Question bank — Direction cosine matrix (DCM) — construction from Euler angles

False. The vehicle turns, but $C_{BN}$ only re-expresses the same physical vector's components in a new frame — the arrow in space never moves. Rotating the axes by $+\theta$ looks numerically like rotating the vector by $-\theta$, which is why signs feel backwards.

True. Each cell is ${\hat{b}}_i\cdot {\hat{n}}_j=\cos ({\theta }_{ij})$ (the grid in Figure s02), and a cosine can never leave $[-1,1]$. If you ever compute a cell of magnitude $>1$, your matrix is corrupted (not orthonormal).

True. Reversing the translation direction inverts the matrix, and for a rotation the inverse is the transpose: $C_{NB}=C_{BN}^{-1}=C_{BN}^T$. Flipping the grid across its diagonal undoes the turn.

False. $C_{BN}^{-1}=C_{BN}^T$ means orthogonal, not symmetric. Most DCMs are not symmetric (a $90^{\circ }$ yaw isn't). You also need $\det C_{BN}=+1$ to exclude reflections.

True. Orthonormality ($C_{BN}{C}_{BN}^T=I$ and $C_{BN}^T{C}_{BN}=I$) forces both rows and columns to be mutually perpendicular unit vectors — one statement is the row version, the other the column version of the same fact.

True. The product of two rotations is a rotation: orthogonality and $\det =+1$ both survive multiplication ($\det (AB)=\det A\det B=1\cdot 1$). This closure is exactly what makes SO(3) a group.

False. Matrix multiplication doesn't commute, so rotation order changes the result. Same numbers, different sequence = different attitude in general.

True, but only to first order. Each tiny turn is $R_k\approx I-ϵ[{\hat{e}}_k\times ]$; multiplying two gives $I-ϵ([{\hat{e}}_a\times ]+[{\hat{e}}_b\times ])+{ϵ}^2(\dots )$. The linear part is a plain sum of skew matrices, and addition commutes, so swapping order only changes the discarded ${ϵ}^2$ block. That surviving sum is exactly $I-[\mathbf{\alpha }\times ]$.

False. The DCM stays a perfectly good rotation with $\det =+1$; it is the Euler-angle parameterization that goes singular, because at pitch $\theta =\pm 90^{\circ }$ two of the three angles rotate about the same physical axis.

Error. Mini-derivation: pitch spins the $3$–$1$ plane, so a point at $(x_3,x_1)$ goes to $x_1^{\prime }=x_{1}c\theta +x_{3}s\theta$, $x_3^{\prime }=-x_{1}s\theta +x_{3}c\theta$ (the standard 2D turn in the $1,3$ order). Because the ordered pair here is $(3,1)$ not $(1,2)$, when you write the matrix in normal $1,2,3$ row order the $-\sin$ lands in the top-right ($C_{13}$) and $+\sin$ in the bottom-left ($C_{31}$) — the opposite of $R_1,R_3$. Figure s03's middle panel shows this plane; always re-derive by asking which two axes span the rotation plane and in what cyclic order.

Error. Each matrix acts on the output of the previous one (Figure s04), so the last physical rotation multiplies first (leftmost). Correct order is $C_{BN}=R_1(\varphi )R_2(\theta )R_3(\psi )$ — yaw ($R_3$), applied first, sits rightmost; roll ($R_1$), applied last, sits leftmost.

Error. Plain $\arctan$ collapses the two quadrants that share a tangent value, so signs of $C_{11},C_{12}$ get lost. Use $\psi =atan2(C_{12},C_{11})$, which reads both signs separately and returns the correct quadrant across the full $360^{\circ }$.

Error — a sign slip. In the 3-2-1 DCM the entry is $C_{13}=-\sin \theta$ (here $\theta$ is the pitch angle), so $\theta =-\arcsin (C_{13})$. Forgetting the minus flips your pitch upside down.

Half-error. Perpendicular unit rows give orthogonality but can still produce a reflection ($\det =-1$, a left-handed frame). You must also ensure right-handedness: ${\hat{b}}_1\times {\hat{b}}_2={\hat{b}}_3$, giving $\det =+1$.

Error. The skew vector orders components by axis number, not rotation sequence: $\mathbf{\alpha }=(\varphi ,\theta ,\psi )$ for axes $1,2,3$ (recall $\varphi \to 1$, $\theta \to 2$, $\psi \to 3$). This ordering is what makes it plug directly into a linearized error state.

Because it is a dot product of two unit vectors, and $\hat{u}\cdot \hat{v}=\cos (\text{angle})$ when both have length $1$. The cosine cleanly measures "how much of one axis points along another" on a fixed $[-1,1]$ scale.

Rotations about different axes don't add like numbers — they interact. Matrix multiplication is precisely the operation that composes "do this transform, then that transform," respecting that the second acts on the result of the first (Figure s04).

Inverting a general $3\times 3$ matrix costs arithmetic and can amplify round-off; transposing is just relabeling entries — exact and instant. On a power- and cycle-limited flight computer that saving, done thousands of times per second, is huge.

$C_{13}$ maps the inertial $3$-axis onto the body $1$-axis. In the $3$-$2$-$1$ build, only the pitch rotation ($R_2(\theta )$) tilts the body $1$-axis out of the inertial horizontal plane; yaw spins within that plane and roll happens after, so neither touches this particular projection. It's a handy sanity anchor.

Euler angles hit the gimbal-lock singularity and their rate equations blow up near pitch $\theta =\pm 90^{\circ }$. Quaternions parameterize the same rotations with no singularity, so the internal state stays well-behaved; humans just prefer reading angles.

$\det =-1$ would encode a reflection — turning a right-handed frame into a left-handed mirror image, which no physical rotation can do. A proper rotation preserves handedness, forcing $\det =+1$.

An infinitesimal rotation of a vector is a tiny cross product, $\delta \mathbf{v}=-\mathbf{\alpha }\times \mathbf{v}$, and the cross product is exactly what the skew-symmetric operator $[\mathbf{\alpha }\times ]$ performs. So $C_{BN}\approx I-[\mathbf{\alpha }\times ]$ is just "identity plus a tiny cross-product nudge."

It is the identity matrix $I$. With all angles zero, every body axis coincides with its inertial counterpart, so ${\hat{b}}_i\cdot {\hat{n}}_j=1$ when $i=j$ and $0$ otherwise — exactly $I$.

In both cases $\cos \theta =0$, so yaw ($R_3$) and roll ($R_1$) turn about the same physical axis and only their combination is recoverable, not each separately. At $\theta =+90^{\circ }$ the DCM depends on yaw and roll only through the difference $\varphi -\psi$; at $\theta =-90^{\circ }$ only through the sum $\varphi +\psi$ — mirror-image versions of the same singularity, since flipping the sign of pitch flips the relative sense of the two collapsed axes.

Yes — $\cos 360^{\circ }=\cos 0^{\circ }=1$ and $\sin 360^{\circ }=\sin 0^{\circ }=0$, giving identical matrices. The DCM is periodic in each angle with period $360^{\circ }$; distinct angle triples can name the same attitude.

Yes — this is Euler-angle non-uniqueness. Different $(\psi ,\theta ,\varphi )$ triples can produce one identical DCM (a $180^{\circ }$ flip is reachable multiple ways), which is why extraction formulas pick one canonical representative.

An improper rotation (rotation combined with a reflection). It is a valid orthogonal matrix but not a DCM, because it flips handedness — no rigid body can physically achieve it.

The $\varphi$ terms collapse ($\cos \varphi \to 1,\sin \varphi \to 0$) and $C_{BN}\to R_2(\theta )R_3(\psi )$ — a clean two-rotation matrix. No singularity here; only pitch $\theta =\pm 90^{\circ }$ triggers gimbal lock, roll does not.