Converting Between Vector Forms

Hello again. In the previous lesson, you reviewed radians and right-triangle trigonometry, including the key pattern that an angled force can be split into horizontal and vertical effects using sine and cosine. This lesson makes that pattern precise through vectors.

Vectors are essential throughout physics: displacement, velocity, acceleration, force, momentum, and electric field all have both magnitude and direction. Here you will learn to translate freely between the two standard descriptions of a two-dimensional vector:

• magnitude-direction form: a size and an angle;
• Cartesian component form: its signed horizontal and vertical changes.

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A vector is a change, not a location

A scalar has magnitude only: $5\mathrm{kg}$, $12\mathrm{s}$, and $300\mathrm{K}$ are scalars. A vector needs direction as well: $8\mathrm{m}$ east, $15m/s$ downward, or a $20\mathrm{N}$ force at an angle.

We can draw a vector as an arrow. Its:

• length represents magnitude;
• arrowhead indicates direction;
• tail is its starting point;
• head is its ending point.

A vector is unchanged if we slide it to another place without rotating it or changing its length. For instance, a velocity of $4m/s$ to the right is the same vector wherever it is drawn.

In a Cartesian coordinate system, we describe a vector $\vec{A}$ using its horizontal and vertical components:

$$\vec{A}=⟨A_x,A_y⟩.$$

The first number, $A_x$, is the signed change in the $x$-direction. The second, $A_y$, is the signed change in the $y$-direction.

• Right gives a positive $x$-component; left gives a negative $x$-component.
• Up gives a positive $y$-component; down gives a negative $y$-component.

Thus,

$$⟨6,-2⟩$$

means “move $6$ units right and $2$ units down.” The pair is not automatically a point’s coordinates. It describes a change. It becomes the coordinates of the arrow’s tip only when the arrow begins at the origin.

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The geometry behind components

Every two-dimensional vector can be viewed as the hypotenuse of a right triangle. Its components form the two perpendicular legs.

Suppose a vector has magnitude $A$ and standard direction angle $\theta$, measured counterclockwise from the positive $x$-axis. The horizontal component is adjacent to $\theta$, and the vertical component is opposite it. From cosine and sine,

$$\cos \theta =\frac{A_x}{A},$$ $$\sin \theta =\frac{A_y}{A}.$$

Multiplying both equations by $A$ gives the central conversion formulas:

$$A_x=A\cos \theta ,$$ $$A_y=A\sin \theta .$$

These formulas include the signs automatically, provided that $\theta$ is the standard angle from the positive $x$-axis.

Example: magnitude and direction to components

A drone has a displacement of magnitude

$$A=50\mathrm{m}$$

at

$$\theta =35^{\circ }$$

above the positive $x$-axis.

The components are

$$A_x=(50\mathrm{m})\cos (35^{\circ })\approx 41.0\mathrm{m},$$ $$A_y=(50\mathrm{m})\sin (35^{\circ })\approx 28.7\mathrm{m}.$$

Therefore,

$$\vec{A}\approx ⟨41.0,28.7⟩\mathrm{m}.$$

The components are both positive because the vector lies in quadrant I: rightward and upward.

A useful interpretation is that the one diagonal displacement has the same net effect as performing two perpendicular displacements: $41.0\mathrm{m}$ right and $28.7\mathrm{m}$ up. It does not mean the drone traveled $69.7\mathrm{m}$; its actual straight-line displacement magnitude is still $50\mathrm{m}$.

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Signs and direction conventions

The formulas do not change from one quadrant to another. What changes are the signs of sine and cosine.

Quadrant	Direction of vector	Sign of $A_x$	Sign of $A_y$
I	right and up	positive	positive
II	left and up	negative	positive
III	left and down	negative	negative
IV	right and down	positive	negative

For example, let

$$A=20\mathrm{N}$$

and

$$\theta =150^{\circ }.$$

Because $150^{\circ }$ is in quadrant II,

$$A_x=(20\mathrm{N})\cos (150^{\circ })\approx -17.3\mathrm{N},$$ $$A_y=(20\mathrm{N})\sin (150^{\circ })=10.0\mathrm{N}.$$

So

$$\vec{A}\approx ⟨-17.3,10.0⟩\mathrm{N}.$$

The negative horizontal component says the force points westward; the positive vertical component says it points northward.

Compass directions: translate the words before calculating

Physics problems often describe direction in forms such as “$30^{\circ }$ north of west” rather than giving a standard angle. Interpret the phrase from the first named direction, turning toward the second.

For a $20\mathrm{N}$ vector directed $30^{\circ }$ north of west:

1. Start pointing west, so the horizontal component must be negative.
2. Turn $30^{\circ }$ toward north, so the vertical component must be positive.
3. The $30^{\circ }$ angle is measured from the horizontal westward direction.

The horizontal and vertical component magnitudes are therefore

$$|A_x|=(20\mathrm{N})\cos (30^{\circ }),$$ $$|A_y|=(20\mathrm{N})\sin (30^{\circ }).$$

Now attach signs from the direction:

$$A_x=-(20\mathrm{N})\cos (30^{\circ })\approx -17.3\mathrm{N},$$ $$A_y=+(20\mathrm{N})\sin (30^{\circ })=10.0\mathrm{N}.$$

This agrees with the standard-angle method, since $30^{\circ }$ north of west is equivalent to a standard angle of $150^{\circ }$.

A brief sketch before calculation is the most reliable way to avoid accidentally assigning the wrong sign or interchanging sine and cosine.

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Components to magnitude and direction

Now take the reverse problem. Suppose we know

$$\vec{F}=⟨F_x,F_y⟩.$$

Because the components are perpendicular, the Pythagorean theorem gives the vector magnitude:

$$F=\sqrt{F_x^2+F_y^2}.$$

Notice that a negative component becomes positive when squared. Magnitude cannot be negative.

To recover direction, the right triangle gives

$$\tan {\theta }_{\text{reference}}=\frac{|F_y|}{|F_x|}.$$

Thus the reference angle, the positive acute angle made with the horizontal, is

$${\theta }_{\text{reference}}={\tan }^{-1}\left(\frac{|F_y|}{|F_x|}\right).$$

The absolute values here deliberately give only the triangle’s acute angle. You then use the component signs to place the vector in the correct quadrant.

Example: components to magnitude and direction

Let

$$\vec{v}=⟨-3,4⟩m/s.$$

First find the magnitude:

$$v=\sqrt{(-3m/s{)}^2+(4m/s{)}^2}.$$ $$v=5m/s.$$

The component signs show that the vector points left and up, so it lies in quadrant II. The reference angle is

$${\theta }_{\text{reference}}={\tan }^{-1}\left(\frac{4}{3}\right)\approx 53.1^{\circ }.$$

There are two clear ways to state the direction:

• Compass-style direction: $53.1^{\circ }$ north of west.
• Standard direction angle: start from the positive $x$-axis and move counterclockwise:

$$\theta =180^{\circ }-53.1^{\circ }=126.9^{\circ }.$$

Thus the vector can be written in magnitude-direction form as either

$$5m/s,53.1^{\circ }\text{ north of west},$$

or

$$5m/s,126.9^{\circ }\text{ counterclockwise from the positive }x\text{-axis}.$$

Both descriptions represent exactly the same vector.

Why inverse tangent alone can be misleading

If you enter

$${\tan }^{-1}\left(\frac{4}{-3}\right)$$

into a basic calculator, it may return approximately

$$-53.1^{\circ }.$$

But this angle lies in quadrant IV, whereas the components $⟨-3,4⟩$ clearly locate the vector in quadrant II. Tangent has the same value at angles separated by $180^{\circ }$, so inverse tangent by itself cannot always choose the right quadrant.

A reliable hand method is:

1. Determine the quadrant from the signs of the components.
2. Find the positive reference angle using component magnitudes.
3. Convert that reference angle into the requested direction.

Many scientific calculators and programming languages provide an atan2 function. It uses the ordered inputs $y,x$ and automatically accounts for the quadrant:

$$\theta =atan2(A_y,A_x).$$

For the vector $⟨-3,4⟩$, atan2(4,-3) returns $126.9^{\circ }$, or the equivalent angle in radians if the calculator is in radian mode.

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A complete conversion routine

Use this compact routine whenever you encounter a two-dimensional vector.

From magnitude and direction to components

Given magnitude $A$ and standard angle $\theta$,

$$\vec{A}=⟨A\cos \theta ,A\sin \theta ⟩.$$

Before accepting the result, check:

• Does the quadrant predict the signs you obtained?
• Is each component’s magnitude no greater than $A$?
• Are the units on both components the same as the original vector’s units?

From components to magnitude and direction

Given

$$\vec{A}=⟨A_x,A_y⟩,$$

calculate

$$A=\sqrt{A_x^2+A_y^2}.$$

Then determine the direction using either atan2 or a reference angle plus a quadrant check.

Finally, verify by converting back:

$$A_x\stackrel{?}{=}A\cos \theta ,$$ $$A_y\stackrel{?}{=}A\sin \theta .$$

This reverse check is especially useful when the vector lies in quadrants II, III, or IV.

Vectors drawn away from the origin

If a displacement arrow begins at

$$(x_i,y_i)$$

and ends at

$$(x_f,y_f),$$

its components are changes in coordinates:

$$\Delta x=x_f-x_i,$$ $$\Delta y=y_f-y_i.$$

Therefore,

$$\Delta \vec{r}=⟨x_f-x_i,y_f-y_i⟩.$$

For example, a particle moving from

$$(2,1)\mathrm{m}$$

to

$$(-4,5)\mathrm{m}$$

has displacement components

$$\Delta \vec{r}=⟨-4-2,5-1⟩\mathrm{m},$$ $$\Delta \vec{r}=⟨-6,4⟩\mathrm{m}.$$

The negative $x$-component means $6\mathrm{m}$ left; the positive $y$-component means $4\mathrm{m}$ up. From there, the same Pythagorean and trigonometric methods give its magnitude and direction.

There are two special cases worth recognizing:

• If $A_x=0$ and $A_y\ne 0$, the vector is vertical, with direction $90^{\circ }$ or $270^{\circ }$.
• The zero vector,

$$⟨0,0⟩,$$

has magnitude zero and no defined direction.

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Wrap-up

A two-dimensional vector can be represented equally well in either form:

$$\text{magnitude-direction}$$

or

$$\text{Cartesian components}.$$

The essential formulas are

$$A_x=A\cos \theta ,$$ $$A_y=A\sin \theta ,$$

and, in reverse,

$$A=\sqrt{A_x^2+A_y^2}.$$

To find direction from components, use the signs to identify the quadrant and then use inverse tangent carefully, or use atan2 when available. A component is a signed horizontal or vertical change, not merely a coordinate.

Next, you will use this component form to add and subtract vectors. That is the reason components are so powerful: once every vector is expressed along the same two perpendicular axes, combining physical effects becomes ordinary algebra.