Visighta

Vector Kinematics in Two and Three Dimensions

3.2 Component Form and Changing Bases

Two Observers, One Ball

Two observers watch the same ball roll across a tilted table. Observer A sets up horizontal and vertical axes --- the standard choice, aligned with gravity. Observer B sets up axes aligned with the table's surface --- one axis along the slope, the other perpendicular to it.

Observer A writes down the ball's velocity: $(2.1, -3.6)$ m/s.

Observer B writes down the ball's velocity: $(4.2, 0)$ m/s.

They look at each other's numbers. The numbers disagree. Neither made a measurement error.

Are they both right?

Before you read on: A velocity vector has components $(3, 4)$ in one coordinate system. Can you find a coordinate system where the same vector has components $(5, 0)$? What about $(0, 5)$? What about $(4, 3)$?

Try to answer all three before continuing.

If you said yes to all three, you are already sensing the main idea. If you are not sure, that is fine --- this section will make it concrete.

What Does It Mean to Write a Vector as Numbers?

In Section 3.1, you met the velocity vector as an arrow: it has a magnitude (speed) and a direction (tangent to the path). The arrow is a physical object. It does not depend on anyone's choice of coordinates.

But arrows are hard to compute with. To do algebra, we need numbers. And to get numbers from an arrow, we need to project the arrow onto some chosen directions.

Here is the procedure, stated plainly: pick two perpendicular directions --- call them $\hat{x}$ and $\hat{y}$. To find the components of a vector $\vec{v}$, drop perpendicular lines from the tip of the arrow onto each axis. The lengths of those projections, with appropriate signs, are the components $v_x$ and $v_y$.

$$\vec{v} = v_x \, \hat{x} + v_y \, \hat{y}$$

This is not the vector. This is the vector expressed in a particular basis. The arrow is the vector. The numbers $(v_x, v_y)$ are a description of the arrow in a language you chose.

An analogy: the same city can be described as "London" in English and "Londres" in French. Different words, same city. Components are the language; the vector is the city.

Descartes introduced the idea of coordinate axes in the 1630s, unifying algebra and geometry. Before that, mathematicians had to reason about length and direction separately. Coordinates made it possible to replace geometric arguments with equations --- the foundation of everything in this course.

Projections, Not Containers

There is a subtle but important distinction to make here. Components are projections, not pieces.

When you decompose $\vec{v}$ into $v_x$ and $v_y$, you are not breaking the vector into two independent parts that were somehow hidden inside it. You are measuring how much of the vector lies along each of your chosen directions. The vector was always a single arrow pointing in a single direction. The two numbers are your description of that arrow, not its ingredients.

This matters because it means the numbers change when you change the directions you project onto. If the components were intrinsic pieces of the vector, they would not change. But they are projections, so they change whenever the "screen" you project onto rotates.

[Video: A fixed arrow on screen. A flashlight shines from different angles, casting the arrow's shadow onto a wall (the $x$-axis) and onto the floor (the $y$-axis). As the flashlight rotates, the shadows change length. The arrow never moves. Caption: "The shadows change. The arrow does not."]

Watch It Happen

Now let's see this directly.

[Interactive: Rotating Axes Explorer. A fixed vector arrow is displayed on screen --- bold, colored, and clearly labeled $\vec{v}$. Two perpendicular axes ($\hat{x}$ and $\hat{y}$) are drawn through the tail of the vector, initially aligned with the standard horizontal/vertical orientation. A rotation slider at the bottom lets the student rotate the axes from $0^\circ$ to $360^\circ$.

As the axes rotate: - Dashed projection lines drop from the tip of $\vec{v}$ onto each axis, updating in real time. - The component values $v_x$ and $v_y$ are displayed numerically and update continuously. - The magnitude $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ is computed and displayed --- it does not change. - The vector arrow itself stays perfectly fixed.

The initial vector has components $(3, 4)$ at $0^\circ$ rotation, giving magnitude $5$.]

Guided exploration:

  1. Start with the axes at $0^\circ$. Confirm the components are $(3, 4)$ and the magnitude is $5$.
  2. Slowly rotate the axes. Watch $v_x$ and $v_y$ change. Does the magnitude change?
  3. Find the rotation angle that makes $v_y = 0$. What are the components at this angle? (You should get something close to $(5, 0)$.)
  4. Keep rotating. Find the angle that makes $v_x = 0$. What are the components now?
  5. Is there an angle where both components are zero simultaneously?

If you found the angle for part 3, you rotated the $x$-axis to align with the vector itself. When the axis points along the arrow, the entire vector is "in the $x$-direction" and there is nothing left over for $v_y$. The components become $(5, 0)$ --- the full magnitude along $x$, zero along $y$.

For part 5: no. There is no rotation angle that makes both components zero, because the vector has nonzero length. The magnitude is invariant. You can shuffle the numbers between $v_x$ and $v_y$, but you cannot make them both vanish.

The Mathematics of Rotation

Let's make the exploration precise. Suppose a vector $\vec{v}$ has components $(v_x, v_y)$ in the original coordinate system. You rotate the axes counterclockwise by an angle $\theta$ to get a new coordinate system with axes $\hat{x}'$ and $\hat{y}'$. The components in the new system are:

$$v_{x'} = v_x \cos\theta + v_y \sin\theta$$ $$v_{y'} = -v_x \sin\theta + v_y \cos\theta$$

This is a rotation transformation. It looks like a formula to memorize, but it is not --- it is the algebraic statement of what you just saw in the interactive. The new components are projections of the same arrow onto rotated directions.

Check it against your exploration: The vector $(3, 4)$ has magnitude $5$. To make $v_{y'} = 0$, we need $-3\sin\theta + 4\cos\theta = 0$, which gives $\tan\theta = 4/3$, so $\theta \approx 53.1°$. At this angle, $v_{x'} = 3\cos(53.1°) + 4\sin(53.1°) = 3(0.6) + 4(0.8) = 1.8 + 3.2 = 5$. The components become $(5, 0)$, exactly as the interactive showed.

Notice what the rotation preserved:

$$v_{x'}^2 + v_{y'}^2 = v_x^2 + v_y^2 = |\vec{v}|^2$$

The magnitude is invariant under rotation. This is the Pythagorean theorem wearing a different hat. No matter how you orient your axes, $v_x^2 + v_y^2$ gives the same number. Components change; the magnitude does not.

The Concept: Basis Independence

Here is the central idea of this section, stated as clearly as possible:

Components are projections onto chosen directions. Different bases give different numbers, but the vector --- the physical quantity --- does not change.

The velocity of the ball on the tilted table is a single arrow. Observer A's numbers $(2.1, -3.6)$ and Observer B's numbers $(4.2, 0)$ are two descriptions of the same arrow in two different coordinate languages. Neither is more "real" than the other. Both are correct. They must be, because the ball does not know or care what axes anyone drew.

This is the same idea you met in Section 1.3, when you learned that choosing a reference frame affects the numbers you write for position. Here, we are extending it: choosing a basis affects the numbers you write for any vector quantity --- velocity, acceleration, force. The physics is invariant. The description is not.

This distinction --- between the physics (which is real) and the description (which is a choice) --- will follow you through the entire course. Every time you set up a dynamics problem, you choose axes. That choice determines your component equations. But the forces, the motion, the outcomes are the same regardless of what you chose.

Variation: Same Vector, Different Bases

Let's drive the point home with a systematic comparison. Consider a force vector $\vec{F}$ with magnitude 10 N, pointing $30°$ above the horizontal.

| Basis | Axes | $F_x$ | $F_y$ | $|\vec{F}|$ | |:---|:---|:---|:---|:---| | Standard | $\hat{x}$ horizontal, $\hat{y}$ vertical | $10\cos 30° = 8.66$ | $10\sin 30° = 5.00$ | 10 | | Rotated 30° | $\hat{x}'$ along $\vec{F}$, $\hat{y}'$ perpendicular | $10$ | $0$ | 10 | | Rotated 45° | $\hat{x}'$ at $45°$ from horizontal | $10\cos(-15°) = 9.66$ | $10\sin(-15°) = -2.59$ | 10 | | Rotated 90° | $\hat{x}'$ vertical, $\hat{y}'$ horizontal | $5.00$ | $-8.66$ | 10 |

What changed? What stayed the same?

The components changed in every row. The magnitude stayed at 10 N in every row. The direction of the physical force did not change --- it still points 30 degrees above the horizontal. The only thing that changed is how we describe that direction using numbers.

Decomposing and Reconstructing

In practice, you need two skills: taking a vector apart (decomposition) and putting it back together (reconstruction).

Decomposition: Given a vector with magnitude $|\vec{v}|$ and direction $\alpha$ measured from the positive $x$-axis:

$$v_x = |\vec{v}| \cos\alpha, \qquad v_y = |\vec{v}| \sin\alpha$$

Reconstruction: Given components $v_x$ and $v_y$:

$$|\vec{v}| = \sqrt{v_x^2 + v_y^2}, \qquad \alpha = \arctan!\left(\frac{v_y}{v_x}\right)$$

with appropriate attention to the quadrant (use atan2 in practice, or check the signs of $v_x$ and $v_y$ to determine which quadrant $\alpha$ falls in).

These two operations are inverses. Decomposition turns an arrow into numbers. Reconstruction turns numbers back into an arrow. Together, they let you move freely between the geometric picture (arrows on a diagram) and the algebraic picture (equations with components).

Before you read on: A displacement vector has magnitude 13 m and direction $\alpha$ such that $\cos\alpha = 5/13$ and $\sin\alpha = 12/13$. What are its components? Now go backward: if the components are $(5, 12)$, what are the magnitude and direction?

Check your answer **Decomposition:** $v_x = 13 \cdot (5/13) = 5$ m, $v_y = 13 \cdot (12/13) = 12$ m. Components: $(5, 12)$. **Reconstruction:** $|\vec{v}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ m. Direction: $\alpha = \arctan(12/5) \approx 67.4°$ from the positive $x$-axis. Since both components are positive, the vector is in the first quadrant, consistent with $\alpha \approx 67.4°$. The key check: decomposition and reconstruction are inverses. You get back what you started with.

Why Basis Choice Matters in Practice

If the physics does not depend on the basis, why does the choice of basis matter?

Because some choices make the math simpler.

Consider a block sliding down a ramp inclined at $30°$. You want to analyze the forces and acceleration.

Choice 1: Standard axes ($\hat{x}$ horizontal, $\hat{y}$ vertical). Gravity is simple: $\vec{F}_g = (0, -mg)$. But the normal force points perpendicular to the ramp, which is not along either axis. You have to decompose the normal force into $x$ and $y$ components, and the constraint "the block stays on the ramp" becomes $a_x \sin 30° = a_y \cos 30°$ --- an awkward coupling between the two directions.

Choice 2: Tilted axes ($\hat{x}'$ along the ramp, $\hat{y}'$ perpendicular to the ramp). Now the normal force is simple: $\vec{N} = (0, N)$ in this basis. The constraint "stays on the ramp" is just $a_{y'} = 0$. The price is that gravity must be decomposed: $\vec{F}_g = (mg\sin 30°,\, -mg\cos 30°)$. But this decomposition is straightforward, and the equations decouple cleanly.

Choice 2 is almost always better for ramp problems. The physics is the same either way, but the algebra in Choice 2 is shorter, cleaner, and less error-prone.

The art of choosing axes is not about finding the "correct" basis (any basis is correct). It is about finding the convenient basis --- the one that simplifies the constraints and decouples the equations. You will practice this judgment call throughout the course.

Extension to Three Dimensions

Everything we have discussed extends naturally to three dimensions. A vector in 3D has three components:

$$\vec{v} = v_x \, \hat{x} + v_y \, \hat{y} + v_z \, \hat{z}$$

and its magnitude is:

$$|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$

Rotations in 3D are more complex --- you can rotate around any of three axes --- but the fundamental principle is the same. Components are projections onto chosen directions. Rotating the basis changes the components but not the vector. The magnitude $|\vec{v}|$ is invariant under any rotation.

We will not need the full 3D rotation machinery in this course, but the conceptual point is important: basis independence is not a 2D trick. It is a general property of vectors in any number of dimensions.

Practice

Layer 1: Concrete

A velocity vector has magnitude 10 m/s and points at $60°$ above the positive $x$-axis.

(a) Find the components $v_x$ and $v_y$.

(b) Verify that $\sqrt{v_x^2 + v_y^2} = 10$ m/s.

Check your answer (a) $v_x = 10\cos 60° = 10 \times 0.5 = 5.0$ m/s. $v_y = 10\sin 60° = 10 \times 0.866 = 8.66$ m/s. (b) $\sqrt{5.0^2 + 8.66^2} = \sqrt{25 + 75} = \sqrt{100} = 10$ m/s. The magnitude is recovered, confirming the decomposition is consistent.

Layer 2: Pattern

A vector $\vec{A}$ has components $(6, 8)$ in the standard basis.

(a) Find the magnitude of $\vec{A}$.

(b) Find the rotation angle $\theta$ that makes the $y'$-component zero. What are the components in this rotated basis?

(c) Find the rotation angle that makes the $x'$-component zero. What are the components in this rotated basis?

(d) Find the components when $\theta = 45°$.

Check your answer (a) $|\vec{A}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. (b) Setting $A_{y'} = -6\sin\theta + 8\cos\theta = 0$ gives $\tan\theta = 8/6 = 4/3$, so $\theta \approx 53.1°$. Then $A_{x'} = 6\cos 53.1° + 8\sin 53.1° = 6(0.6) + 8(0.8) = 3.6 + 6.4 = 10$. Components: $(10, 0)$. The entire vector aligns with the $x'$-axis. (c) Setting $A_{x'} = 6\cos\theta + 8\sin\theta = 0$ gives $\tan\theta = -6/8 = -3/4$, so $\theta \approx -36.9°$ (or equivalently $\theta \approx 143.1°$). At $\theta = 143.1°$: $A_{y'} = -6\sin 143.1° + 8\cos 143.1° = -6(0.6) + 8(-0.8) = -3.6 - 6.4 = -10$. Components: $(0, -10)$. The entire vector aligns with the negative $y'$-axis. (d) At $\theta = 45°$: $A_{x'} = 6\cos 45° + 8\sin 45° = (6+8)/\sqrt{2} = 14/\sqrt{2} \approx 9.90$. $A_{y'} = -6\sin 45° + 8\cos 45° = (-6+8)/\sqrt{2} = 2/\sqrt{2} \approx 1.41$. Check: $\sqrt{9.90^2 + 1.41^2} = \sqrt{98 + 2} = \sqrt{100} = 10$. The magnitude is preserved, as it must be.

Layer 3: Structure

You are solving a projectile problem: a ball launched at $60°$ above the horizontal. A classmate suggests rotating the axes by $45°$ to "split the difference." Why does this make the problem harder, not easier?

Check your answer In the standard axes (horizontal and vertical), gravity points entirely along the $y$-axis: $\vec{g} = (0, -g)$. This means the horizontal motion has zero acceleration and the vertical motion has constant acceleration. The two directions are *independent*, and you can solve them separately --- two simple 1D problems. If you rotate the axes by $45°$, gravity has components along *both* axes: $g_{x'} = g\sin 45°$ and $g_{y'} = -g\cos 45°$. Now both components of motion are accelerated. The equations are still correct, but they are more complicated, and you lose the simplifying independence between horizontal and vertical. The good basis is the one where gravity (or whatever acceleration is present) aligns with one axis. For projectile problems near Earth's surface, that is the standard horizontal/vertical basis. Rotating away from it introduces unnecessary coupling. The lesson: the "best" basis is usually the one that aligns with the forces or constraints of the problem, not an arbitrary geometric choice.

Layer 4: Transfer

In data science, Principal Component Analysis (PCA) takes a cloud of data points and rotates the coordinate axes to align with the directions of greatest variance in the data. The first principal component axis points along the direction where the data is most spread out.

How is this related to the idea of choosing a basis for a vector? Why might aligning your axes with the data's structure be useful?

Check your answer PCA is exactly a basis change. The data points exist in some space, and their positions do not change when you rotate the axes. But the *components* change --- and in the rotated basis, the components carry more meaningful information. In the original basis (say, raw measurements), the variance is spread across all axes, and the axes may not correspond to anything physically meaningful. In the PCA basis, the first axis captures the most variance, the second captures the next most, and so on. If most of the variance is captured by just a few axes, you can *ignore the rest* --- reducing the dimensionality of the data while losing very little information. This is the same logic as choosing tilted axes for a ramp problem: the physics does not change, but the right basis makes the important structure obvious and the unimportant details easy to discard.

Reflection

Does the vector care what coordinate system you use?

Think about what is "real" in this section and what is a choice. The arrow is real --- it has a magnitude and a direction that exist independently of any coordinate system. The components are a choice --- they depend on the basis you picked. The magnitude is real --- it is the same in every basis. The individual component values are not.

Now ask yourself: when you solve a physics problem and get a numerical answer for the $x$-component of velocity, what part of that answer is physics and what part is your coordinate system talking?

What are you least sure about? If something about basis changes still feels slippery, write down specifically what confuses you. Is it the rotation formula? The idea that the components are not intrinsic? The claim that any basis is equally valid? Naming the confusion is the first step to resolving it.

Looking Ahead

You now understand that a single vector can be described by different sets of numbers depending on the basis you choose --- and that the physics does not depend on that choice. This is a powerful idea, and it will shape how you approach every problem from here on.

In the next section, we move from describing individual vectors to describing entire paths. When an object traces a curve through space, its position at each moment in time is a vector $\vec{r}(t) = (x(t), y(t))$. This is a parametric description of the trajectory, and it encodes both the shape of the path and the timing of the motion along it. You already have the tools --- derivatives give velocity, second derivatives give acceleration. What is new is that the path is a curve, not a line, and the relationship between the geometry of the curve and the components of motion will open up everything that follows in this chapter.