2.1 Vectors in the Plane

Big idea

Some quantities are settled by a single number. Mass, temperature and elapsed time are finished once you name the number and the unit. Other quantities are not. Saying that a plane flies at $500$ kilometers per hour leaves out the part that decides where it lands, and saying that a rope is pulled with $200$ newtons of force leaves out the part that decides which way the crate moves. A quantity that needs a size and a direction together is a vector, and this section builds an arithmetic for those quantities so that you can compute with them instead of describing them in words.

The arithmetic rests on two operations. Adding two vectors means laying the second one down where the first one ends and asking where you finished; scaling a vector by a number stretches or shrinks it and, when the number is negative, turns it around. Each operation has a picture and each has a formula. The reason for introducing coordinates is that the formulas are almost invisible: you work one entry at a time, and the pictures follow along without any extra effort.

A vector in the plane therefore has two descriptions, and moving between them is most of the routine work. The geometric description gives a length and an angle. The algebraic description gives a horizontal component and a vertical component. Right-triangle trigonometry converts either description into the other, which is why force problems and navigation problems that sound like geometry can be finished with arithmetic.

Everything here extends to three dimensions in the next section with a third component and no new ideas. That is worth noticing now, because the rest of this course runs on vectors: they carry positions, velocities and rates of change once a function has more than one input, and the two operations you learn here are the ones you will still be using in the final chapter.

Decoder

A vector is determined by its magnitude and its direction alone, so two arrows of the same length pointing the same way represent the same vector no matter where in the plane they are drawn.

That sentence is what makes vector arithmetic work. If arrows had to stay where they were drawn, you could not slide the second one to the tip of the first one to add them, and the parallelogram picture would be meaningless. Because position carries no information, every vector may be redrawn starting at the origin, and that standard position is what lets a vector be recorded as a pair of numbers.

The pair of numbers is the displacement from tail to tip: how far right and how far up. Those are the components. The classic mistake is treating the point $(3, 5)$ and the vector $\langle 3, 5 \rangle$ as the same object. They are recorded the same way and they are not the same thing. A point is a location; a vector is a change of location, and the vector $\langle 3, 5 \rangle$ describes the same motion whether you begin at the origin or somewhere else entirely.

The second common slip is a sign. The vector running from $P$ to $Q$ is found by subtracting the starting point from the ending point, not the other way round. Reversing the order reverses the arrow.

Definitions and results

Vectors and scalars. A scalar is an ordinary real number. A vector has a magnitude and a direction. In print a vector is written $\mathbf{v}$ or $\overrightarrow{PQ}$, where $P$ is the tail and $Q$ is the tip. The zero vector $\mathbf{0}$ has magnitude $0$ and no assigned direction.

Component form. In standard position a plane vector is written $\mathbf{v} = \langle v_1, v_2 \rangle$, where $v_1$ is the horizontal displacement and $v_2$ the vertical one. For the vector from $P(x_1, y_1)$ to $Q(x_2, y_2)$,

$$ \overrightarrow{PQ} = \langle x_2 - x_1, \; y_2 - y_1 \rangle $$

Two vectors are equal exactly when both components match.

Magnitude. The magnitude, or length, follows from the Pythagorean theorem:

$$ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} $$

It is a scalar, it is never negative, and it is zero only for $\mathbf{0}$.

Sum and difference. Add and subtract componentwise: $\mathbf{u} + \mathbf{v} = \langle u_1 + v_1, \; u_2 + v_2 \rangle$. Geometrically the sum is the diagonal of the parallelogram built on the two arrows, and $\mathbf{u} - \mathbf{v}$ is the arrow running from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$.

Scalar multiplication. For a scalar $c$, $c\mathbf{v} = \langle cv_1, cv_2 \rangle$ and $\|c\mathbf{v}\| = |c| \, \|\mathbf{v}\|$. The result points the same way as $\mathbf{v}$ when $c > 0$ and the opposite way when $c < 0$. Two nonzero vectors are parallel exactly when one is a scalar multiple of the other.

Algebraic properties. Addition is commutative and associative, $\mathbf{0}$ is neutral, $-\mathbf{v}$ undoes $\mathbf{v}$, scalars distribute over vector sums, sums of scalars distribute over a vector, and $1\mathbf{v} = \mathbf{v}$. These are the rules of ordinary algebra, so expressions such as $3(\mathbf{u} - 2\mathbf{v})$ expand exactly as they look. Magnitudes, however, do not add: the triangle inequality says $\|\mathbf{u} + \mathbf{v}\| \le \|\mathbf{u}\| + \|\mathbf{v}\|$, with equality only when the two vectors point the same way.

Unit vectors. A unit vector has magnitude $1$. Any nonzero $\mathbf{v}$ has exactly one unit vector in its direction, obtained by dividing out the length:

$$ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} $$

A vector of length $L$ in that direction is then $L\mathbf{u}$. This is how you separate “which way” from “how much”.

Standard basis. With $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$, every plane vector is $\langle v_1, v_2 \rangle = v_1\mathbf{i} + v_2\mathbf{j}$. The two notations mean the same thing and both appear in later sections.

Magnitude and direction form. If $\mathbf{v}$ has magnitude $r$ and makes angle $\theta$ with the positive $x$-axis, then $\mathbf{v} = \langle r\cos\theta, \; r\sin\theta \rangle$. Going the other way, $r = \|\mathbf{v}\|$ and $\theta$ is found from $\tan\theta = v_2/v_1$, with the quadrant of the point $(v_1, v_2)$ deciding which of the two candidate angles is correct.

Worked examples

From two points to components

Find $\overrightarrow{PQ}$ and its magnitude for $P(-1, 4)$ and $Q(3, -2)$.

Subtract the tail from the tip:

$$ \overrightarrow{PQ} = \langle 3 - (-1), \; -2 - 4 \rangle = \langle 4, -6 \rangle $$

$$ \|\overrightarrow{PQ}\| = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} $$

Check the direction by walking it: starting at $P(-1,4)$ and moving $4$ right and $6$ down lands on $(3, -2)$, which is $Q$.

A combination of two vectors

Let $\mathbf{u} = \langle 3, -1 \rangle$ and $\mathbf{v} = \langle -2, 5 \rangle$. Compute $2\mathbf{u} + 3\mathbf{v}$.

Scale first, then add componentwise. Doubling $\mathbf{u}$ gives $\langle 6, -2 \rangle$ and tripling $\mathbf{v}$ gives $\langle -6, 15 \rangle$, so

$$ 2\mathbf{u} + 3\mathbf{v} = \langle 6 + (-6), \; -2 + 15 \rangle = \langle 0, 13 \rangle $$

The result is vertical, which is the horizontal components cancelling exactly. Its magnitude is $13$, while $2\|\mathbf{u}\| + 3\|\mathbf{v}\| = 2\sqrt{10} + 3\sqrt{29} \approx 22.5$. The triangle inequality is comfortably satisfied, as it must be when the pieces partly oppose each other.

A unit vector and a direction angle

Let $\mathbf{w} = \langle -3, \; 3\sqrt{3} \rangle$. Its magnitude is

$$ \|\mathbf{w}\| = \sqrt{9 + 27} = \sqrt{36} = 6 $$

so the unit vector in its direction is

$$ \mathbf{u} = \frac{1}{6}\langle -3, \; 3\sqrt{3} \rangle = \left\langle -\tfrac{1}{2}, \; \tfrac{\sqrt{3}}{2} \right\rangle $$

Confirm that this is a unit vector: $\tfrac{1}{4} + \tfrac{3}{4} = 1$. Since $\tan\theta = -\sqrt{3}$ and the point $(-3, 3\sqrt{3})$ sits in the second quadrant, the direction angle is $\theta = 120$ degrees rather than $-60$ degrees. A vector of length $10$ in the same direction is $10\mathbf{u} = \langle -5, \; 5\sqrt{3} \rangle$.

Resolving a force

A rope pulls a sled with a force of $50$ newtons directed $30$ degrees above the horizontal. Write the force in components.

Use magnitude and direction form with $r = 50$ and $\theta = 30$ degrees:

$$ \mathbf{F} = \langle 50\cos 30^\circ, \; 50 \sin 30^\circ \rangle = \langle 25\sqrt{3}, \; 25 \rangle $$

The horizontal part $25\sqrt{3} \approx 43.3$ newtons is what drags the sled forward, and the vertical part $25$ newtons is what lifts it. As a check, $\|\mathbf{F}\| = \sqrt{1875 + 625} = \sqrt{2500} = 50$, which returns the original magnitude.

Practice

Start with the operations themselves. These ask you to add, subtract and scale vectors, and to combine several of those steps in one expression.

Then move between the two descriptions. These problems go from points to components, from components to magnitude and direction, and back again, including normalizing to a unit vector.

Quiz

Five items on component form, magnitude, sums and scalar multiples, and unit vectors.