Chapter 5 Activities C

Chapter 5 Activities C

Name:

Move 5 up and 5 over4 points

Demo the earlier instructions6 points

Determinants on the trig one6 points

Graph the 60 degree rotation4 points

30 degree rotation8 points

Reflect across the x axis5 points

5.5Isometries and Orthogonal Matrices

Now rigid motions in the plane:

Translations:

Given a path and a distance…perform the translation. Translations are isometries.

Go 45 degrees to the horizontal a distance of 5 cm with the point (1, 3) is the same as saying:

F(x + 5, y + 5) with (1, 3)

Show this:

Show the new endpoints;

Why is the length 5 ?

How do you give these instructions with matrices?

Let’s move

Could we do this to a segment or a triangle? Sure just move the endpoints or vertices one at a time and then reconnect them.

Now what about rotations?

Also an isometry.

We use orthogonal matrices to perform these.

A matrix is said to be ORTHOGONAL if its transpose is equal to its inverse.

This is very special matrix and there are not too many of these around. These are always square matrices, have a non-zero determinant, so they’re in our 2x2 set of invertible matrices

The identity matrix is the easiest case! The identity matrix times negative one is another.

Then we have2 more:

and

The determinant of the first matrix is −1:

Find the determinant of the second matrix above here:

Show your work clearly

The determinant of an orthogonal matrix is always 1 or −1.

And the two matrices with the trig functions perform rotations about the origin.

An orientation reversing rotation holding the origin fixed can be represented as

where A is an orthogonal matrix with −1 as its determinant. These are often called “improper”. Think of it as a reflection across the x axis followed by a rotation.

The matrix to use is

Now let’s do an example with theta = 60 degrees degrees on the point (1, 0)…the vector from (0, 0) to (1, 0). Where will it start? And where will it end?

From special triangles we know that the sides go in the ratio 1: : 2 and that the 60 degree angle is the one side across from the .

We can fill in our matrix with 30 degrees and then fill in those trig values:

Now let’s multiply the second matrix times the tip of our vector (1, 0)

Graph this using about 4 divisions to be a step of size 1.

Do you see that the 60 degree rotation is across from y, the side length that is the square root of 3?

An orientation preserving rotation about the origin is given by

You do one of 30 degrees on the point (1, 0).

First an analysis – what will the 30 degree rotation be across from as a side length in y? then the math (fill in the matrix and the point) and then the picture.

Graph the rotation using 4 marks as a unit step

How would you apply this to a triangle? Rotate it vertex by vertex.

Now let’s look at some similarity transforms and discuss one for the homework.

If you use a matrix that is rather like the identity matrix you can effect stretches and shrinks to values.

For example, to stretch x by 2 and shrink y by ½:

So we can have similarity transforms.

And last we’ll look at reflections.

If you reflect across the y-axis, your x becomes –x.

To reflect the point (2, 3) across the y-axis into Quadrant 2, you would need to come up with a matrix that would give you (−2, 3). And this is not so hard:

As desired.

What matrix will reflect across the x- axis?

First take the point (2, 3) to Quadrant 4. What changed in the reflection? How will you say this in matrix instructions?

Each rigid motion in the Euclidean plane may be represented as a proper or improper rotation about the origin followed by a translation.

Suppose you wish to reflect a point across the line y = 3x + 5. First you rotate the line so it’s horizontal or vertical…moving the point too. The you translate the line tilit’s right over the x or y axis. Then you reflect the point. Then you UNtranslate and UN rotate both the point and the line and VOILA the new point. Now this is a lot of work when you’re doing it by hand so we’ll resist the urge to do it.

Let me just point out that EVERY matrix we’ve looked at in this section is in the 2x2 invertible set so we have inverses for each one. Thus we can rotate, move, unrotate and unmove handily.