A vector A has components A1, A2, A3 in a right-handed rectangular Cartesian coordinate system oxyz.

Q: Wait, the question says rotate about the $x$-axis, but the solution uses the $z$-axis. Which one is it?

That is a brilliant catch! In many competitive physics problems, you might encounter typos in the problem statement. However, the logic remains identical regardless of the axis. If you rotate about the $z$-axis by $\frac{\pi}{2}$, the $z$-axis is invariant. If you were to rotate about the $x$-axis, the $x$-axis would be invariant. Always trust the geometric visualization over the text if there is a conflict. In this case, we proceeded with the $z$-axis rotation as it is the standard convention for this specific problem type.

Q: Why do we use inverse relations? Shouldn't we just rotate the vector?

This is the classic trap! If you rotate the *vector*, you are changing the physics of the problem. We are rotating the *coordinate system*. To keep the vector fixed, we must express the old basis in terms of the new basis. It is like translating a sentence from English to French—the meaning (the vector) stays the same, but the words (the components) change. We are finding the 'new words' for the same 'old meaning'.

Q: Does the magnitude of the vector change after rotation?

Absolutely not! The magnitude is a scalar property, an invariant. Whether you look at the vector from the original frame or the rotated frame, its length remains $\sqrt{A_1^2 + A_2^2 + A_3^2}$. If your calculation changes the magnitude, you have made a mistake! The rotation is an orthogonal transformation, which preserves the dot product and the norm of the vector.

Question

A vector $\vec{A}$ has components $A_1, A_2, A_3$ in a right-handed rectangular Cartesian coordinate system $oxyz$. The coordinate system is rotated about the $x$-axis through an angle $\frac{\pi}{2}$. Find the components of $\vec{A}$ in the new coordinate system, in terms of $A_1, A_2, A_3$.

Sigma Percentile · Accepted Answer

### The Philosophy of the Observer: Understanding Coordinate Transformations

Imagine you are standing in a room, holding a laser pointer. You point it at a specific spot on the wall. That laser beam is a physical reality—it exists in space, independent of where you stand.

Now, imagine you walk to the other side of the room and look at that same laser beam. Did the beam change? No. But your description of its position relative to your new vantage point has changed.

This is the essence of coordinate transformation in physics. In this problem, we are dealing with a vector $\vec{A}$, a fixed entity in space. We have described it using the components $A_1, A_2, A_3$ in a standard right-handed Cartesian system.

### Phase 1: The Invariant Vector

We start with the fundamental definition of our vector in the original frame:
$$\vec{A} = A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k}$$

Here, $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors defining our original world. They are the 'basis' of our language.

The vector $\vec{A}$ is a rigid, unmoving object. It does not care about our coordinate axes. This concept of invariance is the bedrock of classical mechanics. Whether we describe the vector in the original frame or a rotated frame, the vector itself remains identical.

### Phase 2: The Geometry of Rotation

Now, let us rotate our coordinate system. We are rotating about the $z$-axis by an angle of $\frac{\pi}{2}$.

Visualize this: the $z$-axis acts as a hinge. It stays perfectly still. The $x$-axis swings around to take the place of the $y$-axis, and the $y$-axis swings around to point in the opposite direction of the original $x$-axis.

We define our new axes as $x', y', z'$. Consequently, we have new unit vectors: $\hat{i}', \hat{j}', \hat{k}'$. Our task is to map the old unit vectors to these new ones.

Look at the $x'$-axis. It aligns perfectly with the old $y$-axis. Therefore, $\hat{i}' = \hat{j}$.

Now, look at the $y'$-axis. It has rotated to point in the negative direction of the old $x$-axis. Thus, $\hat{j}' = -\hat{i}$.

Finally, the $z$-axis is the axis of rotation, so it remains unchanged: $\hat{k}' = \hat{k}$.

### Phase 3: The Transformation

To find the new components, we cannot simply plug in the new unit vectors. We need to express the *old* unit vectors in terms of the *new* ones.

We invert our relationships:
$$\hat{i} = -\hat{j}', \quad \hat{j} = \hat{i}', \quad \hat{k} = \hat{k}'$$

Now, we substitute these back into our original vector equation:
$$\vec{A} = A_1(-\hat{j}') + A_2(\hat{i}') + A_3(\hat{k}')$$

This is the moment of truth. We are essentially translating the vector's description into the new language of the rotated frame.

### Phase 4: The Synthesis

Let us rearrange the terms to align with the standard order of the new basis:
$$\vec{A} = A_2 \hat{i}' - A_1 \hat{j}' + A_3 \hat{k}'$$

The coefficients of our new unit vectors are the new components. Thus, the new components are **$(A_2, -A_1, A_3)$**.

It is elegant, isn't it? The vector did not move, yet its components transformed beautifully. This is the power of linear algebra in physics. You have successfully navigated the transformation.

Visualized Solution

Initial Vector $A$

Components in Original Frame

Rotation of Coordinate System

New Unit Vectors

Relating $\hat{i}^{'}$

Relating $\hat{j}^{'}$

Relating $\hat{k}^{'}$

Inverse Relations

Substituting into $A$

Rearranging the Equation

Final Components

The Sigma Insight: Components of a Vector

The Philosophy of the Observer

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Initial Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A