Name: ___________________

Math 122 Lab 8: Projections

Introduction

One of the most important uses of the dot product is in computing projections. In this assignment, you will learn the basic rules and see two applications of projections.

What to Turn In

Turn in this worksheet with everything filled in. This will be graded for accuracy and understanding, so write good explanations when you are asked to explain something. For some questions, you may need more space. If so, turn in additional pages typed or neatly written.

Projections

Draw the vectors a = <4,2> and b = <7,0> with both vectors starting at the origin. Drop a perpendicular from a onto b. The resulting vector is called the projection of a onto b. From your drawing, determine what the projection is.

In the above drawing, would anything have changed if b = <9,0>? Briefly explain.

Because the result does not depend on the length of b, the projection is defined in terms of a and a unit vector in the direction of b. Explain why a unit vector in the direction of b is given by b / ||b||.

To define the projection of a onto b, we need to define its magnitude and its direction. The magnitude is given by the dot product a · b / ||b|| and the direction is given by the unit vector b / ||b||. Combined together, the projection is given by

( a · b / ||b|| ) ( b / ||b|| ) = ( a · b / ||b||² ) b

Use this formula to compute the projection of a = <4,3> onto b = <2,−2>. Draw a graph showing all three vectors.

Compute the projection of a = <1,5> onto b = <2,−2>. Draw a graph showing all three vectors. Based on this, comment on the graphical significance of a negative dot product.

Three-dimensional vectors work the same way. Compute the projection of a = <1,2,1> onto b = <3,−1,0>. Draw a graph showing all three vectors.

Banked curves

It is common driving knowledge that you can faster (and safer) around a banked curve than around a flat curve. Race tracks are built with this in mind. For example, the Bristol Speedway is famous for high banks on its turns. Projections help explain the usefulness of banked turns. A car on a curve needs to do two things to stay on the road. It must stay on the ground and it must change directions to follow the curve. Clearly, gravity is the main force that keeps the car on the ground. But, on a banked curve, gravity can also help the car change directions. See the graphic on page 812. A car is on a 10º banked curve. If the car weighs 2000 pounds, then the gravitational force (weight) is a = <0,−2000>. A unit vector parallel to the road is b = <cos10º,−sin10º>. Find the projection of a onto b. This force pushes the car to the right in the direction of the curve.

Find the projection of a onto <sin10º,cos10º>. This is the force that keeps the car on the ground. Compare its magnitude to the car’s weight of 2000 pounds. Has much been lost?

Repeat the above for the same car on a 30º bank.

Computer Graphics

Most of the good computer animation for movies uses vector calculations. We will look at a simple problem to see how it works. A satellite dish has a shape called a paraboloid. The equation of one paraboloid is z = x² + y². We will get a graph on the TI-89. First, change the graphing mode to 3-D. Then enter x² + y² for z1 in the y= list. Then graph it! It will take awhile to compute points. It may not look like much at first, but sketch what you see below. You can rotate it by pressing the arrow keys. Start with the left arrow or right arrow key and press it several times. The rotation should help you get oriented to the graph. You can press 0 to return to the original view.

What the calculator has done is to compute several points on the graph and connect them with lines. Actually, there are too many lines. As we talked about in class with the sphere, some of the lines should be invisible and drawn with dashes (or not drawn at all). The calculator can do this, too. Press the green diamond key and then the vertical bar (to the left of the 7). This will bring up your graphing format options. You can turn axes and labels on or off as you like. You can also choose a different style. You should be getting a “Wire Frame” graph first. Change this to “Hidden Surface” and press Enter to redraw the graph. Briefly describe the difference in the graph and what it looks like as you rotate it.

Our goal is to understand what the calculator is doing. For example, if x = 1 and y = 2, then z = 1+4 = 5 and the calculator should plot the point (1,2,5). How does it decide where to put this dot on the screen and how does it recompute the location when you rotate to change perspectives? Copy down the picture on the board and briefly describe what it represents.

Define a to be the vector from the origin to the middle of the calculator screen and define b to be the vector from the origin to a point on the paraboloid. Let p be the projection of b onto a and let q be the vector such that b = p + q. Sketch a picture showing all of these vectors for a point on the paraboloid.

Explain which vector (p or q) the calculator would “see”. The calculator would use this vector to plot the point.

For the vector (p or q) that the calculator cannot see, explain how knowledge of that vector can help the calculator know whether the point is “close” or “far away” and therefore provide some depth perception.

One remaining detail is how the calculator would be able to sketch in the axes. It can start with the origin. The point (0,0,1) would then identify the direction of the z-axis (up). Briefly describe how the calculator could determine the directions of the x- and y-axes.

To illustrate these ideas, take a = <100,0,0>. Briefly describe where the calculator’s viewpoint is (on the x-axis or on the y-axis or wherever). First, compute the vectors p and q for b = <3,2,1>. Sketch a picture showing all important vectors. Then, compute the vectors p and q for b = <−3,2,1> and sketch a picture for it. Explain how you can tell that the calculator would plot both points in the same place. Explain how these vectors tell the calculator that the point (3,2,1) is closer than the point (−3,2,1). If the calculator was doing a “Hidden Surface” plot, which of these points would be invisible?