# Why is the dot product defined as a scalar and the cross product a normal vector?

These are the definitions of the dot and cross products as I understand them:

The cross product provides a vector perpendicular to both input vectors with magnitude proportional to the area of the parallelogram traced out by the vectors.

a × b = ∥a∥∥b∥sinθ n

where n is the normal vector

The dot product provides a scalar proportional to the magnitudes of both vectors and the cosine of the angle between them (θ = 0° is pure magnitude multiplication whereas θ = 90° is 0).

a.b = ∥a∥∥b∥cosθ

Assuming I understand the definitions correctly, I don't see why the dot product is defined as a scalar and the cross product a vector. Why was the dot product not defined as a vector perpendicular to the input vectors with magnitude proportional to the cosine of the angles between them. Likewise why was the cross product not defined as a scalar representing the area of the parallelogram traced out between them?

For example:

a.b = ∥a∥∥b∥cosθ n

a × b = ∥a∥∥b∥sinθ

• Remember, the cross product is only defined in $\mathbb{R}^3$ while dot product is in $\mathbb{R}^n$. We cannot define the perpendicular in a general $n$-space because with 2 input vectors there are $n-2$ directions that are perpendicular to the inputs. Jun 5, 2016 at 2:18
• I think you have it a little backwards. The dot product is not necessarily defined as being the product of the vectors magnitudes multiplied by the cosine of the angle between them and the cross product is not necessarily defined to be perpendicular etc. Rather, these two things are very nice, geometric, properties of the dot product and cross product. Jun 5, 2016 at 2:19
• I pulled the definitions from wiki so they could be off. I've updated it for clarity. Jun 5, 2016 at 2:28

The dot product is the special case of a more general concept, the inner product. If you have a vector space $V$ over the reals or the complex numbers, then an inner product is a map $f : V \times V \to \mathbb{C}$ or $f : V \times V \to \mathbb{R}$ which is conjugate symmetric, positive definite, and linear in its first argument. We usually write $f(u, v) = \langle u, v \rangle$, in which case these properties can be summed up as follows:

• Conjugate symmetry: $\overline{\langle u, v \rangle} = \langle v, u \rangle$, where $\bar{z}$ denotes complex conjugation. Note that this implies $\langle u, u \rangle$ is always real for any vector $u$.
• Positive definiteness: $\langle v, v \rangle \geq 0$ for any $v \in V$, with equality holding iff $v = 0$.
• Linearity in the first argument: $\langle \alpha u + \beta v, w \rangle = \alpha \langle u, w \rangle + \beta \langle v, w \rangle$ where $u, v, w \in V$ and $\alpha, \beta$ are in the field of scalars.

If $V = \mathbb{R}^n$, then we can fix a basis $B = \{ b_i \in \mathbb{R}, 1 \leq i \leq n \}$ and define $\langle b_i, b_i \rangle = 1$ and $\langle b_i, b_j \rangle = 0$ for $i \neq j$. Extending this to all of $\mathbb{R}^n$ by linearity gives us

$$\left \langle \sum_{k=1}^{n} c_k b_k, \sum_{j=1}^{n} d_j b_j \right \rangle = \sum_{1 \leq k, j \leq n} d_k c_j \langle b_i, b_j \rangle = \sum_{i=1}^{n} c_i d_i$$

where positive definiteness is readily verified. You will recognize this expression as the definition of the dot product. Indeed, if we take our basis $B$ to be the standard basis of $\mathbb{R}^n$, then this inner product is the dot product.

Why is this formalism more powerful? A result about the inner product is the Cauchy-Schwarz inequality, which says that $|\langle u, v \rangle| \leq |u| |v|$ where $|u| = \sqrt{\langle u, u \rangle}$. This tells us that

$$-1 \leq \frac{\langle u, v \rangle}{|u| |v|} \leq 1$$

assuming that our field of scalars is $\mathbb{R}$. We then see that the arccosine of this expression is well-defined, so we can define the angle between nonzero vectors $u$ and $v$ as

$$\theta = \arccos \left( \frac{\langle u, v \rangle}{|u| |v|} \right)$$

The properties we expect to be true are then easily verified. This notion extends to infinite dimensional vector spaces over $\mathbb{R}$, where defining angle is not at all obvious. It is then trivially true that we have $\langle u, v \rangle = |u| |v| \cos(\theta)$, since that is how $\theta$ was defined.

The cross product is an entirely separate concept which allows us to find a vector orthogonal to two given vectors in $\mathbb{R}^3$. In addition, its magnitude also gives the area of the parallelogram spanned by the vectors. These properties can be taken as the definition of the cross product (with appropriate care for orientation), or they can be derived as theorems starting from the algebraic definition.

• Thank you for your insightful response. I have a much clearer understanding now Jun 5, 2016 at 14:13

Well firstly understand what actually a dot and cross product tell us

Dot product gives a product of two input vectors moving in same direction ( how much two vectors are going in same direction) and it's products also will moves in same direction as both inputs are moving so there is no need to define the direction of their products ( because we know direction of their products is always in same direction as input are..)

Cross product gives a product of two input vectors moving perpendicular direction to each other ( how much two vectors are going perpendicular each other ) and it's products will move in direction perpendicular to both inputs vectors directions so there is need to define the direction of their product ( as their product has different direction than inputs )

That's why a dot product is scalar and cross product is vector

Now you can think why area of parallelogram is a cross product and why not a dot product

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