View Full Version : Translation problem

buckaroobill

05-06-2007, 08:19 PM

Okay, so I did the following problem out but I'm not sure if it is right, so if anyone could help, it would be greatly appreciated!

Consider the quadrilateral (namely Q) in R^3 formed by the points

(1, 0, 0), (2, 0, 0), (1, 1, 3), and (2, 1, 3).

a) What should the coordinates be for the figure R we get by rotating Q counterclockwise in the x-y plane by 45 degrees, then dilating it by a factor of 3/2, then translating it along the vector (-2, 1, -1)?

b) What matrix transforms Q into R?

First, I took the matrix

[cosθ, sinθ, 0

-sinθ, cosθ, 0

0, 0, 1]

and I set all the θ equal to 45. This led to the matrix:

[1/√2, 1/√2, 0)

-1/√2, 1/√2, 0

0, 0, √2]

In order to do dilation by 3/2, I multiplied each entry by 3/2 retrieving:

[3/(2√2), 3/(2√2), 0

-3/(2√2), 3/(2√2), 0

0, 0, (3√2)/2]

Then to translate it along the vector (-2, 1, -1), I added -2 to the x coordinates, 1 to the y coordinates, and -1 to the z coordinates.

[3/(2√2) -2, 3/(2√2) +1, -1

-3/(2√2) - 2, 3/(2√2) + 1, -1

-2, 1, (3√2)/(2) - 1]

So the above was my final matrix (and I said that the rows of that matrix were the coordinates for the figure R.

JakeD

05-07-2007, 02:39 AM

Okay, so I did the following problem out but I'm not sure if it is right, so if anyone could help, it would be greatly appreciated!

Consider the quadrilateral (namely Q) in R^3 formed by the points

(1, 0, 0), (2, 0, 0), (1, 1, 3), and (2, 1, 3).

a) What should the coordinates be for the figure R we get by rotating Q counterclockwise in the x-y plane by 45 degrees, then dilating it by a factor of 3/2, then translating it along the vector (-2, 1, -1)?

b) What matrix transforms Q into R?

First, I took the matrix

[cosθ, sinθ, 0

-sinθ, cosθ, 0

0, 0, 1]

and I set all the θ equal to 45. This led to the matrix:

[1/√2, 1/√2, 0)

-1/√2, 1/√2, 0

0, 0, √2]

In order to do dilation by 3/2, I multiplied each entry by 3/2 retrieving:

[3/(2√2), 3/(2√2), 0

-3/(2√2), 3/(2√2), 0

0, 0, (3√2)/2]

Then to translate it along the vector (-2, 1, -1), I added -2 to the x coordinates, 1 to the y coordinates, and -1 to the z coordinates.

[3/(2√2) -2, 3/(2√2) +1, -1

-3/(2√2) - 2, 3/(2√2) + 1, -1

-2, 1, (3√2)/(2) - 1]

So the above was my final matrix (and I said that the rows of that matrix were the coordinates for the figure R.

You are mixing up questions a and b. For b, calculate the matrix M that transforms Q to R. Your first two steps (the rotation and dilation) in calculating M are almost correct (the (3,3) entry should be 1, not sqrt(2)). But handling the translation using matrix multiplication requires that M be a 4x4 matrix and the use of homogeneous coordinates of the form (x,y,z,1). Then for question a, multiply the 4 points defining Q by M to get the 4 points defining R.

What you did was start calculating M. Then you translated the 3 rows of M and called that R. But R is defined by 4 points, not 3, and the points must correspond to the points defining Q.

buckaroobill

05-07-2007, 10:09 AM

You are mixing up questions a and b. For b, calculate the matrix M that transforms Q to R. Your first two steps (the rotation and dilation) in calculating M are almost correct (the (3,3) entry should be 1, not sqrt(2)). But handling the translation using matrix multiplication requires that M be a 4x4 matrix and the use of homogeneous coordinates of the form (x,y,z,1). Then for question a, multiply the 4 points defining Q by M to get the 4 points defining R.

What you did was start calculating M. Then you translated the 3 rows of M and called that R. But R is defined by 4 points, not 3, and the points must correspond to the points defining Q.

Thanks, I did out (a) and wound up with the following coordinates below:

(3/(2√2) - 2, 3/(2√2) + 1, -1),

(3/√2 - 2, 3/√2 + 1, -1),

(-2, 3/√2 + 1, 3.5),

(3/(2√2) - 2, 9/(2√2) + 1, 3.5)

I am a little confused about b, though. Okay, so first I have...

[1/√2, 1/√2, 0)

-1/√2, 1/√2, 0

0, 0, 1]

In order to do dilation by 3/2, I multiplied each entry by 3/2 retrieving:

[3/(2√2), 3/(2√2), 0

-3/(2√2), 3/(2√2), 0

0, 0, 3/2]

After this part is where I get confused - meaning, how would one use a 4 by 4 matrix?

buckaroobill

05-08-2007, 09:32 AM

Sorry, disregard what I said above. I figured out how to do (b).

Okay, so what I did was I used the matrix

cosθ -sinθ 0

sinθ cosθ 0

0 0 1

Then substituted 45 for θ.

After this, multiplied the identity matrix for R^3 by 3/2 and then multiplied it by the matrix with 45 substituted for θ.

Then T(x,y,z) = (-2+3sqrt(2)x/4-3sqrt(2)y/4,1+3sqrt(2)x/4+3sqrt(2)y/4,-1+3z/2).

I substituted each of the quadrilateral points in for T(x, y, z) to come up with the four points and got:

(3/(2√2) - 2, 3/(2√2) + 1, -1),

(3/√2 - 2, 3/√2 + 1, -1),

(-2, 3/√2 + 1, 3.5),

(3/(2√2) - 2, 9/(2√2) + 1, 3.5)

However, I'm still unsure of how to do (a).

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.