Example 10-10. Consider the fixed set of axes x, y and a set of barred axes
x, ̄y ̄where the barred set of axes is rotated about the origin through an angle θas
illustrated below. Consider a general point P having coordinates (x, y )with respect
the unbarred axes. This same point P has coordinates ( ̄x,y ̄) with respect to the
barred set of axes. Let ˆe 1 , eˆ 2 and ˆe ̄ 1 , ˆe ̄ 2 denote unit vectors in the directions of the
x, y and ̄x,y ̄-axes respectively. The position vector r of the point P can be expressed
in either of the forms
r =xˆe 1 +yˆe 2 or r = ̄xˆe ̄ 1 + ̄yˆe ̄ 2
The transformation equations between the coordinates can be obtained by taking
the dot product of r with the unit vectors ˆe ̄ 1 and ˆe ̄ 2 to obtain
r ·eˆ ̄ 1 = ̄x=xˆe 1 ·ˆe ̄ 1 +yˆe 2 ·ˆe ̄ 1 =xcos θ+ysin θ
r ·eˆ ̄ 2 = ̄y=xˆe 1 ·ˆe ̄ 2 =yˆe 2 ·ˆe ̄ 2 =x(−sin θ) + ycos θ
The above transformation equations between
the ( ̄x,y ̄)axes which have been rotated through
an angle θwith respect to a fixed set of (x, y )axes
can be represented by the matrix equation
[
̄x
̄y
]
=
[
cos θ sin θ
−sin θ cos θ
][
x
y
]
or X ̄=AX (10.7)
where X ̄=col( ̄x,y ̄)and X=col(x, y )are column
vectors. Here the coefficient matrix of the above
transformation is A=
[
cos θ sin θ
−sin θ cos θ
]
and its transpose matrix is AT=
[
cos θ −sin θ
sin θ cos θ
]
.
If one calculates the matrix product of Atimes it transpose AT, one finds AA T=I,
the identity matrix. Matrices with this property are called orthogonal matrices.
Left-multiplication of equation (10.7) by A−^1 =AT gives the inverse transformation
ATX ̄=ATAX =IX =Xwhich can be expressed in expanded form
[
x
y
]
=
[
cos θ −sin θ
sin θ cos θ
][
x ̄
y ̄
]
.
The row and column vectors which make up the rows and columns of the matrix A
are called orthogonal vectors.
