Register to reply

Transforming a 3D vector using matrix

by Gajan
Tags: matrix, transformation, vector
Share this thread:
Gajan
#1
Feb18-09, 05:13 AM
P: 3
Hi Friends,

I have a problem in transforming a vector in 3d using a matrix.

I have two points A, B. Assume origin is O, here A,B ,O are in 3d.

First construct a vector AB:

OB = OA + AB
AB = OB - OA
AB = OB + AO

I have a 4*4 matrix that gives the transformation matrix(rotation+translation)

I need to transform vector AB using this matrix.

Currently what I do is transform individual points A , B first and the reconstruct the vector A'B' again (after tranforming)
What I want to do is to without doing the above steps, transform the vector as it is (without transforming the individual points). Is this possible?

How could I do this computation?

I am looking for your reply!

thank you.

/Gajan
Phys.Org News Partner Science news on Phys.org
Physical constant is constant even in strong gravitational fields
Montreal VR headset team turns to crowdfunding for Totem
Researchers study vital 'on/off switches' that control when bacteria turn deadly
HallsofIvy
#2
Feb18-09, 06:36 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,693
Using a single 4x4 matrix to do both rotation and translation, you have to use a "projective" space. That means you are representing the point (x,y,z) as the (column) vector [x y z 1] with the provision that [a b c d] is the same as [a/d b/d c/d 1] (d can never be 0). In that case the matrix that rotates, say, [itex]\theta[/itex] degrees about the y-axis and translates by (tx,ty, tz) is
[tex]\begin{bmatrix} cos(\theta) & 0 & -sin(\theta) & tx \\ 0 & 1 & 0 & ty \\ sin(\theta) & 0 & cos(\theta) & tz \\ 0 & 0 & 0 & 1\end{bmatrix}[/tex]
Notice that in the particular case of [itex]\theta= 0[/itex] where there is no rotation and so a pure translation, this becomes
[tex]\begin{bmatrix} 1 & 0 & 0 & tx \\ 0 & 1 & 0 & ty \\ 0 & 0 & 1 & tz \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix}= \begin{bmatrix} x+ tx \\ y+ ty \\ z+ tz \\ 1\end{bmatrix}[/tex]

While if tx= ty= tz= 0 so there is a pure rotation and no translation it is
[tex]\begin{bmatrix} cos(\theta) & 0 & -sin(\theta) & 0 \\ 0 & 1 & 0 & 0 \\ sin(\theta) & 0 & cos(\theta) & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix}= \begin{bmatrix} xcos(\theta)- zsin(\theta) \\ y \\ xsin(\theta)+ zcos(\theta)\end{bmatrix}[/tex]

I would handle a general rotation as the product of two rotations around coordinate axes.
Gajan
#3
Feb18-09, 07:40 AM
P: 3
Hi,

First of all thank you for your reply.

In my case , I have the matrix in the following form:

Transpose matrix of the transformation : M
vector :V

V * M

still is it the same way the matrix product is done?

HallsofIvy
#4
Feb18-09, 01:19 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,693
Transforming a 3D vector using matrix

Assuming that you are writing V as a row matrix, yes, swapping "row" and "column" is purely a matter of convention.
Gajan
#5
Feb19-09, 05:46 AM
P: 3
Thanking you. I understand it now.


Register to reply

Related Discussions
Diagonal Matrix from vector Linear & Abstract Algebra 8
Transforming a 4x4 matrix from one base to another Linear & Abstract Algebra 3
Transforming base vectors in Matrix. General Math 11
Matrix/Vector differentiation Calculus & Beyond Homework 1
Matrix associated to a vector (how?) Quantum Physics 6