3d Linear Algebra, rotating basis question

[David01010101]

Hi all!

Say I have a shape in 3d space where basis.x=(1, 0, 0), basis.y=(0, 1, 0) and basis.z=(0, 0, 1) and an origin=(0, 0, 0)

This shape is then rotated 90 degrees so that basis.y=(0, 0, 1) with the origin remaining the same.

How can I solve for the new basis.x and basis.z without sheering so that scale and orthogonality are preserved? So I keep the shape intact I mean. Obviously, in this case it is trivial: basis.x=(1, 0, 0) and basis.z=(0, -1, 0) but how do you solve in general?*

Thanks in advance!!

*without converting to angles

 

[HallsofIvy]

Hi all!

Say I have a shape in 3d space where basis.x=(1, 0, 0), basis.y=(0, 1, 0) and basis.z=(0, 0, 1) and an origin=(0, 0, 0)

This shape is then rotated 90 degrees so that basis.y=(0, 0, 1) with the origin remaining the same.

That means that the coordinate system is rotated 90 degrees around the x-axis. That is equivalent to a matrix multiplication \(\displaystyle \begin{bmatrix}1 & 0 & 0 \\0 & 0 &

How can I solve for the new basis.x and basis.z without sheering so that scale and orthogonality are preserved? So I keep the shape intact I mean. Obviously, in this case it is trivial: basis.x=(1, 0, 0) and basis.z=(0, -1, 0) but how do you solve in general?*

Thanks in advance!!

*without converting to angles

\)