# (p.205) Appendix D Oblique axes

# (p.205) Appendix D Oblique axes

Let **x** be a row-vector giving the coordinates of a point relative to a set of *P* orthogonal Cartesian axes. Suppose now, we wish to refer this same point to a set of oblique axes whose directions relative to the Cartesian axes are given by the columns of a matrix **C**. Thus the *k*th column **c** _{k} of **C** gives the direction cosines of the *k*th oblique axis. There are at least two ways of representing coordinates relative to oblique axes. These are shown in Figure D.1, below.

In the first version (a) one projects orthogonally onto the oblique axes, so that the oblique coordinates **y** are given by (C.3) as:

*y*

_{1}= 3 and

*y*

_{2}= 3.

In the second version (b), the *parallel axes system*, **x** is the vector-sum of the coordinates (*z* _{1}, *z* _{2}, *z* _{3}, …, *z* _{p}). Thus **x** = *z* _{1} **c** _{1} + *z* _{2} **c** _{2} + *z* _{3} **c** _{3} + … + *z* _{p} **c** _{p} so that *x* _{k} = *z* _{1} *c* _{1k} + *z* _{2} *c* _{2k} + *z* _{3} *c* _{3k} +…+ *z* _{p} *c* _{pk} and **x** = **zC′** giving:

*z*

_{1}= 2 and

*z*

_{2}= 1.

It follows from (D.1) and (D.2) that:

**C′C**, which has a unit diagonal. When the oblique axes are themselves orthogonal then the cosines are zero so that

**C′C**=

**I**, and

**C**is orthogonal. In this case

**y**and

**z**coincide, as is obvious from the geometry of Figure D.1.

Using (D.1) we can evaluate the distance between two points given by their projection coordinates **y** _{1} and **y** _{2}. From (D.1), their Cartesian coordinates are **y** _{1} **C** ^{−1}, **y** _{2} **C** ^{−1} so that their squared distance is given by:

**z**

_{1}and

**z**

_{2}is: