What Is Vector Subtraction?
Vector subtraction is the operation of finding the difference between two vectors, written $\vec{a} - \vec{b}$. It is defined not as a brand-new rule but as addition in disguise: subtracting $\vec{b}$ means adding the negative of $\vec{b}$.
$$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$$
The negative of a vector, $-\vec{b}$, has the same magnitude as $\vec{b}$ but points in the opposite direction. So subtraction reverses the second vector, then adds. That one reframing lets every tool you already have for vector addition carry straight over to subtraction, with a single flip.
Two conditions must hold for the operation to make sense. Both quantities must be vectors, never a vector and a scalar, and they must represent the same physical quantity, two velocities or two forces, never a velocity minus a force. Reversing $\vec{b}$ is exactly a scaling by $-1$, the simplest case of multiplying a vector by a scalar.
What Is the Vector Subtraction Formula?
In component form, subtract like-for-like: x from x, y from y, z from z. For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$:
$$\vec{a} - \vec{b} = (a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j} + (a_3 - b_3)\hat{k}$$
The same rule in column form:
$$\begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix} - \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix} = \begin{bmatrix} a_1 - b_1 \ a_2 - b_2 \ a_3 - b_3 \end{bmatrix}$$
Variable key:
Symbol | Meaning |
|---|---|
$\vec{a}, \vec{b}$ | The two vectors being subtracted |
$a_1, a_2, a_3$ | Components of $\vec{a}$ along $\hat{i}, \hat{j}, \hat{k}$ |
$-\vec{b}$ | The negative of $\vec{b}$: same length, opposite direction |
$\vec{a} - \vec{b}$ | The difference vector, pointing from the tip of $\vec{b}$ to the tip of $\vec{a}$ |
Working in components is the least error-prone method, because it reduces the whole operation to ordinary subtraction, one axis at a time.
How Do You Subtract Vectors Geometrically?
Two drawing methods give the same result, and seeing both is worth it because each makes a different feature obvious.
The triangle (tip-to-tip) method. Draw $\vec{a}$ and $\vec{b}$ from the same starting point. The difference $\vec{a} - \vec{b}$ is the vector drawn from the tip of $\vec{b}$ to the tip of $\vec{a}$. This is the fastest way to see the direction, and it answers the question students ask most: which way does $\vec{a} - \vec{b}$ point? Toward the tip of $\vec{a}$, starting at the tip of $\vec{b}$.
The parallelogram method. Reverse $\vec{b}$ to get $-\vec{b}$, then add $\vec{a}$ and $-\vec{b}$ tip-to-tail. Complete the parallelogram and the diagonal is $\vec{a} - \vec{b}$. This method makes the "subtraction is addition of the negative" definition visible on the page.
Which way does $\vec{a} - \vec{b}$ point? From the tip of $\vec{b}$ to the tip of $\vec{a}$ when both are co-initial. Swap the order and the arrow reverses, which is the geometric face of subtraction not being commutative.
What Are the Properties of Vector Subtraction?
The properties follow from treating subtraction as "add the negative," and the most important one is a property it lacks.
Not commutative. $\vec{a} - \vec{b} \ne \vec{b} - \vec{a}$ in general. In fact $\vec{b} - \vec{a} = -(\vec{a} - \vec{b})$: same length, opposite direction. Order matters.
Not associative in the naive sense. $(\vec{a} - \vec{b}) - \vec{c} \ne \vec{a} - (\vec{b} - \vec{c})$, so brackets change the answer.
Subtracting a vector from itself gives zero. $\vec{a} - \vec{a} = \vec{0}$, the zero vector, which has no direction.
Distributes over scalar multiplication. $k(\vec{a} - \vec{b}) = k\vec{a} - k\vec{b}$ for any scalar $k$.
Magnitude is bounded. The length $|\vec{a} - \vec{b}|$ satisfies the triangle inequality: $\big||\vec{a}| - |\vec{b}|\big| \le |\vec{a} - \vec{b}| \le |\vec{a}| + |\vec{b}|$, so the difference is never longer than the two magnitudes added, nor shorter than their gap.
The non-commutativity is the property that separates vectors from ordinary numbers, and it is where most errors are born.
Examples of Vector Subtraction
The set builds from a plain component subtraction to a relative-velocity problem, with one wrong turn in the middle.
Example 1
Subtract $\vec{b} = 2\hat{i} + \hat{j}$ from $\vec{a} = 5\hat{i} + 4\hat{j}$.
Subtract component by component:
$$\vec{a} - \vec{b} = (5 - 2)\hat{i} + (4 - 1)\hat{j}$$
$$\vec{a} - \vec{b} = 3\hat{i} + 3\hat{j}$$
Final answer: $3\hat{i} + 3\hat{j}$.
Example 2
Given $\vec{a} = \begin{bmatrix} 6 \ 2 \end{bmatrix}$ and $\vec{b} = \begin{bmatrix} 1 \ 5 \end{bmatrix}$, find $\vec{a} - \vec{b}$.
The tempting move, borrowing from ordinary numbers where $6 - 1$ and $1 - 6$ feel interchangeable, is to subtract in whichever order is convenient and write $\vec{b} - \vec{a}$:
$$\begin{bmatrix} 1 - 6 \ 5 - 2 \end{bmatrix} = \begin{bmatrix} -5 \ 3 \end{bmatrix}$$
Check the direction against the picture. $\vec{a} - \vec{b}$ should run from the tip of $\vec{b}$ to the tip of $\vec{a}$; the vector above runs the opposite way. It is $\vec{b} - \vec{a}$, the negative of what we wanted. Subtraction is not commutative, so order cannot be swapped.
Subtract in the correct order:
$$\vec{a} - \vec{b} = \begin{bmatrix} 6 - 1 \ 2 - 5 \end{bmatrix} = \begin{bmatrix} 5 \ -3 \end{bmatrix}$$
Final answer: $\begin{bmatrix} 5 \ -3 \end{bmatrix}$.
Example 3
Find $\vec{a} - \vec{b}$ for $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - 4\hat{k}$.
$$\vec{a} - \vec{b} = (3 - 1)\hat{i} + (-2 - 1)\hat{j} + (1 - (-4))\hat{k}$$
$$\vec{a} - \vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$$
Final answer: $2\hat{i} - 3\hat{j} + 5\hat{k}$.
Example 4
Find the magnitude of $\vec{a} - \vec{b}$ when $\vec{a} - \vec{b} = 3\hat{i} + 4\hat{j}$.
Apply the magnitude formula:
$$|\vec{a} - \vec{b}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Final answer: the magnitude is 5.
Example 5
Two boats leave a dock. Boat A has velocity $\vec{v}_A = 8\hat{i} + 6\hat{j}$ km/h; boat B has $\vec{v}_B = 3\hat{i} + 2\hat{j}$ km/h. Find the velocity of A relative to B.
Relative velocity is $\vec{v}_A - \vec{v}_B$:
$$\vec{v}_A - \vec{v}_B = (8 - 3)\hat{i} + (6 - 2)\hat{j} = 5\hat{i} + 4\hat{j}$$
Final answer: A moves at $5\hat{i} + 4\hat{j}$ km/h relative to B.
Example 6
Given $\vec{a} = 4\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$, find both $\vec{a} - \vec{b}$ and $\vec{b} - \vec{a}$, and state how they relate.
$$\vec{a} - \vec{b} = (4 - 1)\hat{i} + (1 - 1)\hat{j} = 3\hat{i}$$
$$\vec{b} - \vec{a} = (1 - 4)\hat{i} + (1 - 1)\hat{j} = -3\hat{i}$$
They are equal in length and opposite in direction: $\vec{b} - \vec{a} = -(\vec{a} - \vec{b})$.
Final answer: $\vec{a} - \vec{b} = 3\hat{i}$ and $\vec{b} - \vec{a} = -3\hat{i}$, exact opposites.
Where Is Vector Subtraction Used?
"Not its speed over the ground, but its speed relative to me."
Vector subtraction exists because so many real questions are about differences and relative quantities, and those cannot be answered by scalar arithmetic. The moment a quantity has a direction, comparing two of them means subtracting vectors, a framing physicists formalised as they built the graphical methods that OpenStax Physics still teaches today.
Where it does real work:
Relative motion. The velocity of one object as seen from another, $\vec{v}_A - \vec{v}_B$, drives air-traffic separation, radar, and even how a passing train appears to move from a moving car.
Displacement. The straight-line change from position $\vec{r}_1$ to position $\vec{r}_2$ is $\vec{r}_2 - \vec{r}_1$, the backbone of navigation and robotics path planning.
Net force and change in momentum. The change in a quantity, $\Delta\vec{p} = \vec{p}{\text{final}} - \vec{p}{\text{initial}}$, is a subtraction, and it is what impulse and Newton's second law are built on.
What Are the Most Common Mistakes With Vector Subtraction?
Mistake 1: Treating subtraction as commutative
Where it slips in: any problem where the order of the two vectors is not read carefully.
Don't do this: computing $\vec{b} - \vec{a}$ when the question asks for $\vec{a} - \vec{b}$.
The correct way: keep the order the question gives, since $\vec{a} - \vec{b} = -(\vec{b} - \vec{a})$. The first instinct, carried over from ordinary numbers, is that order does not matter; with vectors it flips the whole direction. Read which vector is the "start" and which is being taken away before subtracting.
Mistake 2: Adding components instead of subtracting the second
Where it slips in: rushing the component step, especially with negative components.
Don't do this: writing $(a_1 + b_1)\hat{i}$ for the difference.
The correct way: subtract the second vector's components, $(a_1 - b_1)$. When $b_2$ is already negative, subtracting it adds, so $a_2 - (-4) = a_2 + 4$. The rusher who skips the sign bookkeeping lands here most often.
Mistake 3: Getting the geometric direction backwards
Where it slips in: the triangle method, when the arrow is drawn from $\vec{a}$ to $\vec{b}$ instead of the reverse.
Don't do this: drawing $\vec{a} - \vec{b}$ from the tip of $\vec{a}$ to the tip of $\vec{b}$.
The correct way: $\vec{a} - \vec{b}$ runs from the tip of $\vec{b}$ to the tip of $\vec{a}$. A quick way to remember it: the arrow points toward the vector that stays positive, $\vec{a}$.
Conclusion
Vector subtraction is defined as $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$, adding the reversed second vector.
In components, subtract axis by axis; geometrically, $\vec{a} - \vec{b}$ runs from the tip of $\vec{b}$ to the tip of $\vec{a}$.
The triangle and parallelogram methods give the same difference vector two ways.
Vector subtraction is not commutative: $\vec{b} - \vec{a} = -(\vec{a} - \vec{b})$.
It powers relative velocity, displacement, and change-in-momentum calculations.
To go deeper into vectors with a teacher, explore Bhanzu's geometry tutor or a high school math tutor, or see math tutoring options.
Practice the six examples, then redraw each with the triangle method so the direction becomes something you see rather than memorise. If the order ever feels ambiguous, come back to the tip-of-b-to-tip-of-a rule. To work through vectors with a Bhanzu trainer, book a free demo class.
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