0.3 Adding Vectors: Parallelograms and Boxes

Two displacements performed one after the other are a single displacement. Walk along $\mathbf{u}$ , then along $\mathbf{v}$ : you end up somewhere, and the arrow from where you started to where you ended is the sum $\mathbf{u} + \mathbf{v}$ . Drawn, that is the head-to-tail rule — slide $\mathbf{v}$ so its tail sits on the head of $\mathbf{u}$ , and the sum runs from the tail of $\mathbf{u}$ to the head of $\mathbf{v}$ . Walking $\mathbf{v}$ first and then $\mathbf{u}$ lands you at the same place, so $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ , and the two routes together trace a parallelogram whose sides are copies of $\mathbf{u}$ and $\mathbf{v}$ . The sum is its diagonal from the shared tail.

Parallelograms live in a plane, so the first figure is drawn in one: two vectors on four sliders, their sum, and the parallelogram they span.

Interactive figureThe parallelogram law of vector additionDrag the 4 sliders below the figure.

XYZ Graph · viewer build 5edf91b

The parallelogram spanned by

\mathbf{u} = (a, b)

in red and

\mathbf{v} = (c, d)

in blue. The purple diagonal from the origin is

\mathbf{u} + \mathbf{v}

; the dotted gray diagonal from the tip of

\mathbf{v}

to the tip of

\mathbf{u}

\mathbf{u} - \mathbf{v}

Explore the figure

At the starting values $\mathbf{u} = (3, 1)$ and $\mathbf{v} = (1, 2)$ , read the far corner: it sits at $(4, 3)$ , which is $(3 + 1, \; 1 + 2)$ . Adding the components adds the arrows. As a bonus, the sum has length $\sqrt{16 + 9} = 5$ .
Follow the dashed blue copy of $\mathbf{v}$ that starts at the tip of $\mathbf{u}$ . It ends at the far corner: that is head-to-tail addition. The dashed red copy of $\mathbf{u}$ , starting at the tip of $\mathbf{v}$ , ends at the same corner — the other order, the same sum.
Drag $c$ to $-3$ . Now $\mathbf{v} = (-3, 2)$ and the far corner is at $(0, 3)$ : the parallelogram leans backward and the sum points straight up. A sum can be shorter than either of its parts.
Set $c = -3$ and $d = -1$ , so that $\mathbf{v} = (-3, -1) = -\mathbf{u}$ . The parallelogram flattens to a segment and the purple arrow disappears: $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$ .
Return to the starting values and look at the dotted gray arrow from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ . Head minus tail gives $(3 - 1, \; 1 - 2) = (2, -1)$ , which is $\mathbf{u} - \mathbf{v}$ : the other diagonal of the same parallelogram, and the arrow you would add to $\mathbf{v}$ to reach $\mathbf{u}$ .

Adding by components

The corner did the algebra for you. Sliding $\mathbf{v}$ to start at the head of $\mathbf{u}$ moves its head by exactly the components of $\mathbf{u}$ , so the head of the sum is at $(u_1 + v_1, \; u_2 + v_2)$ . In $\mathbb{R}^n$ the same sentence has $n$ clauses:

$\mathbf{u} + \mathbf{v} = (u_1 + v_1, \; u_2 + v_2, \; \dots, \; u_n + v_n)$

Subtraction is addition of the opposite: $\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v})$ , where $-\mathbf{v}$ has every component negated, so the difference is computed component by component too. Geometrically $\mathbf{u} - \mathbf{v}$ is the arrow from the head of $\mathbf{v}$ to the head of $\mathbf{u}$ — head minus tail once more — which is why it appears as the second diagonal.

Because addition happens one component at a time, it inherits the arithmetic of ordinary numbers. For all vectors $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{w}$ of $\mathbb{R}^n$ :

$\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ , and $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$ , so a sum of several vectors needs no parentheses and no order.
$\mathbf{u} + \mathbf{0} = \mathbf{u}$ , and $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$ .

Each is checked by looking at one component: the $i$ th component of $\mathbf{u} + \mathbf{v}$ is $u_i + v_i$ , which equals $v_i + u_i$ , the $i$ th component of $\mathbf{v} + \mathbf{u}$ . Four more rules arrive in the next section once scalars join in, and the eight together are the rules of the game for the whole book.

Three vectors, one diagonal

Head-to-tail addition is not a plane fact — it works for arrows anywhere in space, and it works for more than two arrows. The second figure adds three: $\mathbf{u} = (2, 1, 0)$ and $\mathbf{v} = (-1, 2, 0)$ lie in the floor, and $\mathbf{w} = (1, 1, h)$ climbs by a slider. Three vectors that are not all in one plane span a parallelepiped, a box with parallelogram faces, and their sum is its long diagonal.

A slanted box, a parallelepiped, standing on a horizontal grid with one corner at the origin. Three thick shaded arrows leave the origin: one, u, to (2, 1, 0) and another, v, to (-1, 2, 0), both flat on the floor and at right angles to each other, and a third, w, rising to (1, 1, h), where h is set by a slider that starts at 2. The three faces that meet at the origin are shaded translucent, and the remaining nine edges are thin gray lines. A thinner copy of v starts at the tip of u, a thinner copy of w starts where that copy ends, and the head-to-tail path finishes at the far corner of the box, at (2, 4, h). A fourth thick arrow runs from the origin straight to that corner: the diagonal, which is the sum u + v + w. Dragging h raises or lowers the top of the box, and the diagonal arrow tilts to keep its tip on the moving corner; at h = 0 the box flattens onto the floor and the diagonal lies flat with it, and at negative h the box hangs below the floor. — The box spanned by $\mathbf{u} = (2, 1, 0)$ , $\mathbf{v} = (-1, 2, 0)$ and $\mathbf{w} = (1, 1, h)$ , with the three faces at the origin shaded. The thin copies run head to tail — $\mathbf{u}$ , then $\mathbf{v}$ , then $\mathbf{w}$ — and finishes at the far corner, which the fourth arrow, the diagonal $\mathbf{u} + \mathbf{v} + \mathbf{w}$ , reaches directly.

Explore the box

Orbit the box at the starting value $h = 2$ . The thin copy of $\mathbf{v}$ leaves the tip of $\mathbf{u}$ , the thin copy of $\mathbf{w}$ leaves the tip of that, and the path ends at the box's far corner. Read the corner: $(2, 4, 2)$ , which is $(2 - 1 + 1, \; 1 + 2 + 1, \; 0 + 0 + 2)$ .
The diagonal arrow goes straight from the origin to that corner, so it is the sum. Every one of the box's other diagonals is a sum too, with different signs: the one from the tip of $\mathbf{u}$ to the opposite corner is $\mathbf{v} + \mathbf{w} - \mathbf{u}$ .
Drag $h$ to $0$ . All three vectors now lie in the floor, the box collapses to a flat parallelogram, and the diagonal lies flat with it, ending at $(2, 4, 0)$ . Three vectors in one plane span no volume — a fact Chapter 4 will measure with a determinant.
Drag $h$ to $-2$ . The box hangs below the floor and the diagonal dips to $(2, 4, -2)$ . Only the third component of the sum has changed, because only $\mathbf{w}$ 's third component did.

Example 1 Sums and differences by components

With $\mathbf{u} = (2, 1, 0)$ , $\mathbf{v} = (-1, 2, 0)$ and $\mathbf{w} = (1, 1, 2)$ as in the figure at $h = 2$ , compute $\mathbf{u} + \mathbf{v}$ , $\mathbf{u} + \mathbf{v} + \mathbf{w}$ , and $\mathbf{u} - \mathbf{w}$ .

Component by component: $\mathbf{u} + \mathbf{v} = (2 - 1, \; 1 + 2, \; 0 + 0) = (1, 3, 0)$ , the corner where the thin copy of $\mathbf{w}$ begins. Adding $\mathbf{w}$ gives $(1 + 1, \; 3 + 1, \; 0 + 2) = (2, 4, 2)$ , the far corner. And $\mathbf{u} - \mathbf{w} = (2 - 1, \; 1 - 1, \; 0 - 2) = (1, 0, -2)$ , the arrow from the tip of $\mathbf{w}$ to the tip of $\mathbf{u}$ , which runs along one of the box's face diagonals.

\begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}

The first two vectors, added ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The vectors $\mathbf{u}$ and $\mathbf{v}$ of the box, stacked as columns and added by a computer algebra system: $(1, 3, 0)$ , the corner of the floor parallelogram.

\begin{bmatrix}2 \\ 4 \\ 2\end{bmatrix}

Then the third ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The same sum with $\mathbf{w}$ added at $h = 2$ : $(2, 4, 2)$ , the far corner of the box. Head-to-tail addition is associative, so it does not matter that this was done in two steps.

\begin{bmatrix}1 \\ 0 \\ -2\end{bmatrix}

A difference ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The difference $\mathbf{u} - \mathbf{w}$ , computed component by component: $(1, 0, -2)$ .

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.