Adding Vectors

Picture a swimmer crossing a river. She swims straight across, aiming for the point directly opposite her on the far bank. But the river's current is quietly pushing her downstream the whole time. She isn't doing one thing or the other — she's doing both at once. Her actual path through the water is neither her swimming alone nor the current alone, but something new: a single combined push that is the sum of the two.

That combining process is vector addition, and it works the same way whether you're adding a swimmer's effort to a current, two forces pulling on a hook, or two gusts of wind. Walk along \vec{u}, and from wherever you end up, walk along \vec{v}. The single arrow from your start to your finish is the sum \vec{u} + \vec{v}. This is the tip-to-tail rule: slide \vec{v} so its tail sits on the tip of \vec{u}, and join up the ends.

In components it could not be simpler — just add the matching entries:

\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}.

Tip to tail

Below, \vec{u} is drawn from the origin, then \vec{v} is carried to its tip. The bold arrow back to the origin is the sum. Drag the components and confirm the rule: the sum's entries are just the entries added.

Order doesn't matter — and here's the second picture

Walking \vec{u} then \vec{v} lands you in exactly the same place as \vec{v} then \vec{u} — the two routes are the two sides of a parallelogram with the sum as its diagonal. So vector addition is commutative, just like adding ordinary numbers:

\vec{u} + \vec{v} = \vec{v} + \vec{u}.

The figure below draws both routes at once on the very same pair of vectors: go \vec{u} then \vec{v} along the bottom and right sides of the parallelogram, or \vec{v} then \vec{u} along the left and top sides. Either way you arrive at exactly the same corner — the diagonal is the one true sum, seen two ways.

Vector addition is associative too, and adding the \vec{0} vector changes nothing — the same friendly rules arithmetic already taught you, now working on arrows.

Worked example: adding by components

Add (3, 1) and (2, 4). There's no picture needed at all — just line up the matching components and add:

(3, 1) + (2, 4) = (3 + 2,\ 1 + 4) = (5, 5).

That's the whole method: horizontal with horizontal, vertical with vertical. If you like, sketch it tip-to-tail as a check — an arrow to (3,1), then a second arrow added on from there reaching 2 further across and 4 further up, should land you exactly on (5, 5).

Worked example: crossing a river

Back to our swimmer. Relative to the still water, she swims straight across the river at 2 m/s — call "across" the y-direction, so her swimming velocity is \vec{s} = (0, 2). The current flows downstream at 1.5 m/s — call "downstream" the x-direction, so the current's velocity is \vec{c} = (1.5, 0).

Her actual velocity over the ground is the sum of the two:

\vec{s} + \vec{c} = (0 + 1.5,\ 2 + 0) = (1.5, 2).

How fast is she really moving, and in what direction? The magnitude comes from Pythagoras, and the direction from the arctangent of "sideways over across":

|\vec{s}+\vec{c}| = \sqrt{1.5^2 + 2^2} = \sqrt{6.25} = 2.5 \text{ m/s}, \qquad \theta = \arctan\!\left(\frac{1.5}{2}\right) \approx 36.9^\circ.

So although she is only trying to swim straight across, her true path over the ground is 2.5 m/s, angled about 37^\circ downstream of straight across. If she wants to actually land directly opposite her starting point, she has to aim upstream of it to compensate — exactly the correction pilots and sailors make every day.

Worked example: three legs of a journey

Vector addition doesn't stop at two vectors — you can chain as many as you like, adding one component at a time. Suppose a hiker walks three legs of a journey, measured in kilometres east (x) and north (y):

Add all three, component by component, to find the net displacement — the single straight-line arrow from start to finish:

(2+1+3,\ 3-1+2) = (6, 4).

However winding the walk, the hiker ends up exactly where a single arrow of (6, 4) would have taken her — about \sqrt{6^2+4^2} = \sqrt{52} \approx 7.2 km from where she began, in a straight line she never actually walked.

Three pictures, one idea

By now you've seen vector addition drawn three different ways, and it's worth holding all three in your head together, because each one is the right tool for a different job:

They always agree, because they're really the same fact seen from three angles. When a picture gets confusing, drop down to components and add; when a number feels abstract, sketch it tip-to-tail and watch it happen.

It's a very natural mistake to think that if \vec{u} has length 5 and \vec{v} has length 3, then \vec{u} + \vec{v} must have length 8. It almost never does. Lengths only add directly when both vectors point in exactly the same direction; point them differently and the combined length shrinks, sometimes a lot (two vectors pointing in exactly opposite directions can even cancel to length zero).

The safe method is always the same: add the components first, then find the length of whatever comes out — |\vec{u}+\vec{v}| = \sqrt{(u_x+v_x)^2 + (u_y+v_y)^2}, never |\vec{u}| + |\vec{v}|.

The second trap is in the drawing itself: tip-to-tail only works when the tail of \vec{v} sits exactly on the tip (the arrowhead) of \vec{u} — nose to tail, like train carriages. Drawing them overlapping, or both starting from the origin, does not give you the sum; it just gives you two separate arrows.

Every student pilot learns to draw what's nicknamed the wind triangle — three vectors added tip-to-tail: the plane's heading through the air, the wind's velocity, and the plane's actual path over the ground. If a crosswind blows from the side, a pilot who simply points the nose where she wants to go will drift sideways and miss the airport entirely.

The fix is pure vector addition, done in reverse: knowing the desired ground path and the wind, she works out exactly how far into the wind to angle the nose so that her heading vector plus the wind vector adds up to the straight path she actually wants. It's the very same river-crossing correction from the worked example above, just with an aeroplane instead of a swimmer — and it's flown, precisely, thousands of times a day.

A tug-of-war rope shows the idea running the other way. Two teams each pull with their own force vector; the rope stays put exactly when those two vectors add to the \vec{0} vector — equal in size, opposite in direction, cancelling perfectly. The moment one side's vector sum edges away from zero, the rope starts to move.

See it explained