Vectors — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Many quantities need two pieces of information: not just "how much" but "which way". A wind of 20 km/h from the west is different from a wind of 20 km/h from the north. A vector is a quantity with a magnitude (size) and a direction. You draw it as an arrow; you compute with it through its components.

Components

An arrow from $(x_1, y_1)$ to $(x_2, y_2)$ moves $x_2 - x_1$ across and $y_2 - y_1$ up. Those two numbers are its components:

\mathbf{v} = \left\langle x_2 - x_1, y_2 - y_1 \right\rangle.

Only the move matters, not where the arrow starts. The arrow from $(2, 1)$ to $(1, 4)$ and the arrow from $(0, 0)$ to $(-1, 3)$ are the same vector, $\left\langle -1, 3 \right\rangle$ . Another way to write $\left\langle a, b \right\rangle$ is $a\mathbf{i} + b\mathbf{j}$ , where $\mathbf{i} = \left\langle 1, 0 \right\rangle$ and $\mathbf{j} = \left\langle 0, 1 \right\rangle$ .

Magnitude and direction

The magnitude is the arrow's length, from the Pythagorean theorem: $\left\| \left\langle a, b \right\rangle \right\| = \sqrt{a^2 + b^2}$ . Going the other way, a vector with magnitude $m$ pointing at angle $\theta$ from the positive $x$ -axis has components $\left\langle m\cos\theta, m\sin\theta \right\rangle$ , the same idea as a point on a circle of radius $m$ .

Worked example 1. A boat heads $30^\circ$ north of east at $12$ km/h, and a current pushes it south at $4$ km/h. Find its actual velocity vector and speed.

Heading: $\left\langle 12\cos 30^\circ, 12\sin 30^\circ \right\rangle$ $= \left\langle 6\sqrt{3}, 6 \right\rangle$ , using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$ .

Current: $\left\langle 0, -4 \right\rangle$ (south is the negative $y$ direction).

Sum: $\left\langle 6\sqrt{3}, 2 \right\rangle$ . Speed: $\sqrt{(6\sqrt{3})^2 + 2^2}$ $= \sqrt{108 + 4}$ $= \sqrt{112}$ $= 4\sqrt{7}$ km/h, a little over $10.5$ km/h.

Adding, subtracting and scaling

To add $\mathbf{u} + \mathbf{v}$ on a picture, put the tail of $\mathbf{v}$ at the tip of $\mathbf{u}$ ; the sum runs from the tail of $\mathbf{u}$ to the tip of $\mathbf{v}$ . In components you just add matching entries.

Vector u = <3, 1> drawn from the origin to (3, 1) in green, vector v = <1, 2> drawn from the tip of u at (3, 1) to (4, 3) in blue, and their sum u + v = <4, 3> drawn from the origin to (4, 3) in vermillion. The sum has length 5. — **Figure 1.** $\left\langle 3, 1 \right\rangle + \left\langle 1, 2 \right\rangle = \left\langle 4, 3 \right\rangle$ , tip to tail. LevlPrep original diagram.

Operation	Components	Picture
$\mathbf{u} + \mathbf{v}$	$\left\langle a + c, b + d \right\rangle$	tip to tail
$\mathbf{u} - \mathbf{v}$	$\left\langle a - c, b - d \right\rangle$	add the reversed $\mathbf{v}$
$k\mathbf{u}$	$\left\langle ka, kb \right\rangle$	$\|k\|$ times as long; reversed if $k < 0$
$\frac{\mathbf{u}}{\left\\| \mathbf{u} \right\\|}$	each component over the length	same direction, length $1$

The last row is the unit vector in the direction of $\mathbf{u}$ . For $\left\langle 5, -12 \right\rangle$ , the length is $\sqrt{25 + 144} = 13$ , so the unit vector is $\left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle$ .

The dot product and angles

The dot product of $\mathbf{u} = \left\langle a, b \right\rangle$ and $\mathbf{v} = \left\langle c, d \right\rangle$ is the number $\mathbf{u} \cdot \mathbf{v} = ac + bd$ . It is also $\left\| \mathbf{u} \right\| \left\| \mathbf{v} \right\| \cos\theta$ , where $\theta$ is the angle between the two arrows when they start at the same point. So

\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\left\| \mathbf{u} \right\| \left\| \mathbf{v} \right\|}.

Worked example 2. Find the angle between $\left\langle 1, 0 \right\rangle$ and $\left\langle 1, \sqrt{3} \right\rangle$ .

Dot product: $1(1) + 0(\sqrt{3}) = 1$ . Lengths: $1$ and $\sqrt{1 + 3} = 2$ .

$\cos\theta = \frac{1}{1 \cdot 2} = \frac{1}{2}$ , so $\theta = \frac{\pi}{3}$ .

A dot product of $0$ means $\cos\theta = 0$ : the vectors are perpendicular. $\left\langle 2, 3 \right\rangle \cdot \left\langle -3, 2 \right\rangle$ $= -6 + 6$ $= 0$ . A positive dot product means the angle is less than $\frac{\pi}{2}$ ; a negative one, more.