# AP Precalculus Vectors

> AP Precalculus · Unit 4: Functions Involving Parameters, Vectors, and Matrices
> Source: https://www.owlsprep.com/study/ap-precalculus-u4-vectors/

This guide covers core vector concepts for AP Precalculus Unit 4, including notation, components, magnitude, direction, vector operations, dot product, projections, and applications to displacement and motion, aligned to College Board CED.

**Prerequisites:** Cartesian coordinate geometry; Right triangle trigonometry; Basic algebraic operations

## Learning objectives

- Define vectors and distinguish vectors from scalars
- Calculate component form, magnitude, and direction of planar vectors
- Perform vector addition and scalar multiplication
- Compute the dot product and find angles between vectors
- Calculate vector projections and test for orthogonality
- Apply vectors to solve displacement and planar motion problems

## Core Vector Definitions

A vector is a mathematical quantity with both magnitude (size) and direction, unlike a scalar, which only has magnitude. Common examples include displacement, velocity, and force. Note: Unit 4 is not assessed on the AP Precalculus Exam (the exam covers Units 1–3 only); vectors are taught at teacher discretion for enrichment and as a foundation for later STEM courses.

**Vector** — A quantity with both magnitude and direction. Two vectors are equal if and only if they have the same magnitude and direction, regardless of starting point. A position vector has its tail anchored at the origin $(0,0)$.

*Notation:* Boldface $\mathbf{u}, \mathbf{v}$ or $\vec{v}$ (handwritten); magnitude $|\vec{v}|$

*Example:* Displacement of 5 miles northeast is a vector; 5 miles is a scalar.

## Components, Magnitude, and Direction

Any planar vector can be split into horizontal ($x$) and vertical ($y$) components, which give displacement along each axis. For a vector from initial point $P(x_1,y_1)$ to terminal point $Q(x_2,y_2)$, component form is calculated as terminal minus initial coordinates:

$$\vec{v} = \langle x_2 - x_1, y_2 - y_1 \rangle$$

Magnitude (length) of $\vec{v} = \langle a, b \rangle$ comes from the Pythagorean theorem:

$$|\vec{v}| = \sqrt{a^2 + b^2}$$

Direction is given as the standard position angle $\theta$, measured counterclockwise from the positive $x$-axis. To find $\theta$ from components: $\tan\theta = \frac{b}{a}$, so $\theta = \arctan\left(\frac{b}{a}\right)$ plus quadrant correction: add $180^\circ$ (or $\pi$ radians) for Quadrants II/III, add $360^\circ$ for negative angles in Quadrant IV. If given magnitude $r=|\vec{v}|$ and direction $\theta$, components are $a = r\cos\theta, b = r\sin\theta$.

**Worked example:** Find the component form, magnitude, and direction of the vector from initial point $P(-2, 5)$ to terminal point $Q(4, -3)$. Give direction as an angle in degrees from the positive $x$-axis, rounded to one decimal place.

1. Calculate components by subtracting initial coordinates from terminal:
2. $$x = 4 - (-2) = 6 \\ y = -3 - 5 = -8$$
3. So component form is $\vec{v} = \langle 6, -8 \rangle$.
4. Calculate magnitude using the Pythagorean theorem:
5. $$|\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
6. Find the raw arctangent result: $\tan\theta = \frac{-8}{6} = -\frac{4}{3}$, so $\arctan\left(-\frac{4}{3}\right) \approx -53.1^\circ$.
7. Correct for quadrant: $x>0$ and $y<0$ places $\vec{v}$ in Quadrant IV, so add $360^\circ$:
8. $$\theta = -53.1^\circ + 360^\circ = 306.9^\circ$$
9. Final result: $\vec{v} = \langle 6, -8 \rangle$, $|\vec{v}| = 10$, $\theta = 306.9^\circ$.

> **Exam tip:** Always sketch a quick rough plot of your vector on a coordinate grid before reporting the direction angle to catch quadrant correction mistakes.

## Vector Addition and Scalar Multiplication

Vector addition produces a resultant vector from two input vectors. Algebraically, addition is always component-wise for vectors in component form:

$$\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$$

Scalar multiplication multiplies a vector by a real scalar $k$, scaling its magnitude and reversing direction if $k$ is negative. This is also component-wise:

$$k\vec{u} = \langle k u_1, k u_2 \rangle$$

A unit vector is a vector with magnitude 1, used to represent direction only. To get a unit vector in the direction of non-zero $\vec{v}$, divide by its magnitude: $\hat{v} = \frac{1}{|\vec{v}|}\vec{v}$. The standard unit vectors are $\hat{i} = \langle 1, 0 \rangle$ and $\hat{j} = \langle 0, 1 \rangle$, so any vector $\langle a, b \rangle$ can be written as $a\hat{i} + b\hat{j}$.

**Worked example:** Let $\vec{u} = 3\hat{i} - 2\hat{j}$ and $\vec{v} = -\hat{i} + 5\hat{j}$. Find $2\vec{u} - 3\vec{v}$, then find a unit vector in the direction of $2\vec{u} - 3\vec{v}$.

1. Rewrite vectors in component form: $\vec{u} = \langle 3, -2 \rangle$, $\vec{v} = \langle -1, 5 \rangle$.
2. Perform scalar multiplication for each term:
3. $$2\vec{u} = \langle 6, -4 \rangle \\ -3\vec{v} = \langle 3, -15 \rangle$$
4. Add component-wise to get the result:
5. $$2\vec{u} - 3\vec{v} = \langle 6 + 3, -4 - 15 \rangle = \langle 9, -19 \rangle$$
6. Calculate the magnitude of the resulting vector:
7. $$|\langle 9, -19 \rangle| = \sqrt{9^2 + (-19)^2} = \sqrt{442}$$
8. Divide by the magnitude to get the unit vector, rationalizing the denominator:
9. $$\hat{w} = \left\langle \frac{9\sqrt{442}}{442}, \frac{-19\sqrt{442}}{442} \right\rangle$$

> **Exam tip:** Rewrite subtraction of scalar multiples as addition of the negative scalar multiple, like $2\vec{u} - 3\vec{v} = 2\vec{u} + (-3)\vec{v}$, to avoid sign errors when distributing.

## Dot Product and Vector Projections

The dot product (or scalar product) is an operation that takes two vectors and returns a scalar (not a vector). For $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$:

$$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$$

The dot product relates to the angle $\theta$ (between $0^\circ$ and $180^\circ$) between two vectors via the identity:

$$\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$$

Two key uses of the dot product are: 1) testing for orthogonality (perpendicularity): vectors are orthogonal if and only if their dot product is zero; 2) finding the projection of one vector onto another, which is the shadow of $\vec{u}$ along the line containing $\vec{v}$. The formula for the vector projection of $\vec{u}$ onto $\vec{v}$ is:

$$\text{proj}_{\vec{v}} \vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v}$$

**Worked example:** Let $\vec{u} = \langle 4, 2 \rangle$ and $\vec{v} = \langle -3, 6 \rangle$. (a) Confirm the vectors are orthogonal, (b) Find the vector projection of $\vec{u}$ onto $\vec{v}$.

1. For part (a), calculate the dot product:
2. $$\vec{u} \cdot \vec{v} = (4)(-3) + (2)(6) = -12 + 12 = 0$$
3. Since the dot product is zero, the vectors are confirmed to be orthogonal.
4. For part (b), first calculate $|\vec{v}|^2$:
5. $$|\vec{v}|^2 = (-3)^2 + 6^2 = 9 + 36 = 45$$
6. Substitute into the projection formula:
7. $$\text{proj}_{\vec{v}} \vec{u} = \left(\frac{0}{45}\right)\langle -3, 6 \rangle = \langle 0, 0 \rangle$$
8. This result makes geometric sense: if two vectors are orthogonal, no part of $\vec{u}$ lies along the direction of $\vec{v}$, so the projection is the zero vector.

> **Exam tip:** Remember that vector projection uses $|\vec{v}|^2$ in the denominator, while scalar projection uses $|\vec{v}|$. Write both formulas down before solving to avoid mixing them up.

## AP-Style Concept Check

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. Which of the following is a unit vector orthogonal to $\vec{v} = \langle 2, -5 \rangle$?

   - A) $\left\langle \frac{2}{\sqrt{29}}, \frac{-5}{\sqrt{29}} \right\rangle$
   - B) $\left\langle \frac{5}{\sqrt{29}}, \frac{2}{\sqrt{29}} \right\rangle$
   - C) $\langle 5, 2 \rangle$
   - D) $\left\langle \frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle$

   *Answer:* B) $\left\langle \frac{5}{\sqrt{29}}, \frac{2}{\sqrt{29}} \right\rangle$

   *Why:* Correct. This vector has a dot product of 0 with $\vec{v}$ and magnitude 1, meeting both requirements. Check: $\vec{v} \cdot \langle 5/\sqrt{29}, 2/\sqrt{29} \rangle = (2)(5/\sqrt{29}) + (-5)(2/\sqrt{29}) = 0$, and $(5/\sqrt{29})^2 + (2/\sqrt{29})^2 = 29/29 = 1$.

## Common pitfalls

- **Wrong:** For a vector $\langle -3, 4 \rangle$ in Quadrant II, you report $\arctan(4/-3) = -53.1^\circ$ as the final direction.
  - Why it fails: Arctangent only outputs values between $-90^\circ$ and $90^\circ$, so it does not automatically account for vectors in Quadrants II and III.
  - Correct: Always plot the vector to check its quadrant, then add $180^\circ$ for Quadrants II/III and $360^\circ$ for negative angles in Quadrant IV.
- **Wrong:** When calculating $2\vec{u} - 3\vec{v}$ for $\vec{u} = \langle 1, 2 \rangle$ and $\vec{v} = \langle 3, 4 \rangle$, you get $\langle -1, -8 \rangle$ by only distributing the negative to the first component.
  - Why it fails: Students often forget to distribute the negative sign to both components of $3\vec{v}$ when subtracting.
  - Correct: Calculate each scalar multiple separately, including the negative sign, before adding component-wise.
- **Wrong:** You calculate the dot product of $\langle 2,3 \rangle$ and $\langle 4,5 \rangle$ as $\langle 8, 15 \rangle$, a vector instead of a scalar.
  - Why it fails: Students confuse the dot product with component-wise multiplication.
  - Correct: Always add the products of corresponding components, so the result is a single real number, not a vector.
- **Wrong:** When finding a vector from $P(x_1,y_1)$ to $Q(x_2,y_2)$, you calculate components as $\langle x_1 - x_2, y_1 - y_2 \rangle$.
  - Why it fails: Students mix up the order of subtraction for initial vs terminal points.
  - Correct: Remember the rule: components are always terminal minus initial coordinates.
- **Wrong:** You calculate the vector projection of $\vec{u}$ onto $\vec{v}$ as $\left(\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}\right)\vec{v}$.
  - Why it fails: Students mix up the formulas for scalar projection and vector projection.
  - Correct: Label both formulas clearly on your scratch work before substituting values.
- **Wrong:** You conclude two vectors are equal because they have the same magnitude, even though they point in different directions.
  - Why it fails: Students forget the full definition of equal vectors.
  - Correct: Always check that both magnitude and direction match before confirming two vectors are equal.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Component from two points | $\vec{v} = \langle x_2 - x_1, y_2 - y_1 \rangle$ | $(x_1,y_1)$ = initial, $(x_2,y_2)$ = terminal; always terminal minus initial |
| Magnitude | $\|\langle a,b \rangle\| = \sqrt{a^2 + b^2}$ | Always non-negative, equal to the length of the vector |
| Components from magnitude/direction | $a = r\cos\theta, \, b = r\sin\theta$ | $\theta$ = standard angle counterclockwise from positive $x$-axis |
| Vector Addition | $\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$ | Performed component-wise |
| Scalar Multiplication | $k\vec{u} = \langle k u_1, k u_2 \rangle$ | Negative $k$ reverses direction; new magnitude = $\|k\|\|\vec{u}\|$ |
| Unit Vector | $\hat{v} = \frac{1}{\|\vec{v}\|}\vec{v}$ | Has magnitude 1, same direction as original vector |
| Dot Product | $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2$ | Result is a scalar, not a vector |
| Angle between two vectors | $\cos\theta = \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\|\vec{v}\|}$ | $\theta$ ranges from $0^\circ$ (parallel same direction) to $180^\circ$ (parallel opposite direction) |
| Orthogonal Vectors | $\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0$ | Applies to perpendicular vectors |
| Vector Projection | $\text{proj}_{\vec{v}} \vec{u} = \left(\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^2}\right)\vec{v}$ | Projection of $\vec{u}$ onto $\vec{v}$ is a vector along $\vec{v}$ |

## What's next

Vectors are the foundation for all remaining topics in AP Precalculus Unit 4. Next, you will apply vector addition and component operations to model the position of moving objects as functions of time with parametric equations. Without a solid understanding of vector components and magnitude, you will not be able to calculate velocity, speed, or acceleration for parametrically defined motion. Vectors also lay the groundwork for matrix transformations of the plane, where vectors represent points to be transformed.

- [Unit 4 Full Overview](https://www.owlsprep.com/study/ap-precalculus-u4-overview/)
- [Vector-valued functions](https://www.owlsprep.com/study/ap-precalculus-u4-vector-valued-functions/)
- [Matrices](https://www.owlsprep.com/study/ap-precalculus-u4-matrices/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-precalculus-u4-vectors/