3.2 Unit Vectors: How to Add Vectors by Components

If you can add vectors only by drawing triangles, every force, velocity, and field problem becomes slow and error-prone. Unit vectors and component addition turn geometry into algebra and give you the speed AP, A-Level, IB and IPhO problems demand.

Key Insights

• Unit vectors \(\hat{i}, \hat{j}, \hat{k}\) have magnitude 1 and point along +x, +y, +z in a right-handed system.
• Any vector can be written as \(\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\) with vector components \(a_x\hat{i}\) and scalar components \(a_x\).
• Vector addition is done axis by axis: \(r_x=a_x+b_x\), \(r_y=a_y+b_y\), \(r_z=a_z+b_z\).
• Rotating the coordinate system changes \(a_x, a_y\) but never changes the vector \(\vec{a}\) itself.
• One vector equation \(\vec{r}=\vec{a}+\vec{b}\) represents three scalar equations and works for subtraction as \(\vec{d}=\vec{a}-\vec{b}\).

Unit Vectors

A unit vector is a vector that has a magnitude of exactly 1 and points in a particular direction. It lacks both dimension and unit. Its sole purpose is to point — that is, to specify a direction. The unit vectors in the positive directions of the \(x, y,\) and \(z\) axes are labeled \(\hat{i}, \hat{j},\) and \(\hat{k}\), where the hat \(^\wedge\) is used instead of an overhead arrow as for other vectors.

The arrangement of axes is said to be a right-handed coordinate system. The system remains right-handed if it is rotated rigidly. We use such coordinate systems exclusively in this book.

Unit vectors are very useful for expressing other vectors; for example, we can express \(\vec{a}\) and \(\vec{b}\) as

\(\vec{a}=a_x\hat{i}+a_y\hat{j}\)

\(\vec{b}=b_x\hat{i}+b_y\hat{j}\)

Right-handed coordinate system with unit vectors i along x, j along y, k along z, and vector a decomposed into x vector component ax i and y vector component ay j forming rectangle

These two equations are illustrated in the typical diagrams for vector components. The quantities \(a_x\hat{i}\) and \(a_y\hat{j}\) are vectors, called the vector components of \(\vec{a}\). The quantities \(a_x\) and \(a_y\) are scalars, called the scalar components of \(\vec{a}\) (or, as before, simply its components).

In three dimensions we extend this to:

\(\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\)

in which \(a_x\hat{i}, a_y\hat{j},\) and \(a_z\hat{k}\) are the vector components of \(\vec{a}\) and \(a_x, a_y, a_z\) are its scalar components. The hat notation tells you: this vector only carries direction information, the scalar out front carries magnitude and sign.

Vector Component vs Scalar Component:\(a_x\hat{i}\) is a vector along \(x\) with magnitude \(|a_x|\). \(a_x\) is a number (with unit) that can be positive or negative. Negative means the vector component points opposite to \(\hat{i}\). Mixing them up is the most common algebra error in AP Physics C.

Term	Symbol	What It Is	Example
Unit vector	\(\hat{i}, \hat{j}, \hat{k}\)	Magnitude 1, direction +x,+y,+z	\(\hat{i}\) points +x
Vector component	\(a_x\hat{i}\)	Vector along one axis	\(4.2\,\text{m}\,\hat{i}\)
Scalar component	\(a_x\)	Signed number, projection on axis	\(4.2\,\text{m}\)

Exam Tip:Write every vector you are given in \(\hat{i},\hat{j}\) form immediately. A magnitude-angle like \(5\,\text{m}\) at \(30^{\circ}\) becomes \(5\cos30\hat{i}+5\sin30\hat{j}\). It prevents head-to-tail drawing errors.

Exam Trap:\(\hat{i}, \hat{j}, \hat{k}\) have no units. The units live on the scalar components. So \(\vec{a}=(4.2\,\text{m})\hat{i}+(-1.5\,\text{m})\hat{j}\) is correct, but \((4.2\hat{i})\,\text{m}\) is sloppy and can lose marks in IB HL.

Adding Vectors by Components

We can add vectors geometrically on a sketch or directly on a vector-capable calculator. A third way is to combine their components axis by axis. This is the method you will use in every A-Level, AP, and IPhO solution because it is exact and fast.

To start, consider the statement

\(\vec{r}=\vec{a}+\vec{b}\)

which says that the vector \(\vec{r}\) is the same as the vector \((\vec{a}+\vec{b})\). Thus, each component of \(\vec{r}\) must be the same as the corresponding component of \((\vec{a}+\vec{b})\):

\(r_x = a_x + b_x\)

\(r_y = a_y + b_y\)

\(r_z = a_z + b_z\)

In other words, two vectors must be equal if their corresponding components are equal. The equations tell us that to add vectors \(\vec{a}\) and \(\vec{b}\), we must:

1. Resolve the vectors into their scalar components.
2. Combine these scalar components, axis by axis, to get the components of the sum \(\vec{r}\).
3. Combine the components of \(\vec{r}\) to get \(\vec{r}\) itself. We have a choice in step 3. We can express \(\vec{r}\) in unit-vector notation or in magnitude-angle notation.

This procedure for adding vectors by components also applies to vector subtractions. Recall that a subtraction such as \(\vec{d}=\vec{a}-\vec{b}\) can be rewritten as an addition \(\vec{d}=\vec{a}+(-\vec{b})\). To subtract, we add \(\vec{a}\) and \(-\vec{b}\) by components, to get

\(d_x = a_x - b_x,\quad d_y = a_y - b_y,\quad d_z = a_z - b_z\)

where

\(\vec{d}=d_x\hat{i}+d_y\hat{j}+d_z\hat{k}\)

So subtraction is just adding the opposite: flip the sign of every scalar component of the vector being subtracted.

Three vectors a, b, c on x-y grid with projections ax, ay etc added to give resultant r with rx=2.6m and ry=-2.3m arranged head-to-tail showing r = rx i + ry j

Checkpoint Analysis:Given two vectors \(\vec{d}_1\) pointing mostly up-right and \(\vec{d}_2\) pointing down-right from same origin: \(d_{1x}\) is positive, \(d_{2x}\) is positive, both x components point +x. \(d_{1y}\) is positive, \(d_{2y}\) is negative. For \(\vec{d}_1+\vec{d}_2\), x components add positive+positive=positive, so \((\vec{d}_1+\vec{d}_2)_x\) is positive; y is positive+negative and magnitude of negative may dominate, so sign depends but in typical textbook diagram net y is still positive small or near zero. Always check signs before calculating.

Exam Tip:When adding three or more vectors, make a table: columns \(a_x, b_x, c_x\) then sum. Do x separately, y separately. It prevents sign errors and is the expected working for A-Level and IB HL.

Exam Trap:Students forget \(c_x=0\) when \(\vec{c}=-3.7\,\text{m}\,\hat{j}\). Zero is a component too. If you omit it, \(r_x\) is wrong. Write \(0\) explicitly.

Worked Example: Adding Vectors by Components

The following three vectors are given:

\(\begin{aligned} \vec{a} &= (4.2\,\text{m})\hat{i}+(-1.5\,\text{m})\hat{j} \\ \vec{b} &= (-1.6\,\text{m})\hat{i}+(2.9\,\text{m})\hat{j} \\ \vec{c} &= (-3.7\,\text{m})\hat{j} \end{aligned}\)

What is their vector sum \(\vec{r}\)?

To add these vectors, find their net \(x\) component and their net \(y\) component.

For the \(x\) axis, we add the \(x\) components of \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), to get the \(x\) component of the vector sum \(\vec{r}\):

\(r_x = a_x + b_x + c_x = 4.2\,\text{m} -1.6\,\text{m} + 0 = 2.6\,\text{m}\)

Similarly, for the \(y\) axis:

\(r_y = a_y + b_y + c_y = -1.5\,\text{m} +2.9\,\text{m} -3.7\,\text{m} = -2.3\,\text{m}\)

We then combine these components of \(\vec{r}\) to write the vector in unit-vector notation:

\(\vec{r}=(2.6\,\text{m})\hat{i}-(2.3\,\text{m})\hat{j}\)

where \((2.6\,\text{m})\hat{i}\) is the vector component of \(\vec{r}\) along the \(x\) axis and \((-2.3\,\text{m})\hat{j}\) is that along the \(y\) axis. One way to arrange these vector components head-to-tail forms \(\vec{r}\). You can also sketch the other order.

We can also answer the question by giving the magnitude and an angle for \(\vec{r}\). From the magnitude-angle relations, the magnitude is

\(r = \sqrt{(2.6\,\text{m})^2 + (-2.3\,\text{m})^2} \approx 3.5\,\text{m}

and the angle, measured from the \(+x\) direction, is

\(\theta = \tan^{-1}\left(\dfrac{-2.3\,\text{m}}{2.6\,\text{m}}\right) = -41^{\circ}\)

where the minus sign means clockwise from \(+x\), which is consistent with a negative \(y\) component.

Operation	\(x\) Rule	\(y\) Rule	Unit-Vector Result
Addition \(\vec{r}=\vec{a}+\vec{b}\)	\(r_x=a_x+b_x\)	\(r_y=a_y+b_y\)	\(r_x\hat{i}+r_y\hat{j}\)
Subtraction \(\vec{d}=\vec{a}-\vec{b}\)	\(d_x=a_x-b_x\)	\(d_y=a_y-b_y\)	\(d_x\hat{i}+d_y\hat{j}\)
Zero check	\(0 = a_x + (-a_x)\)	\(0 = a_y + (-a_y)\)	\(\vec{0}\)

Vectors and the Laws of Physics

So far, in every figure that includes a coordinate system, the \(x\) and \(y\) axes are parallel to the edges of the page. Thus, when a vector \(\vec{a}\) is included, its components \(a_x\) and \(a_y\) are also parallel to the edges. The only reason for that orientation of the axes is that it looks proper; there is no deeper reason.

We could, instead, rotate the axes (but not the vector \(\vec{a}\)) through an angle \(\phi\), in which case the components would have new values, call them \(a'_x\) and \(a'_y\). Since there are an infinite number of choices of \(\phi\), there are an infinite number of different pairs of components for \(\vec{a}\).

Which then is the right pair of components? The answer is that they all are equally valid because each pair, with its axes, just gives us a different way of describing the same vector \(\vec{a}\); all produce the same magnitude and direction for the vector. In that rotated system we have

\(a = \sqrt{a_x^2 + a_y^2} = \sqrt{{a'_x}^2 + {a'_y}^2}\)

and

\(\theta = \theta' + \phi\)

Same vector a in standard x-y system with components ax, ay and angle theta, and in rotated x'-y' system rotated by phi showing new components a'x', a'y' and angles theta' with relation theta = theta' + phi and magnitude invariance

The point is that we have great freedom in choosing a coordinate system, because the relations among vectors do not depend on the location of the origin or on the orientation of the axes. This is also true of the relations of physics; they are all independent of the choice of coordinate system. Add to that the simplicity and richness of the language of vectors and you can see why the laws of physics are almost always presented in that language: One equation, like \(\vec{r}=\vec{a}+\vec{b}\), can represent three (or even more) relations, like \(r_x=a_x+b_x\), \(r_y=a_y+b_y\), and \(r_z=a_z+b_z\).

Why Physics Uses Vectors:A single vector equation \(\vec{F}_{net}=m\vec{a}\) encodes three independent scalar equations in 3D. Nature does not care how you oriented your paper. The law is invariant; only its component description changes.

Exam Tip:For projectile problems, choose axes along the slope or horizontal/vertical whichever makes one acceleration component zero. You are free to rotate. Just state your coordinate choice and stick to it. Examiners award marks for clever axis choice.

Exam Trap:After rotating axes, students think magnitude changed because \(a'_x \neq a_x\). It did not. Magnitude is invariant: \(a\) is same in all frames. Only components change. Always check \(a = \sqrt{{a'_x}^2+{a'_y}^2}\) to verify.

Practice Questions

MCQs (Multiple Choice Questions)

01. Unit vector \(\hat{i}\) has magnitude:

1. 0
2. 1
3. x
4. variable

Correct: B
Explanation: By definition unit vector magnitude exactly 1, dimensionless, only direction.

02. \(\vec{a}=a_x\hat{i}+a_y\hat{j}\). \(a_x\hat{i}\) is called:

1. Scalar component
2. Unit vector
3. Vector component
4. Magnitude

Correct: C
Explanation: \(a_x\hat{i}\) is vector along x, \(a_x\) alone is scalar component.

03. Right-handed system means:

1. \(x\) cross \(y\) gives \(z\)
2. Left-handed after rotation
3. \(\hat{i}\) opposite \(\hat{j}\)
4. 2D only

Correct: A
Explanation: \(\hat{i}\times\hat{j}=\hat{k}\), rotation keeps handedness if rigid.

04. \(\vec{r}=\vec{a}+\vec{b}\). Then \(r_x=\)

1. \(a_x b_x\)
2. \(a_x+b_x\)
3. \(a_y+b_y\)
4. \(a_x-b_x\)

Correct: B
Explanation: Add components axis by axis.

05. To subtract \(\vec{b}\) from \(\vec{a}\), you:

1. Add components \(a_x+b_x\)
2. Add \(a_x-b_x\) as \(a_x+(-b_x)\)
3. Multiply
4. Cannot subtract vectors

Correct: B
Explanation: \(\vec{a}-\vec{b}=\vec{a}+(-\vec{b})\), so \(d_x=a_x-b_x\).

06. Given \(\vec{a}=4.2\hat{i}-1.5\hat{j}\), \(\vec{b}=-1.6\hat{i}+2.9\hat{j}\), \(\vec{c}=-3.7\hat{j}\). \(r_x=\)

1. \(2.6\)
2. \(1.0\)
3. \(-2.3\)
4. \(3.5\)

Correct: A
Explanation: \(4.2-1.6+0=2.6\).

7. Same vectors, \(r_y=\)

1. \(2.6\)
2. \(-2.3\)
3. \(0\)
4. \(-1.5\)

Correct: B
Explanation: \(-1.5+2.9-3.7=-2.3\).

08. Rotating coordinate axes by \(\phi\) (not vector):

1. Changes vector magnitude
2. Changes components but not vector
3. Destroys right-handedness
4. Changes physics law

Correct: B
Explanation: Vector itself unchanged, description changes. Laws independent of coordinate choice.

09. Magnitude invariance after rotation is expressed as:

1. \(a=\sqrt{a_x^2+a_y^2}=\sqrt{{a'_x}^2+{a'_y}^2}\)
2. \(a_x=a'_x\)
3. \(\theta=\theta'\)
4. \(a=0\)

Correct: A
Explanation: Magnitude same in both systems.

10. Angle relation after rotation by \(\phi\) is:

1. \(\theta=\theta'+\phi\)
2. \(\theta=\theta'-\phi^2\)
3. \(\theta=\phi\)
4. \(\theta=0\)

Correct: A
Explanation: Original angle from original x equals new angle plus rotation.

FRQs (Free Response Questions)

FRQ 1: (AP/IB)
Convert \(\vec{a}\) with magnitude 10 m at \(60^{\circ}\) above +x to unit-vector notation, then add \(\vec{b}=-2\hat{i}+3\hat{j}\).

Rubric:

• \(a_x=10\cos60=5\), \(a_y=10\sin60=8.66\) (2 pts)
• \(\vec{a}=5\hat{i}+8.66\hat{j}\) (1 pt)
• \(\vec{r}=(5-2)\hat{i}+(8.66+3)\hat{j}=3\hat{i}+11.66\hat{j}\) (2 pts)

FRQ 2: (A-Level)
\(\vec{d}_1\) has positive x and y, \(\vec{d}_2\) has positive x negative y. Determine signs of \((\vec{d}_1+\vec{d}_2)_x\) and \((\vec{d}_1+\vec{d}_2)_y\) and explain.

Rubric:

• \(x\): pos+pos=pos (2 pts)
• \(y\): pos+neg sign depends on magnitudes, could be pos, neg, zero (2 pts)
• must add scalar components (1 pt)

FRQ 3: (College)
Given \(\vec{a}= (4.2\hat{i}-1.5\hat{j})\,\text{m}\), \(\vec{b}=(-1.6\hat{i}+2.9\hat{j})\,\text{m}\), \(\vec{c}=-3.7\hat{j}\,\text{m}\), find \(\vec{r}\) magnitude and direction. Show alternative arrangement head-to-tail.

Rubric:

• \(r_x=2.6\), \(r_y=-2.3\) (2 pts)
• \(\vec{r}=2.6\hat{i}-2.3\hat{j}\) (1 pt)
• \(r=3.5\,\text{m}\) (1 pt)
• \(\theta=-41^{\circ}\) clockwise from +x (1 pt)

FRQ 4: (USAPhO)
Prove that \(\vec{r}=\vec{a}+\vec{b}\) is equivalent to three scalar equations and explain why this shows physics laws are coordinate-independent.

Rubric:

• Component equality \(r_x=a_x+b_x\) etc (2 pts)
• three relations (1 pt)
• rotating axes changes components but vector relation holds (2 pts)
• freedom of coordinate choice (1 pt)

FRQ 5: (IB HL extension)
Vector \(\vec{a}\) has components \((3,4)\) in standard frame. Frame rotated \(30^{\circ}\) CCW. Find new components \(a'_x, a'_y\) using \(\theta=\theta'+\phi\).

Rubric:

• Original \(\theta=\tan^{-1}4/3=53.13^{\circ}\) (1 pt)
• \(\theta'=\theta-\phi=23.13^{\circ}\) (2 pts)
• \(a'_x=5\cos23.13=4.60\), \(a'_y=5\sin23.13=1.96\) (2 pts)

Frequently Asked Questions

What is a unit vector?

A vector with magnitude exactly 1, no units or dimensions, that only indicates direction. \(\hat{i}, \hat{j}, \hat{k}\) point along +x, +y, +z in a right-handed system and are used to build other vectors.

How do you write a vector in unit-vector notation?

\(\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\) where \(a_x,a_y,a_z\) are scalar components (signed numbers) and \(a_x\hat{i}\) is the x vector component.

How do you add vectors by components?

Add x components together, y together, z together: \(r_x=a_x+b_x\), \(r_y=a_y+b_y\), \(r_z=a_z+b_z\). Then \(\vec{r}=r_x\hat{i}+r_y\hat{j}+r_z\hat{k}\).

What is the difference between \(a_x\) and \(a_x\hat{i}\)?

No. The vector stays same; only its scalar components change to \(a'_x, a'_y\). Magnitude stays invariant: \(a=\sqrt{a_x^2+a_y^2}=\sqrt{{a'_x}^2+{a'_y}^2}\).

What is a right-handed coordinate system?

A system where \(\hat{i}\times\hat{j}=\hat{k}\) and orientation stays right-handed if rotated rigidly. Physics uses right-handed systems exclusively to keep cross product signs consistent.

Why are vector laws independent of coordinate choice?

Because vectors represent physical quantities themselves, not their description. Relations like \(\vec{r}=\vec{a}+\vec{b}\) hold regardless of origin location or axis orientation; only component numbers change.

Exam Mapping

Concept	A-Level Physics	AP Physics 1, 2 & C	IB Physics HL	College Physics 1 & 2	USAPhO/IPhO
Unit vectors \(\hat{i},\hat{j},\hat{k}\), magnitude 1	Vector notation, forces	Unit-vector notation in C	Topic 1.3, HL vector notation	Ch 3 Vectors	Fundamental tool
Writing \(\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\)	Resolving forces	Converting magnitude-angle to components	Component form	Vector components	Required
Vector vs scalar component distinction	Distinguish	Distinguish in free-body diagrams	Scalar projection vs vector component	Component definitions	Conceptual check
Adding by components \(r_x=a_x+b_x\)	Resultant force, equilibrium	Net force, displacement	Adding velocities/forces	Lab data addition	Core method
Subtraction \(d_x=a_x-b_x\)	Relative velocity	Relative motion	Relative velocity	Subtraction	Relative vectors
Right-handed system, rigid rotation	Coordinate choice	Coordinate independence	Coordinate systems	Right-hand rule intro	Invariance
Rotating axes changes components not vector	Choice of axes along plane	Choosing clever axes	Invariance of physical laws	Freedom of coordinates	Symmetry/invariance
Sample problem \(r=(2.6\hat{i}-2.3\hat{j})\,\text{m}\), \(r\approx3.5\,\text{m}\), \(\theta=-41^{\circ}\)	Numerical resultant	Magnitude-angle conversion	Magnitude and direction	Calculation	Computational
\(a=\sqrt{a_x^2+a_y^2}\), \(\theta=\theta'+\phi\)	Magnitude invariant	Vector magnitude	Magnitude and angle relation	Pythagorean + angle addition	Formal

Conclusion

You now own the algebraic language of vectors. Unit vectors \(\hat{i}, \hat{j}, \hat{k}\) give every vector a clean form \(\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\) where \(a_x\hat{i}\) is the vector component and \(a_x\) the scalar component. Adding and subtracting reduces to \(r_x=a_x+b_x\), \(r_y=a_y+b_y\), \(d_x=a_x-b_x\), and you can present the answer either as \(\vec{r}=(2.6\,\text{m})\hat{i}-(2.3\,\text{m})\hat{j}\) or as \(3.5\,\text{m}\) at \(-41^{\circ}\) from \(+x\). Most powerfully, you see that rotating the coordinate system by \(\phi\) gives new components \(a'_x, a'_y\) but the same vector, with \(a=\sqrt{a_x^2+a_y^2}=\sqrt{{a'_x}^2+{a'_y}^2}\) and \(\theta=\theta'+\phi\). Physics does not depend on how you oriented your paper — one vector equation captures three scalar relations. Practice converting instantly between notations, and two-dimensional problems become one-dimensional bookkeeping.

References

Fundamentals of Physics by Halliday, Resnick, and Walker, 12th Edition, Volume 1, Chapter 3.2 – Unit Vectors, Adding Vectors by Components, Vectors and the Laws of Physics – primary source of truth.