$ with a force $\mathbf{F}$, is given by the dot product: $W=\mathbf{F}\cdot \mathbf{s}$, i.e. the force acting in the direction of the motion (=projection) multiplied with the distance it moved (=length of second vector).

Cross product

Even when not thinking of applications, if you want to define a mapping from two vectors to a vector, that has simple and meaningful properties, you would come up with something like the cross product $\mathbf{c}=\mathbf{a}\times\mathbf{b}$ which has the following properties:

• $\mathbf{c}$ is perpendicular to $\mathbf{a}$ and $\mathbf{b}$; Another option could be to have $\mathbf{c}$ halfway between $\mathbf{a}$ and $\mathbf{b}$, but that would seem more constructed than a perpendicular vector. Chosing $\mathbf{c}$ perpendicular fixes the plane in which $\mathbf{a}$ and $\mathbf{b}$ are lying. The actual choice is given by the right hand rule but could as well have been in opposite direction (left hand).

• $\mathbf{c}$ has a length that depends on the lengths of $\mathbf{a}$ and $\mathbf{b}$ and the angle between them. Thus all properties of the input (plane of $\mathbf{a}$ and $\mathbf{b}$, angle between them, lengths of $\mathbf{a}$ and $\mathbf{b}$) will turn up in the result. The actual choice , $||\mathbf{c}||=||\mathbf{a}|| ||\mathbf{b}||\sin\theta$ also seems natural, as it depends symmetrically on the lengths of $\mathbf{a}$ and $\mathbf{b}$ and together with $\sin\theta$ this is the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. Importantly, if both vectors point in the same direction, the direction of $\mathbf{c}$ (perpendicular) would not be well defined. In that case $\sin\theta=0$ makes the product equal to $0$, so there is no problem with the direction of $\mathbf{c}$. Also, if $\mathbf{a}$ and $\mathbf{b}$ are perpendicular (a special case) this product becomes maximal.

Also the cross product has some nice mathematical properties such as: anticommutative, distributive, associative, ...

An example from physics:

A particle of charge $q$ travelling with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ will be deflected by the Lorentz force $$\mathbf{F}=q\mathbf{v}\times\mathbf{B}$$

Note that this expression has many properties which you would expect from symmetry considerations, intuition and experiments:

• the magnitude of force is linear in both the magnitude of the velocity and the magnetic field.Typically equations in physics are like Taylor expansions, i.e. they start with low order powers (in this case first) and possible corrections are then of higher order in the parameters.
• the force is minimal if velocity and field are parallel and maximal if they are perpendicular
• reversing velocity or magnetic field reverses the direction of the force
• the force is perpendicular to velocity and magnetic field.

Answer 2

The dot product is defining the component of a vector in the direction of another, when the second vector is normalized. As such, it is a scalar multiplier.

The cross product is actually defining the directed area of the parallelogram defined by two vectors. In three dimensions, one can specify a directed area its magnitude and the direction of the vector normal to its plane, and the cross product accordingly spits out a unit vector in this direction scaled by the area's magnitude. In higher dimensions than 3, this doesn't work - i.e. you need more than one unit normal vector to specify the orientation of an area element and the cross product does not work. In these cases it is replaced by the wedge product and its output is a 2-form, not a vector.

At another level, the answer to your question is simply that this is how Gibbs and Heaviside defined these operations, and they thought that this was simpler than using quaternions and their multiplication as developed by Hamilton in his Versor Calculus to represent vector operations. In fact, using dot and cross products, one can define quaternions as objects $(a,\,\vec{b})$ with a scalar part $a$ and vector part $\vec{b}$ and with product defined by:

$$(a,\,\vec{b}) \bullet (c,\,\vec{d}) = (a\,c-\vec{b}\cdot\vec{d},\,a\,\vec{d}+c\,\vec{b} + \vec{b}\times\vec{d})$$

and you can check that the objects thus defined together with addition and the product operation above are indeed a skew field (division algebra) wherein all nonzero objects have multiplicative inverses given by:

$$(a,\vec{b})^{-1} = \frac{1}{a^2+\vec{b}\cdot\vec{b}} \,(a,\,-\vec{b})$$

Answer 3

The dot product of two vectors $\vec a$ and $\vec b$ is defined as $$\vec a\cdot \vec b=a_1b_1+a_2b_2+a_3b_3$$ As you can see, this is a number, not a vector.

The cross product of the same vectors is defined as $$\vec a\times \vec b= \begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$$ As you can see, this is a vector, not a number.

Answer 4

In his answer, @Photon correctly gave the definitions of the dot product and the cross product. The simple answer to your question is that the dot product is a scalar and the cross product is a vector because they are defined that way.