written 2.1 years ago by | • modified 2.1 years ago |
Solution:
Co-ordinate frames:
Let $\mathrm{p}$ be a vector in $\mathrm{Rn}$
Let $\mathrm{X}=\{\mathrm{x} 1, \mathrm{x} 2, \mathrm{x} 3, \ldots \mathrm{xn}\}$ be a complete orthonormal set for $\mathrm{Rn}$.
The coordinates of p w.r.t X are denoted as $[p]^X$
Can be defined as:
$ p=\sum_{k=1}^n[p]_k^X x^k \\ $
Coordinate Transformation:
$ \begin{aligned}\\ &F=\{f 1, f 2, f 3, \ldots f n\} \\\\ &M=\{m 1, m 2, m 3, \ldots . m n\}\\ \end{aligned}\\ $
F and M are coordinate frames for $\mathbf{R n}$
$ \begin{aligned}\\ {[P]_k^F } &=P . f^k\left(k^{\text {th }} \text { coordinate of } P \text { w.r.t. frame } F\right) \\\\ &=\left(\sum_{j=1}^n[P]_j^M m^j\right) \cdot f^k \\\\ &=\sum_{j=1}^n[P]_j^M\left(m^j \cdot f^k\right) \\\\ &=\sum_{j=1}^n\left(m^j \cdot f^k\right)[P]_j^m \\\\ &=\sum_{j=1}^n\left(f^k \cdot m^j\right)[P]_j^m \\\\ &=\sum_{j=1}^n A_{k j}[P]_j^m \\\\ \therefore[P]^F &=A[P]^{m 1}\\ \end{aligned}\\ $
Inverse Coordinate Transformation:
$\mathrm{F}$ and $\mathrm{M}=$ orthonormal coordinate frames in $\mathrm{Rn} $
$\mathrm{A}=\mathrm{CTM}$ maps $\mathrm{M}$ coordinates into $\mathrm{F}$ coordinates
$\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}} $
$ \begin{aligned}\\ \left(A^{-1}\right)_{k j} &=m^k \cdot f^j \\\\ &=f^j \cdot m^k \\\\ &=A_{j k} \\\\ &=\left(A^T\right)_{k j}\\ \end{aligned}\\ $