# PlanetPhysics/Direction Cosine Matrix

A direction cosine matrix (DCM) is a transformation matrix that transforms one coordinate reference frame to another. If we extend the concept of how the three dimensional direction cosines locate a vector, then the DCM locates three unit vectors that describe a coordinate reference frame. Using the notation in equation 1, we need to find the matrix elements that correspond to the correct transformation matrix.

${\displaystyle DCM=\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]}$

The first unit vector of the second coordinate frame can be located in the first frame by normal vector notation. See figure 1 for relationship.

${\displaystyle {\hat {y}}_{1}=A_{11}{\hat {x}}_{1}+A_{12}{\hat {x}}_{2}+A_{13}{\hat {x}}_{3}}$

\medskip \begin{figure} \includegraphics[scale=0.78]{DCM.eps} \end{figure} \medskip

Similarily, the other two unit vectors can be described by

${\displaystyle {\hat {y}}_{2}=A_{21}{\hat {x}}_{1}+A_{22}{\hat {x}}_{2}+A_{23}{\hat {x}}_{3}}$ ${\displaystyle {\hat {y}}_{3}=A_{31}{\hat {x}}_{1}+A_{32}{\hat {x}}_{2}+A_{33}{\hat {x}}_{3}}$

It is easy to see how equation 1 works as a transformation matrix through simple matrix multiplication.

${\displaystyle \left[{\begin{matrix}{\hat {y}}_{1}\\{\hat {y}}_{2}\\{\hat {y}}_{3}\end{matrix}}\right]=\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]\left[{\begin{matrix}{\hat {x}}_{1}\\{\hat {x}}_{2}\\{\hat {x}}_{3}\end{matrix}}\right]}$

Once this transformation matrix is found, it can be used to transform vectors from the second frame to the first frame and vice versa. Equation 2 transforms the x frame to the y frame and can be denoted as ${\displaystyle R_{1-2}}$. In order to get ${\displaystyle R_{2-1}}$, which transforms the y frame to the x frame, we use a property of transformation matrices of orthonormal reference frames (a frame that is described by unit vectors and are perpindicular to each other). See the entry on a transformation matrix for more info on its properties. We use the properties that

${\displaystyle R_{1-2}^{-1}=R_{1-2}^{T}=R_{2-1}}$ ${\displaystyle R_{1-2}R_{1-2}^{T}=\left[{\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}}\right]}$

so using these properties and rearranging equation 2 ${\displaystyle {\hat {y}}=R_{1-2}{\hat {x}}}$ yields

${\displaystyle R_{1-2}^{-1}{\hat {y}}=R_{1-2}^{-1}R_{1-2}{\hat {x}}}$

giving the transformation of the y frame to the x frame

${\displaystyle {\hat {x}}=R_{2-1}{\hat {y}}}$

So to extend this concept to transform vectors from one frame to another a closer examination of a vector being represented in both frames is needed. If we denote the second frame as the prime (${\displaystyle \prime }$) frame, then a vector expressed in each of these is given by

${\displaystyle v=v_{1}{\hat {x}}_{1}+v_{2}{\hat {x}}_{2}+v_{3}{\hat {x}}_{3}}$

${\displaystyle v=v_{1}\prime {\hat {y}}_{1}+v_{2}\prime {\hat {y}}_{2}+v_{3}\prime {\hat {y}}_{3}}$

Since both equations describe the same vector, let us set them equal to each other so

${\displaystyle v_{1}{\hat {x}}_{1}+v_{2}{\hat {x}}_{2}+v_{3}{\hat {x}}_{3}=v_{1}\prime {\hat {y}}_{1}+v_{2}\prime {\hat {y}}_{2}+v_{3}\prime {\hat {y}}_{3}}$

This notation is clumsy so we want to represent it in matrix notation. This is simple enough if you have an understanding of multiplying a column vector by a row vector. This allows us to describe equations 3 and 4 by

${\displaystyle v=\left[{\begin{matrix}v_{1}&v_{2}&v_{3}\end{matrix}}\right]\left[{\begin{matrix}{\hat {x}}_{1}\\{\hat {x}}_{2}\\{\hat {x}}_{3}\end{matrix}}\right]}$ ${\displaystyle v=\left[{\begin{matrix}v_{1}\prime &v_{2}\prime &v_{3}\prime \end{matrix}}\right]\left[{\begin{matrix}{\hat {y}}_{1}\\{\hat {y}}_{2}\\{\hat {y}}_{3}\end{matrix}}\right]}$

Setting them equal and substituting equation 2 in for the second coordinate frame yields

${\displaystyle v=\left[{\begin{matrix}v_{1}&v_{2}&v_{3}\end{matrix}}\right]\left[{\begin{matrix}{\hat {x}}_{1}\\{\hat {x}}_{2}\\{\hat {x}}_{3}\end{matrix}}\right]=\left[{\begin{matrix}v_{1}\prime &v_{2}\prime &v_{3}\prime \end{matrix}}\right]\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]\left[{\begin{matrix}{\hat {x}}_{1}\\{\hat {x}}_{2}\\{\hat {x}}_{3}\end{matrix}}\right]}$

Then by inspection (or go through the matrix manipulation to cancel the x frame)

${\displaystyle \left[{\begin{matrix}v_{1}&v_{2}&v_{3}\end{matrix}}\right]=\left[{\begin{matrix}v_{1}\prime &v_{2}\prime &v_{3}\prime \end{matrix}}\right]\left[{\begin{matrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{matrix}}\right]}$

Representing the transformation matrix as ${\displaystyle R_{1-2}}$ as the transformation from the first frame to the second frame and transposing the previous equation gives

${\displaystyle \left[{\begin{matrix}v_{1}&v_{2}&v_{3}\end{matrix}}\right]=(\left[{\begin{matrix}v_{1}\prime &v_{2}\prime &v_{3}\prime \end{matrix}}\right]R_{1-2})^{T}}$

Performing the transposition and using a transposition property for two matrices A and B such that

${\displaystyle (AB)^{T}=B^{T}A^{T}}$

${\displaystyle \left[{\begin{matrix}v_{1}\\v_{2}\\v_{3}\end{matrix}}\right]=R_{1-2}^{T}\left[{\begin{matrix}v_{1}\prime \\v_{2}\prime \\v_{3}\prime \end{matrix}}\right]}$
${\displaystyle {\vec {v}}=R_{2-1}{\vec {v\prime }}}$