Objectives
See applications of multi-variable calculus
Section2.6Projections and Their Applications
Suppose a heavy box needs to be lowered down a ramp. The box exerts a downward force of 200 Newtons, which we will write in vector notation as \(\vec F=\left\lt 0,-200\right>\text{.}\) The ramp was placed so that the box needs to be moved right 6 m, and down 3 m, so we need to get from the origin \((0,0)\) to the point \((6,-3)\text{.}\) This displacement can be written as \(\vec d=\left\lt 6,-3\right>\text{.}\) The force \(F\) acts straight down, which means the ramp takes some of the force. Our goal is to find out how much of the 200N the ramp takes, and how much force must be applied to prevent the box from sliding down the ramp (neglecting friction). We are going to break the force \(\vec F\) into two components, one component in the direction of \(\vec d\text{,}\) and another component orthogonal to \(\vec d\text{.}\)
Exercise2.6.1
Read the preceding paragraph.
We want to write \(\vec F\) as the sum of two vectors: \(\vec F = \vec w+\vec n\)
where \(\vec w\) is parallel to \(\vec d\)
and \(\vec n\) is orthogonal to \(\vec d\)
Since \(\vec w\) is parallel to \(\vec d\text{,}\) we can write \(\vec w=c\vec d\) for some unknown scalar \(c\text{.}\)
(a)
Rewrite \(\vec F\) in terms of \(\vec d\)
(b)
Take the dot-product of both sides with \(\vec d\)
(c)
Since \(\vec n\) is orthogonal to \(\vec d\) we know that \(\vec n \cdot \vec d =\) ?
(d)
Substitute and solve for the unknown \(c\)
The solution to the previous problem gives us the definition of a projection.
Definition2.6.1
The projection of \(\vec F\) onto \(\vec d\text{,}\) written \(\proj_{\vec d}\vec F\text{,}\) is defined as
\begin{equation*} \proj_{\vec d}\vec F = \left(\frac{\vec F\cdot \vec d}{\vec d\cdot \vec d}\right)\vec d. \end{equation*}Subsection2.6.1Applications of Projections
Exercise2.6.2
Let \(\vec u=(-1,2)\) and \(\vec v=(3,4)\text{.}\) Compute the \(\proj_{\vec v}\vec u\text{.}\) Draw \(\vec u,\) \(\vec v\text{,}\) and \(\proj_{\vec v}\vec u\) all on one set of axis. Then draw a line segment from the head of \(\vec u\) to the head of the projection.
Now let \(\vec w=(-2,0)\text{.}\) Compute \(\proj_{\vec v}\vec w\text{.}\) Draw \(\vec u,\) \(\vec v\text{,}\) and \(\proj_{\vec v}\vec w\text{.}\) Then draw a line segment from the head of \(\vec w\) to the head of the projection.
One application of projections pertains to the concept of work. Work is the transfer of energy. If a force \(F\) acts through a displacement \(d\text{,}\) then the most basic definition of work is \(W=Fd\text{,}\) the product of the force and the displacement. This basic definition has a few assumptions.
The force \(F\) must act in the same direction as the displacement.
The force \(F\) must be constant throughout the entire displacement.
The displacement must be in a straight line.
Before the semester ends, we will be able to remove all 3 of these assumptions. The next problem will show you how dot products help us remove the first assumption.
Recall the set up to Exercise 1. We want to lower a box down a ramp (which we will assume is frictionless). Gravity exerts a force of \(\vec F=\left\lt 0,-200\right>\) N. If we apply no other forces to this system, then gravity will do work on the box through a displacement of \(\left\lt 6,-3\right>\) m. The work done by gravity will transfer the potential energy of the box into kinetic energy (remember that work is a transfer of energy). How much energy is transferred?
Exercise2.6.3Projection Application: Work
Find the amount of work done by the force \(\vec F=\left\lt 0,-200\right>\) through the displacement \(\vec d=\left\lt 6,-3\right>\text{.}\) Find this by doing the following:
(a)
Find the projection of \(\vec F\) onto \(\vec d\text{.}\) This tells you how much force acts in the direction of the displacement. Find the magnitude of this projection.
(b)
Since work equals \(W=Fd\text{,}\) multiply your answer above by \(|\vec {d}|\text{.}\)
(c)
Now compute \(\vec F\cdot \vec d\text{.}\)
You have just shown that \(W=\vec F\cdot \vec d\) when \(\vec F\) and \(\vec d\) are not in the same direction.
Exercise2.6.4Projection Application: Planes
Consider the points \(P=(2,4,5)\text{,}\) \(Q=(1,5,7)\text{,}\) and \(R=(-1,6,8)\text{.}\)
(a)
What is the area of the triangle \(PQR\text{.}\)
(b)
Give a normal vector to the plane through these three points.
(c)
What is the distance from the point \(A=(1,2,3)\) to the plane \(PQR\text{.}\)
Hint
Compute the projection of \(\vec {PA}\) onto \(\vec n\text{.}\) How long is it?
Subsection2.6.2Computational Practice
These are provided to help you achieve better skills in basic computational answers