example of Euler angles: elementary axis rotations and passive rotation matrices

Euler Angles: Elementary Axis Rotations and Passive Rotation Matrices Examples, Exercises, and Solutions

This entry is the self study companion to Euler Angles: elementary axis rotations and passive rotation matrices.

The goal is to turn the three elementary passive rotation matrices into working tools. The exercises emphasize geometric interpretation, sign checking, orthogonality, inverse transformations, and the way these elementary factors assemble into larger Euler Angle formulas.

All exercises are stated first. Complete worked solutions follow afterward.

1 Convention summary

PhysicsLibrary uses passive coordinate transformations between right handed orthonormal frames.

If a physical vector has coordinate columns ^Av and ^Bv in frames A and B, then

Bv = BCA Av.

(1)

The elementary passive matrices are

⌊1 0 0 ⌋ ⌈ ⌉ C1 (λ) = 0 cos λ sin λ , 0 − sin λ cosλ

(2)

⌊ ⌋ cos λ 0 − sinλ C2 (λ) = ⌈ 0 1 0 ⌉ , sin λ 0 cosλ

(3)

and

⌊ ⌋ cos λ sin λ 0 C3 (λ) = ⌈− sin λ cosλ 0⌉ . 0 0 1

(4)

For a generic intrinsic i-j-k Euler sequence,

B CA = Ck(γ)Cj (β)Ci(α).

(5)

2 Visual reference sheet

The three figures from the parent article are included here as quick visual reference.

Figure. Passive frame rotation about axis 1, the x axis.

Figure. Passive frame rotation about axis 2, the y axis.

Figure. Passive frame rotation about axis 3, the z axis.

3 Exercises

Interpret the map.
Explain in words what the equation

$Bv = BCA Av$
means.
What stays physically unchanged and what changes?
Identify the unchanged coordinate.
For a passive frame rotation about axis 1, which coordinate component is unchanged?
Repeat the question for rotations about axis 2 and axis 3.
Write the three elementary matrices.
Write the matrices C₁(λ), C₂(λ), and C₃(λ).
A $∘ 90$ check for $C1$ .
Use C₁(π∕2) to compute the new coordinates of

$⌊0 ⌋ A ⌈ ⌉ v = 1 . 0$
A $90∘$ check for $C2$ .
Use C₂(π∕2) to compute the new coordinates of

$⌊ ⌋ A 0 v = ⌈0 ⌉. 1$
A $90∘$ check for $C3$ .
Use C₃(π∕2) to compute the new coordinates of

$⌊ ⌋ 1 Av = ⌈0 ⌉. 0$
A negative angle check.
Evaluate C₃(−π∕2) and apply it to

$⌊ ⌋ 1 Av = ⌈0 ⌉. 0$
Compare the result with Exercise 6.
Orthogonality.
Show directly that

$C1 (λ )TC1(λ ) = I.$
State the analogous result for C₂ and C₃.
Inverse and transpose.
Show that

$Ci(λ)− 1 = Ci(λ)T = Ci(− λ)$
for i = 1, 2, 3.
determinant.
Find det C₂(λ).
What should the determinant be for each elementary passive rotation matrix?
Active versus passive.
If R_i(λ) denotes the active vector rotation matrix for the same positive geometric angle about the same axis, what is the relation between R_i(λ) and C_i(λ)?
Columns as transformed basis vectors.
What do the columns of ^BC_A represent in the passive convention?
Use this to interpret the columns of C₃(λ).
Small angle form.
Write the first order approximation of C₂(λ) for small |λ|.
Build a 3-2-1 passive Euler matrix.
Write the passive intrinsic 3-2-1 matrix product in terms of yaw, pitch, and roll angles (ψ,𝜃,ϕ).
Zero angle specialization.
Starting from the passive intrinsic 3-2-1 product, set

$ϕ = 0, 𝜃 = 0.$
Which elementary matrix remains?
Reverse map.
If

$B CA = C1 (λ),$
write the matrix ^AC_B explicitly.
Sign debugging.
A student proposes that the passive rotation about axis 3 is

$⌊ ⌋ cos λ − sinλ 0 C^3 (λ) = ⌈sin λ cosλ 0⌉ . 0 0 1$
Give one quick test showing that this is not the PhysicsLibrary passive matrix for positive frame rotation about +z.
composition order.
Why does the product

$BCA = Ck (γ )Cj(β)Ci(α )$
still describe the chronological intrinsic sequence i → j → k?

4 Solutions

Solution 1: interpret the map

The equation

Bv = BCA Av

means that the physical vector is the same geometric object in space, but its coordinate description changes when we switch from frame A to frame B.

Under a passive transformation, the vector stays fixed and the coordinate frame changes.

Solution 2: identify the unchanged coordinate

For rotation about axis 1, the first coordinate is unchanged.

For rotation about axis 2, the second coordinate is unchanged.

For rotation about axis 3, the third coordinate is unchanged.

This is visible directly in the matrices and geometrically because the rotation axis itself is common to both frames.

Solution 3: write the three elementary matrices

The three elementary passive matrices are

⌊ ⌋ 1 0 0 C1 (λ) = ⌈0 cos λ sin λ⌉ , 0 − sin λ cosλ

(6)

⌊cos λ 0 − sinλ⌋ ⌈ ⌉ C2 (λ) = 0 1 0 , sin λ 0 cosλ

(7)

and

⌊ cos λ sin λ 0⌋ ⌈ ⌉ C3 (λ) = − sin λ cosλ 0 . 0 0 1

(8)

Solution 4: a $90 ∘$ check for $C1$

At λ = π∕2,

⌊ ⌋ ( ) 1 0 0 C1 π- = ⌈ 0 0 1⌉ . 2 0 − 1 0

Thus

( )⌊ 0⌋ ⌊ 0 ⌋ B π- ⌈ ⌉ ⌈ ⌉ v = C1 2 1 = 0 . 0 − 1

(9)

Solution 5: a $90 ∘$ check for $C2$

At λ = π∕2,

⌊ ⌋ ( ) 0 0 − 1 C2 π- = ⌈ 0 1 0 ⌉ . 2 1 0 0

Therefore

( )⌊ 0⌋ ⌊− 1⌋ B π- ⌈ ⌉ ⌈ ⌉ v = C2 2 0 = 0 . 1 0

(10)

Solution 6: a $90 ∘$ check for $C3$

At λ = π∕2,

⌊ ⌋ ( ) 0 1 0 C3 π- = ⌈ − 1 0 0⌉ . 2 0 0 1

Hence

( )⌊ 1⌋ ⌊ 0 ⌋ B π- ⌈ ⌉ ⌈ ⌉ v = C3 2 0 = − 1 . 0 0

(11)

Solution 7: a negative angle check

Because

( ) ( ) cos − π- = 0, sin − π- = − 1, 2 2

we obtain

( ) ⌊ 0 − 1 0⌋ π- ⌈ ⌉ C3 − 2 = 1 0 0 . 0 0 1

Applying it to the same vector gives

⌊ ⌋ ⌊ ⌋ ( π) 1 0 Bv = C3 − -- ⌈ 0⌉ = ⌈ 1⌉ . 2 0 0

(12)

This is the opposite coordinate change from the positive angle case in Exercise 6.

Solution 8: orthogonality

We have

⌊ 1 0 0 ⌋ T ⌈ ⌉ C1(λ) = 0 cosλ − sin λ . 0 sinλ cos λ

Multiplying gives

⌊ ⌋ 1 0 0 C1(λ )T C1(λ) = ⌈ 0 cos2λ + sin2λ 0 ⌉ = I. 0 0 cos2 λ + sin2 λ

(13)

The analogous results are

T T C2 (λ ) C2(λ ) = I, C3 (λ ) C3(λ ) = I.

(14)

Solution 9: inverse and transpose

For every orthogonal matrix,

Ci(λ)−1 = Ci (λ )T.

Also, replacing λ by −λ reverses the signs of the sine terms but leaves the cosine terms unchanged. That is exactly what the transpose does for these matrices. Hence

C (λ)−1 = C (λ)T = C (− λ ), i = 1,2,3. i i i

(15)

Solution 10: determinant

Expanding det C₂(λ) along the second row gives

2 2 det C2(λ ) = cos λ + sin λ = 1.

(16)

Each elementary passive rotation matrix should have determinant +1 because it is a proper orthogonal matrix.

Solution 11: active versus passive

If R_i(λ) is the active vector rotation matrix for the same positive geometric angle, then

Ri(λ) = Ci(λ )T = Ci(− λ).

(17)

The passive map changes coordinates by the inverse of the active vector rotation.

Solution 12: columns as transformed basis vectors

The columns of ^BC_A are the coordinates of the basis vectors of frame A expressed in frame B.

Therefore the columns of C₃(λ) are

⌊ ⌋ ⌊ ⌋ ⌊ ⌋ cosλ sin λ 0 Bˆa1 = ⌈ − sin λ⌉ , Bˆa2 = ⌈ cosλ⌉ , Bˆa3 = ⌈ 0⌉ . 0 0 1

So the old x and y basis vectors are expressed in the new rotated frame by the first two columns, while the z basis vector is unchanged.

Solution 13: small angle form

For small |λ|,

cosλ ≈ 1, sinλ ≈ λ.

Thus

⌊ ⌋ 1 0 − λ C2(λ) ≈ ⌈ 0 1 0 ⌉ . λ 0 1

(18)

Solution 14: build a 3-2-1 passive Euler matrix

For the PhysicsLibrary passive intrinsic 3-2-1 sequence,

α = ψ, β = 𝜃, γ = ϕ.

Therefore

B CA = C1(ϕ)C2 (𝜃)C3(ψ).

(19)

The rightmost factor acts first on a coordinate column.

Solution 15: zero angle specialization

Starting from

BC = C (ϕ)C (𝜃)C (ψ), A 1 2 3

set ϕ = 0 and 𝜃 = 0. Since

C1 (0 ) = I, C2 (0 ) = I,

the product reduces to

BCA = C3 (ψ ).

(20)

So only the yaw transformation remains.

Solution 16: reverse map

B CA = C1 (λ),

then

ACB = (BCA )T = C1(λ)T = C1 (− λ).

(21)

Explicitly,

⌊ ⌋ 1 0 0 ACB = ⌈ 0 cosλ − sin λ⌉ . 0 sin λ cos λ

(22)

Solution 17: sign debugging

A quick test is to evaluate the student’s matrix at λ = π∕2.

This gives

⌊ ⌋ ( ) 0 − 1 0 ^C3 π- = ⌈ 1 0 0⌉ . 2 0 0 1

Applying it to

⌊1⌋ ⌈ ⌉ 0 0

produces

⌊ 0⌋ ⌈ ⌉ 1 , 0

whereas the correct PhysicsLibrary passive matrix should send that vector to

⌊ ⌋ 0 ⌈− 1⌉ 0

for positive frame rotation about +z.

So the student’s matrix is the active rotation matrix or, equivalently, the passive matrix with the angle sign reversed.

Solution 18: composition order

A coordinate column is multiplied from the right.

Thus in

B CA = Ck(γ)Cj (β)Ci(α),

the first factor acting on ^Av is C_i(α).

The result is then acted on by C_j(β), and finally by C_k(γ).

So the matrix product still represents the chronological intrinsic sequence i → j → k.

5 Compact review

The essential formulas reinforced in this exercise companion are

Bv = BCA Av,

(23)

⌊ ⌋ 1 0 0 C1 (λ) = ⌈0 cos λ sin λ⌉ , 0 − sin λ cosλ

(24)

⌊cos λ 0 − sinλ⌋ ⌈ ⌉ C2 (λ) = 0 1 0 , sin λ 0 cosλ

(25)

⌊ ⌋ cos λ sin λ 0 C3 (λ) = ⌈− sin λ cosλ 0⌉ , 0 0 1

(26)

−1 T Ci(λ) = Ci(λ ) = Ci(− λ),

(27)

and

B CA = Ck(γ)Cj (β)Ci(α).

(28)

These formulas are the building blocks for all later Euler angle matrix derivations.

6 References and further reading

Henderson provides a classic engineering presentation of Euler angles, quaternions, and transformation matrices.

Moore gives a modern passive reference frame treatment.

Diebel provides a compact comparison of attitude representations.

References

[1] D. M. Henderson, Euler Angles, Quaternions, and Transformation Matrices: Working Relationships, JSC-12960, NASA Johnson Space Center, 1977. NASA Technical Reports Server

[2] J. K. Moore, Learn Multibody Dynamics, chapter “Orientation of Reference Frames,” 2026 edition. Licensed CC BY 4.0. Orientation of Reference Frames

[3] J. Diebel, “Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors,” Stanford University, 2006. Online PDF

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Euler Angles: Elementary Axis Rotations and Passive Rotation Matrices Examples, Exercises, and Solutions

1 Convention summary

2 Visual reference sheet

3 Exercises

4 Solutions

Solution 1: interpret the map

Solution 2: identify the unchanged coordinate

Solution 3: write the three elementary matrices

Solution 4: a check for

Solution 5: a check for

Solution 6: a check for

Solution 7: a negative angle check

Solution 8: orthogonality

Solution 9: inverse and transpose

Solution 10: determinant

Solution 11: active versus passive

Solution 12: columns as transformed basis vectors

Solution 13: small angle form

Solution 14: build a 3-2-1 passive Euler matrix

Solution 15: zero angle specialization

Solution 16: reverse map

Solution 17: sign debugging

Solution 18: composition order

5 Compact review

6 References and further reading

References

License

Solution 4: a $90 ∘$ check for $C1$

Solution 5: a $90 ∘$ check for $C2$

Solution 6: a $90 ∘$ check for $C3$