Linear transformation: Rotations in R3

[Mynona]

Homework Statement

A robot arm in a xyz coordinate system is doing three consecutive rotations, which are as follows:
1) Rotates (Pi/4) rad around the z axis
2) Rotates (Pi/3) rad around the y axis
3) Rotations -(Pi/6) rad around the x axis

Find the standard matrix for the (combined) transformation T.

Homework Equations

The Attempt at a Solution

I (think) I have gotten as far as finish part 1. By projecting the robot arm down to the xy plane, and by applying trigonometry, I find that the standard matrix for the rotation in 1) to be as follows (it's a 3x3 matrix, I don't know how to format properly):
[cos (Pi/4) | -sin(Pi/4) | 0]
[sin (Pi/4) | cos(Pi/4) | 0]
[0 | 0 | 1] [||1]

which is:
[1/2\sqrt{}2 | -1/2\sqrt{}2 | 0]
[1/2\sqrt{}2 | 1/2\sqrt{}2 | 0]
[0 | 0 | 1]

Hopefully that is the right answer to question 1), but my answer really is how do I go from here? I've found the first standard matrix, but how do I go forth in trying to find the standard matrix for the entire set of 3 rotations?

 

[Donaldos]

You need to find the two other rotation matrices using the exact same method.

The standard matrix for T is found by simply multiplying the three matrices together.

 

[Mynona]

You need to find the two other rotation matrices using the exact same method.

The standard matrix for T is found by simply multiplying the three matrices together.

I am partially with you. But do I rotate the already rotated arm (ie, finding standard matrix for the rotated arm), or do I find the two other standard matrices from rotating based on the e1, e2, e3 vectors?

 

[cronxeh]

I am partially with you. But do I rotate the already rotated arm (ie, finding standard matrix for the rotated arm), or do I find the two other standard matrices from rotating based on the e1, e2, e3 vectors?

You have 3 matrices:


Rotation in z = [cos(pi/4) -sin(pi/4) 0; sin(pi/4) cos(pi/4) 0; 0 0 1]
Rotation in y = [cos(pi/3) 0 sin(pi/3); 0 1 0; -sin(pi/3) 0 cos(pi/3)]
Rotation in x = [1 0 0; 0 cos(-pi/6) -sin(-pi/6); 0 sin(-pi/6) cos(-pi/6)]


[ 0.3536 -0.9186 0.1768 ]
[ 0.3536 0.3062 0.8839 ]
[-0.8660 -0.2500 0.4330 ]