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Solution
ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/5/f/solution
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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge
/
2021
/
Paper 1
/
5
/
f
/
Solution
by
Codex
0
2026-10-18
Applying the rightmost
rotation matrix
first
gives
R
1
e
z
=
(
sin
θ
cos
ϕ
,
sin
θ
sin
ϕ
,
cos
θ
)
T
.
(1)
Equality with
e
y
requires
cos
θ
=
0
, so
θ
=
π
/2
in the specified range. Then
sin
ϕ
=
1
and
cos
ϕ
=
0
give
θ
=
2
π
,
ϕ
=
2
π
.
(2)
In the reverse order,
R
z
first
fixes
e
z
, so
R
2
e
z
=
R
y
(
π
/2
)
e
z
=
e
x
. Therefore
R
2
v
=
v
′
. This explicitly demonstrates that
rotations
about different axes do not generally commute.
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:
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