Question:medium
Which ONE or MORE among the following homogeneous representations in robotics involve(s) rotation about Y-axis by an angle \( \theta \)?
Which ONE or MORE among the following homogeneous representations in robotics involve(s) rotation about Y-axis by an angle \( \theta \)?
Show Hint
A pure Y-axis rotation matrix keeps the middle row and column as 0, 1, 0; a translation column does not change which axis the rotation happens about.
Updated On: Aug 5, 2026
- \[ \begin{pmatrix} \cos\theta & -\sin\theta & 0 & 0 \\ \sin\theta & \cos\theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
- \[ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta & 0 \\ 0 & \sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
- \[ \begin{pmatrix} \cos\theta & 0 & \sin\theta & 0 \\ 0 & 1 & 0 & 0 \\ -\sin\theta & 0 & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
- \[ \begin{pmatrix} \cos\theta & 0 & \sin\theta & 3 \\ 0 & 1 & 0 & 4 \\ -\sin\theta & 0 & \cos\theta & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
Show Solution
The Correct Option is C, D
Solution and Explanation
Step 1: A Different Way to Test Each Matrix:
Instead of memorizing the standard rotation matrix pattern, we can test each option by seeing what it does to the three basis directions $\hat{i} = (1,0,0)$, $\hat{j} = (0,1,0)$, and $\hat{k} = (0,0,1)$.
A rotation about the Y-axis is the only kind of rotation that leaves the Y-direction, $\hat{j}$, completely unmoved, while both $\hat{i}$ and $\hat{k}$ get mixed together by $\cos\theta$ and $\sin\theta$ terms.
Step 2: Applying the Test to Each Option:
In option (A), multiplying the rotation block by $\hat{j} = (0,1,0)$ gives $(-\sin\theta, \cos\theta, 0)$, which is not equal to $\hat{j}$ unless $\theta = 0$, so $\hat{j}$ moves, this rules out a Y-axis rotation here (it is in fact a Z rotation, since $\hat{k}=(0,0,1)$ stays fixed instead).
In option (B), multiplying by $\hat{i} = (1,0,0)$ leaves it unchanged, meaning the X-direction is the one that stays fixed, so this is an X-axis rotation, not Y.
In option (C), multiplying the rotation block by $\hat{j} = (0,1,0)$ gives exactly $(0,1,0)$ back, unchanged, while $\hat{i}$ and $\hat{k}$ pick up $\cos\theta, \sin\theta$ terms. This confirms option (C) is a pure rotation about Y, with no shift since the translation entries are all zero.
In option (D), the rotation part is identical to (C), so $\hat{j}$ again stays fixed under the rotation block, the extra column $(3,4,2)$ only shifts the origin and does not change which axis the rotation happens about. So (D) also involves a Y-axis rotation, combined with a translation.
Final Answer:
Testing which basis vector stays fixed under each rotation block shows that $\hat{j}$ (the Y-direction) is preserved only in options (C) and (D). \[ \boxed{\text{C and D are the Y-axis rotations}} \]
Instead of memorizing the standard rotation matrix pattern, we can test each option by seeing what it does to the three basis directions $\hat{i} = (1,0,0)$, $\hat{j} = (0,1,0)$, and $\hat{k} = (0,0,1)$.
A rotation about the Y-axis is the only kind of rotation that leaves the Y-direction, $\hat{j}$, completely unmoved, while both $\hat{i}$ and $\hat{k}$ get mixed together by $\cos\theta$ and $\sin\theta$ terms.
Step 2: Applying the Test to Each Option:
In option (A), multiplying the rotation block by $\hat{j} = (0,1,0)$ gives $(-\sin\theta, \cos\theta, 0)$, which is not equal to $\hat{j}$ unless $\theta = 0$, so $\hat{j}$ moves, this rules out a Y-axis rotation here (it is in fact a Z rotation, since $\hat{k}=(0,0,1)$ stays fixed instead).
In option (B), multiplying by $\hat{i} = (1,0,0)$ leaves it unchanged, meaning the X-direction is the one that stays fixed, so this is an X-axis rotation, not Y.
In option (C), multiplying the rotation block by $\hat{j} = (0,1,0)$ gives exactly $(0,1,0)$ back, unchanged, while $\hat{i}$ and $\hat{k}$ pick up $\cos\theta, \sin\theta$ terms. This confirms option (C) is a pure rotation about Y, with no shift since the translation entries are all zero.
In option (D), the rotation part is identical to (C), so $\hat{j}$ again stays fixed under the rotation block, the extra column $(3,4,2)$ only shifts the origin and does not change which axis the rotation happens about. So (D) also involves a Y-axis rotation, combined with a translation.
Final Answer:
Testing which basis vector stays fixed under each rotation block shows that $\hat{j}$ (the Y-direction) is preserved only in options (C) and (D). \[ \boxed{\text{C and D are the Y-axis rotations}} \]
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