Projections tell us how much of one vector lies in the direction of another and are important in physical applications.

### Projections and components

#### Projections

One of the major uses of the dot product is to let us *project* one vector in
the direction of another. Conceptually, we are looking at the “shadow” of
one vector projected onto another, sort of like in the case of a sundial.

While this is good starting point for understanding orthogonal projections, now we need the definition.

**orthogonal projection**of vector in the direction of vector is a new vector denoted

To compute the projection of one vector along another, we use the dot product.

#### Components

Scalar components compute “how much” of a vector is pointing in a particular direction.

**scalar component**in the direction of of vector is denoted

#### Orthogonal decomposition

Given any vector in , we can always write it as for some real numbers and . Here
we’ve broken into the sum of two orthogonal vectors — in particular, vectors
parallel to and . In fact, given a vector and another vector you can always
break into a sum of two vectors, one of which is parallel to and another
that is perpendicular to . Such a sum is called an *orthogonal decomposition*.
Move the point around to see various orthogonal decompositions of vector .

**orthogonal decomposition**of in terms of is the sum where means that “ is parallel to ” and means that “ is perpendicular to ”.

We conclude this section with a physical example where orthogonal decomposition is useful.

What will happen to the box after it is placed and let go? Since there is no friction, the box will slide down the ramp. Objects initially at rest will only start to move if there is an unbalanced force, so there must be a force parallel to the ramp. We know that the force of gravity is pointing straight down, so part of the force due to gravity is certainly directed along the ramp.

Furthermore, the box is confined to slide along the ramp; it does not fall through the
ramp, nor does it jump off of the ramp. This means that the net force in the
direction perpendicular to the ramp must be . There is a component of gravity in this
direction too, so there must be a force that balances this component. In
physics, this force is referred to as the *normal force*, which we will denote by
.

We can find both the force gravity exerts on the box in the direction of the ramp, and the normal force from the orthogonal decomposition of the gravitational force, which we will denote by . Since we are given the weight of the box, we have .

*mass*, but the pound is a measure of

*weight*. These quantities are proportional, and the acceleration due to gravity is the constant of proportionality; that is

Thus, there is no need to multiply the weight by the acceleration due to gravity.

Let’s now find the orthogonal decomposition of in terms of .

Now, the normal force must balance the perpendicular force that gravity exerts on the ramp; that is

since the box is confined to move along the ramp. We can use this to find that the normal force is .

where is the the coefficient of kinetic friction and is an intrinsic property of the material from which the ramp is made. Roughly, this measures how much the surface impedes motion along it; for instance, ice has a much lower coefficient of kinetic friction than dry concrete.