What is the difference between dot and cross product?

Question

1. What is the difference between dot product and cross product?

2. Why do we use cross product to find torque, why can't we use dot product?

3. Also we use dot product to find work done and not cross product?

Answer 1

They are completely different things. One returns a scalar and the other returns vector; one is commutative, the other anticommutative. You shouldn't even compare them (the question why one and not the other is similar to why apples and not pianos).

A derivation of a physical law doesn't guess what to use but actually uses a certain reasoning along the way. You need to look into theoretic linear algebra, not just physics. Each of them has a geometric meaning. The dot product involves projection, extraction of components of vectors, involves things that need to be parallel... the quantity it produces is a scalar and thus orientation-independent. It doesn't have a direction. Work is a scalar quantity. It concerns how much force is used to move an object in the direction of the force (hence, the projection). Dot product is actually very ubiquitous, most of the relationships between quantities you know involve some kind of work with vector's components. In fact, this connection to projection runs very deeply: it works in any number in dimensions, and even in abstract spaces, such as spaces of functions and spectra... the concept of projection is needed everywhere.

On the other hand, the cross product is antisymmetric - it is sensitive to orinetation of the space. It involves chiral phenomena, circulation & rotation, inherently three-dimensional phenomena that use up all three spatial dimensions. It's about perpendicular things instead of parallel. Its result is a vector - something orientation dependent. Torque involves rotation: it distinguishes clockwise and counterclockwise. It has an axis, perpendicular to the plane of motion, it has a direction... the force and the distance to the axis must be non-parallel if you want a torque at all. Similar examples here are everything involving magnetic field (here, the consequences are very deep and fundamental, all the way back to Maxwell's equations).

I deliberately went with the essay approach. You've got equations and stuff covered by other posters.

Answer 2

If we have a force $\vec{F}$ which acts upon a test particle along a curve $C$, then the work done is in the general case a line integral, given by,

$$W=\int_{C} \vec{F} \cdot \mathrm{d}\vec{r}$$

We may think of the integral as a summation over the contributions of the force along an infinitesimally small line element. The work done $W$ is a scalar quantity, and employing a cross product would not be sensible. On the other hand, torque is usually described as a vector, given by,

$$T = \vec{r} \times \vec{F}$$

If we employed the dot product, we would retrieve a scalar rather than a vector. In addition, recall torque is a description for the tendency for a system to rotate; a scalar quantity cannot fully capture this.

Answer 3

Both scalar and vector products are definitions.

$$ {\bf a} \cdot {\bf b} = a b \cos \theta\,\,$$ where $a$ and $b$ are the magnitudes of the two vectors and $\theta$ is the angle between them. This quantity only has a magnitude and hence no direction and is therefore a scalar.

If you apply a (let's say constant for the sake of simplicity) vector force ${\bf F}$ to an object, such that it is displaced by a vector ${\bf r}$. High school physics tells you that work done is force times distance, but in this case you need to resolve the force in the direction of the displacement. But this resolved component is $F \cos \theta$ and so the work done is $F r \cos \theta$ and is the same as ${\bf F} \cdot {\bf r}$.

The definition of vector product is $$ {\bf a} \times {\bf b} = ab \sin \theta\, {\bf u}\, ,$$ where ${\bf u}$ is a unit vector that is perpendicular to both ${\bf a}$ and ${\bf b}$.

From high school physics we know that a torque or moment about a pivot is given by force $F$ times the perpendicular distance to the pivot. This perpendicular distance is given by $r \sin \theta$, where $\theta$ is the angle between the force vector ${\bf F}$ and the displacement ${\bf r}$ between where it is applied and the pivot. So that mean the magnitude of the torque is $rF \sin \theta$ - which is the same as the magnitude of the vector product ${\bf r} \times {\bf F}$; but what about the vector nature?

When you apply a torque to an object (free to move) and it turns, it turns about an axis that is almost uniquely defined. As torque is a vector then the only vector direction that can sensibly be assigned to it, is in this almost unique direction, which is perpendicular to the force and the displacement from the pivot. Why almost? Well, there is a 180 degree ambiguity, which is only resolved by definition. Hence torque is ${\bf r} \times {\bf F}$ and not ${\bf F} \times {\bf r}$ because we use a "right hand" rule to define the direction of ${\bf u}$ rather than a "left hand" rule.

Answer 4

Dot product gives you a scalar (a number), while a cross product gives you a vector. Work done is indeed a real number, while an efficient description of torque is given by a vector.