Subtracting Vectors
A light aircraft is cruising due east when a gust of wind shoves it off course. Before the gust
its velocity was one vector; after the gust it's a different vector. What was the gust itself —
as a vector? It's exactly the arrow that would get you from the first velocity to the
second. That "get-you-there" arrow is what vector subtraction computes.
Subtraction is addition with a flipped arrow. The vector
-\vec{v} is just \vec{v} turned around to
point the opposite way, so we define
\vec{u} - \vec{v} = \vec{u} + (-\vec{v}).
In components, subtract the matching entries — no flipping needed once you're working with numbers:
\begin{bmatrix} u_x \\ u_y \end{bmatrix} - \begin{bmatrix} v_x \\ v_y \end{bmatrix} = \begin{bmatrix} u_x - v_x \\ u_y - v_y \end{bmatrix}.
The arrow from one tip to the other
There's a lovely picture for the difference. Draw \vec{u} and
\vec{v} from the same origin. Then
\vec{u} - \vec{v} is the arrow that runs from the tip of
\vec{v} to the tip of \vec{u} —
the displacement that takes you from \vec{v} to
\vec{u}. (Memory hook: it points to the first one named.)
Drag the four sliders below and watch the teal arrow constantly redraw itself so it always
connects the tip of v to the tip of u —
"connect the tips" is the whole rule. Notice it never starts at the origin like
u and v do; it's a free-floating
arrow that just happens to be drawn between their tips for the picture. You could slide that
exact same arrow anywhere in the plane and it would still represent
u - v.
Worked example: plain numbers
Let \vec{u} = (5, 2) and \vec{w} = (3, 4).
Subtract component by component:
\vec{u} - \vec{w} = (5 - 3,\ 2 - 4) = (2, -2).
That's it — no diagram required to get the answer, though drawing it confirms the arrow really
does run from the tip of \vec{w} to the tip of
\vec{u}, pointing right and slightly down.
Worked example: the displacement from A to B
Say point A sits at (2, 1) and point
B sits at (7, 4). "How far, and in what
direction, do I have to walk to get from A to
B?" is exactly the question
\vec{B} - \vec{A} answers:
\vec{B} - \vec{A} = (7 - 2,\ 4 - 1) = (5, 3).
Walk 5 units right and 3 units up, and you land exactly on B. This is
the single most common thing you'll ever do with vectors — finding the displacement between two
positions — and it comes up everywhere from
machine learning
(measuring the error between a prediction and a target) to plotting a course between two
waypoints.
Worked example: how much did the velocity change?
Back to the aircraft. Suppose its velocity before the gust was
\vec{v}_i = (120, 0) km/h (120 east, 0 north — flying due east), and
after the gust it was \vec{v}_f = (90, 40) km/h (slower
eastward, and now drifting north). The change in velocity is
\Delta\vec{v} = \vec{v}_f - \vec{v}_i = (90 - 120,\ 40 - 0) = (-30, 40).
That single vector, (-30, 40), is the gust's fingerprint: it says the
wind shoved the plane 30 km/h backwards (west) and 40 km/h sideways (north). Whatever pushed the
aircraft had to point in that exact direction — \Delta\vec{v} and the
force that caused it always point the same way.
Chaining subtractions around a triangle
Subtraction gets even more useful once you have three points instead of two. Take a
triangle with corners at A = (0, 0),
B = (5, 1) and C = (2, 4). The three sides,
written as displacement vectors, are
\vec{AB} = B - A = (5, 1), \qquad \vec{BC} = C - B = (-3, 3), \qquad \vec{CA} = A - C = (-2, -4).
Add all three side-vectors together and something neat happens:
(5,1) + (-3,3) + (-2,-4) = (0,0) — the zero vector. That's not a
coincidence. Walk all the way around any triangle — or any closed shape at all — and you
end up exactly back where you started, so the displacements must cancel out. This little
fact is the seed of a whole family of
vector geometry
proofs, where subtracting position vectors turns statements about shapes into
statements about arithmetic.
Subtraction undoes addition
Every subtraction fact is secretly an addition fact in disguise, and that's what makes it so
useful for solving problems. If you know that
\vec{u} + \vec{v} = \vec{w}, and you already know
\vec{v} and \vec{w}, you can peel
\vec{v} back off by subtracting it from both sides:
\vec{u} = \vec{w} - \vec{v}.
For example, suppose two pushes combine to give \vec{w} = (9, 7), and
you know one of them was \vec{v} = (4, 5). What was the other push,
\vec{u}?
\vec{u} = \vec{w} - \vec{v} = (9 - 4,\ 7 - 5) = (5, 2).
That's the same (5,2) from our very first worked example — subtraction
is exactly how you'd recover it if it had gone missing from an addition sum. Whenever a vector
equation has one unknown vector added to or from known ones, subtraction is the tool that isolates
it.
A special case: subtracting from zero
What happens if you subtract a vector from the zero vector, \vec{0} = (0,0)?
\vec{0} - \vec{v} = (0 - v_x,\ 0 - v_y) = (-v_x, -v_y) = -\vec{v}.
So negating a vector — the flip we used right at the top of this page — is really just a
subtraction in disguise: it's what you get by subtracting the vector from nothing.
That's a nice sanity check that the two ideas fit together perfectly: subtraction is built from
addition and negation, and negation turns out to be a subtraction from zero.
Arrows didn't always come with algebra attached. In the early 1800s, mathematicians like Hermann
Grassmann and William Rowan Hamilton were independently trying to build a proper arithmetic for
directed quantities, and Hamilton actually invented a whole four-dimensional number system —
quaternions — partly to make sense of subtracting one direction from another in 3D space.
It was too clumsy for everyday use. Later in the century, Josiah Willard Gibbs and Oliver
Heaviside stripped the idea down to the clean, component-by-component vector algebra you're
using on this very page — proof that sometimes the simplest notation wins out over the cleverest
one.
Open a maps app while walking to a destination and watch the "distance remaining" tick down.
Under the hood, the app knows your current position vector
\vec{p} (from GPS) and the destination's position vector
\vec{d}. Every second or so it recomputes
\vec{d} - \vec{p} — the displacement still standing between you and
your goal — and reports its length as "distance remaining" and its direction as the arrow telling
you which way to turn. Take one step, and the whole vector quietly updates again. You'll see
exactly how to turn that arrow into a plain distance number when you meet
the magnitude of a
vector.
Yes — hugely. \vec{u} - \vec{v} and
\vec{v} - \vec{u} are not the same arrow; they point
in exactly opposite directions, because
\vec{v} - \vec{u} = -(\vec{u} - \vec{v}). Flip the order, flip the
answer. Using the numbers from above,
\vec{u} - \vec{w} = (2, -2) but
\vec{w} - \vec{u} = (-2, 2) — same length, opposite way.
The classic mix-up is displacement: "the vector from
A to B" is
\vec{B} - \vec{A}, not
\vec{A} - \vec{B}. It's easy to write the letters in the order you say
them ("A to B" → "A minus B") and get it backwards. Say it as "where I'm going, minus where I
started" instead, and the order falls into place: destination first, start second.
Because it basically is. In the velocity-change example above,
\Delta\vec{v} = \vec{v}_f - \vec{v}_i is a plain vector subtraction —
but physicists give it a starring role. Divide it by how long the gust lasted and you get
acceleration; multiply that by the plane's mass and you get the net force that must have
acted on it. That chain — subtract the velocities, then scale — is exactly
Newton's second law
written in vectors, and it's why the direction of a force can be read straight off a change in
velocity.
Next time an enemy sprints straight at your character in a game, thank vector subtraction. The
game engine knows the player's position \vec{P} and the enemy's
position \vec{E} every single frame. The direction the enemy should
run is simply \vec{P} - \vec{E} — the arrow from the enemy's tip to
the player's tip, recomputed dozens of times a second so the chase stays locked on. Get the order
backwards in the code and the monster confidently sprints away from you instead.