Vectors — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Vector notation, component form, magnitude, direction, vector addition, scalar multiplication, the dot product, vector projections, and vector applications to displacement and planar motion, aligned to AP Precalculus CED Unit 4 learning objectives.

You should already know: Coordinate geometry of the Cartesian plane, right triangle trigonometry, basic algebraic operations on real numbers.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Vectors?

A vector is a mathematical quantity that has both magnitude (size) and direction, unlike a scalar, which only has magnitude. Common examples relevant to AP Precalculus include displacement, velocity, and force. Vectors are represented as directed line segments in the plane, or written in component form relative to the $xy$-coordinate system. AP Precalculus CED allocates 10-12% of total exam weight to Unit 4, and vectors account for roughly a third of that unit, or 3-4% of the overall exam score. Vectors appear on both the multiple-choice (MCQ) and free-response (FRQ) sections of the exam, most commonly as standalone MCQs or 3-4 point sub-questions on longer FRQs.

Standard notation you must know: Vectors are written as boldface letters ($\mathbf{u},\mathbf{v}$) or with an arrow above the letter ($\vec{v}$) in handwritten work. The magnitude of a vector $\vec{v}$ is written $|\vec{v}|$, analogous to absolute value for scalars. Two vectors are equal if and only if they have the same magnitude and same direction, regardless of their starting point. A position vector is a vector with its tail anchored at the origin $(0,0)$, so its head is at the coordinates that match its components.

2. Components, Magnitude, and Direction of Vectors

Any vector in the plane can be broken into horizontal ($x$) and vertical ($y$) components, which describe the displacement of the vector along each axis. If a vector $\vec{v}$ goes from initial point $P(x_1,y_1)$ to terminal point $Q(x_2,y_2)$, its component form is: $$\vec{v}=⟨x_2-x_1,y_2-y_1⟩$$ If $\vec{v}$ is a position vector starting at the origin with terminal point $(a,b)$, its component form simplifies to $⟨a,b⟩$.

To find the magnitude (length) of $\vec{v}=⟨a,b⟩$, we use the Pythagorean theorem, since the vector forms the hypotenuse of a right triangle with legs equal to the absolute values of the components: $$|\vec{v}|=\sqrt{a^2+b^2}$$

The direction of a vector is most commonly given as the standard position angle $\theta$, measured counterclockwise from the positive $x$-axis to the vector. To find $\theta$, we use inverse tangent with a quadrant correction: $$\tan \theta =\frac{b}{a}⟹\theta =\arctan \left(\frac{b}{a}\right)+\text{quadrant correction}$$ Add $180^{\circ }$ (or $\pi$ radians) for vectors in Quadrants II or III, and add $360^{\circ }$ (or $2\pi$ radians) for negative angles in Quadrant IV.

If given magnitude $|\vec{v}|=r$ and direction $\theta$, you can derive components using right triangle trigonometry: $a=r\cos \theta$, $b=r\sin \theta$, so $\vec{v}=⟨r\cos \theta ,r\sin \theta ⟩$.

Worked Example

Find the component form, magnitude, and direction of the vector that goes 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: $x=4-(-2)=6$, $y=-3-5=-8$, so component form is $\vec{v}=⟨6,-8⟩$.
2. Calculate magnitude: $|\vec{v}|=\sqrt{6^2+(-8{)}^2}=\sqrt{36+64}=\sqrt{100}=10$.
3. Find the raw arctangent result: $\tan \theta =\frac{-8}{6}=-\frac{4}{3}$, so $\arctan \left(-\frac{4}{3}\right)\approx -53.1^{\circ }$.
4. Correct for quadrant: $x>0$ and $y<0$ places $\vec{v}$ in Quadrant IV, so add $360^{\circ }$: $\theta =-53.1+360=306.9^{\circ }$.

Final result: $\vec{v}=⟨6,-8⟩$, $|\vec{v}|=10$, $\theta =306.9^{\circ }$.

Exam tip: Always sketch a quick rough plot of your vector on a coordinate grid before reporting the direction angle. This will instantly catch quadrant correction mistakes, the most common error on this problem type.

3. Vector Addition and Scalar Multiplication

Vector addition combines two vectors to produce a resultant vector. Geometrically, the triangle rule states that if you place the tail of the second vector at the head of the first, the resultant goes from the tail of the first to the head of the second. Algebraically, addition is always component-wise for vectors in component form: $$\vec{u}+\vec{v}=⟨u_1+v_1,u_2+v_2⟩$$ for $\vec{u}=⟨u_1,u_2⟩$ and $\vec{v}=⟨v_1,v_2⟩$.

Scalar multiplication multiplies a vector by a real number (scalar) $k$, scaling its magnitude and reversing direction if $k$ is negative. This operation is also component-wise: $$k\vec{u}=⟨k{u}_1,k{u}_2⟩$$ The magnitude of $k\vec{u}$ is $|k|\cdot |\vec{u}|$, which matches the geometric interpretation of scaling. A unit vector is a vector with magnitude 1, used to represent direction only. To get a unit vector in the direction of any non-zero $\vec{v}$, divide by its magnitude: $\hat{v}=\frac{1}{|\vec{v}|}\vec{v}$. The standard unit vectors are $\hat{i}=⟨1,0⟩$ and $\hat{j}=⟨0,1⟩$, so any vector $⟨a,b⟩$ can also be written as $a\hat{i}+b\hat{j}$, a common alternative notation.

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: $\vec{u}=⟨3,-2⟩$, $\vec{v}=⟨-1,5⟩$.
2. Perform scalar multiplication: $2\vec{u}=⟨6,-4⟩$, $-3\vec{v}=⟨3,-15⟩$.
3. Add component-wise: $2\vec{u}-3\vec{v}=⟨6+3,-4-15⟩=⟨9,-19⟩$.
4. Calculate magnitude: $|⟨9,-19⟩|=\sqrt{9^2+(-19{)}^2}=\sqrt{442}$.
5. Divide by magnitude to get the unit vector: $\hat{w}=⟨\frac{9}{\sqrt{442}},\frac{-19}{\sqrt{442}}⟩=⟨\frac{9\sqrt{442}}{442},\frac{-19\sqrt{442}}{442}⟩$.

Exam tip: Rewrite subtraction of scalar multiples as addition of the negative scalar multiple, like $2\vec{u}-3\vec{v}=2\vec{u}+(-3)\vec{v}$. This ensures you distribute the negative sign to both components, avoiding common sign errors.

4. The 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 $\vec{u}=⟨u_1,u_2⟩$ and $\vec{v}=⟨v_1,v_2⟩$, the dot product is calculated as: $$\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2$$

The dot product has two core uses on the AP Precalculus exam: finding the angle between two vectors and testing for orthogonality (perpendicularity). The relationship between the dot product and the angle $\theta$ (between $0^{\circ }$ and $180^{\circ }$) between two vectors is: $$\vec{u}\cdot \vec{v}=|\vec{u}||\vec{v}|\cos \theta$$ Rearranged to solve for $\theta$: $\cos \theta =\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$. Two vectors are orthogonal if and only if $\vec{u}\cdot \vec{v}=0$, because $\cos 90^{\circ }=0$.

A second key application is the projection of one vector onto another. The vector projection of $\vec{u}$ onto $\vec{v}$ is the vector along $\vec{v}$ that represents the "shadow" of $\vec{u}$ on the line containing $\vec{v}$. The formula is: $${\text{proj}}_{\vec{v}}\vec{u}=\left(\frac{\vec{u}\cdot \vec{v}}{|\vec{v}{|}^2}\right)\vec{v}$$

Worked Example

Let $\vec{u}=⟨4,2⟩$ and $\vec{v}=⟨-3,6⟩$. (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: $\vec{u}\cdot \vec{v}=(4)(-3)+(2)(6)=-12+12=0$. Since the dot product is zero, the vectors are confirmed to be orthogonal.
2. For part (b), first calculate $|\vec{v}{|}^2=(-3{)}^2+6^2=9+36=45$.
3. Substitute into the projection formula: ${\text{proj}}_{\vec{v}}\vec{u}=\left(\frac{0}{45}\right)⟨-3,6⟩=⟨0,0⟩$.
4. 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 on your scratch paper before solving to avoid mixing them up.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For a vector $⟨-3,4⟩$ in Quadrant II, you report direction as $\arctan (4/-3)=-53.1^{\circ }$ as the final direction. Why: Students forget arctangent only outputs values between $-90^{\circ }$ and $90^{\circ }$, so it does not automatically account for vectors in Quadrants II and III. Correct move: 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 move: When calculating $2\vec{u}-3\vec{v}$ for $\vec{u}=⟨1,2⟩$ and $\vec{v}=⟨3,4⟩$, you get $⟨2-3,4-12⟩=⟨-1,-8⟩$. Why: Students distribute the negative only to the first component when subtracting scalar multiples. Correct move: Calculate each scalar multiple separately, including the negative sign, before adding component-wise.
• Wrong move: You calculate the dot product of $⟨2,3⟩$ and $⟨4,5⟩$ as $⟨8,15⟩$, a vector instead of a scalar. Why: Students confuse the dot product with component-wise multiplication, which is not tested on the exam. Correct move: Always add the products of components, so the result is a single real number.
• Wrong move: When finding a vector from $P(x_1,y_1)$ to $Q(x_2,y_2)$, you calculate components as $⟨x_1-x_2,y_1-y_2⟩$. Why: Students mix up the order of subtraction for initial vs terminal points. Correct move: Repeat the phrase "terminal minus initial" to yourself before calculating components every time.
• Wrong move: 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: Students mix up the formulas for scalar projection and vector projection. Correct move: Label both formulas clearly on your scratch work before substituting values.
• Wrong move: You conclude two vectors are equal because they have the same magnitude, even though they point in different directions. Why: Students forget the full definition of equal vectors. Correct move: Always check that both magnitude and direction match before confirming two vectors are equal.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is a unit vector orthogonal to $\vec{v}=⟨2,-5⟩$? A) $⟨\frac{2}{\sqrt{29}},\frac{-5}{\sqrt{29}}⟩$ B) $⟨\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}⟩$ C) $⟨5,2⟩$ D) $⟨\frac{-2}{\sqrt{29}},\frac{5}{\sqrt{29}}⟩$

Worked Solution: We need a vector that meets two requirements: it is orthogonal to $\vec{v}$ (dot product zero) and it has magnitude 1. First, any orthogonal vector $⟨a,b⟩$ must satisfy $2a-5b=0$, so $2a=5b$. Option A has a non-zero dot product and is parallel, not orthogonal, so eliminate A. Option B has a dot product of $2(\frac{5}{\sqrt{29}})+(-5)(\frac{2}{\sqrt{29}})=\frac{10-10}{\sqrt{29}}=0$, so it is orthogonal. Check its magnitude: $(\frac{5}{\sqrt{29}}{)}^2+(\frac{2}{\sqrt{29}}{)}^2=\frac{25+4}{29}=1$, so it is a unit vector. Option C is orthogonal but has magnitude $\sqrt{29}\ne 1$, and D has a non-zero dot product. Correct answer is B.

Question 2 (Free Response)

Let $\vec{u}$ be a vector with magnitude 6, direction $120^{\circ }$ from the positive $x$-axis. Let $\vec{v}$ be the vector from point $(1,-2)$ to point $(-3,4)$. (a) Write $\vec{u}$ in component form. (b) Find the magnitude of $\vec{u}+\vec{v}$, rounded to the nearest tenth. (c) Find the angle between $\vec{u}$ and $\vec{v}$, rounded to the nearest tenth of a degree.

Worked Solution: (a) For magnitude $r=6$ and direction $\theta =120^{\circ }$: $u_x=6\cos (120^{\circ })=6(-\frac{1}{2})=-3$, $u_y=6\sin (120^{\circ })=6(\frac{\sqrt{3}}{2})=3\sqrt{3}$. So $\vec{u}=⟨-3,3\sqrt{3}⟩$. (b) Calculate $\vec{v}$ components: $v_x=-3-1=-4$, $v_y=4-(-2)=6$, so $\vec{v}=⟨-4,6⟩$. Add: $\vec{u}+\vec{v}=⟨-3-4,3\sqrt{3}+6⟩=⟨-7,6+3\sqrt{3}⟩$. Magnitude: $|\vec{u}+\vec{v}|=\sqrt{(-7{)}^2+(6+3\sqrt{3}{)}^2}=\sqrt{49+36+36\sqrt{3}+27}=\sqrt{112+36\sqrt{3}}\approx 13.2$. (c) Dot product: $\vec{u}\cdot \vec{v}=(-3)(-4)+(3\sqrt{3})(6)=12+18\sqrt{3}\approx 43.18$. $|\vec{u}|=6$, $|\vec{v}|=\sqrt{(-4{)}^2+6^2}=\sqrt{52}\approx 7.21$. So $\cos \theta =\frac{43.18}{6\cdot 7.21}\approx 0.998$, so $\theta \approx 3.6^{\circ }$.

Question 3 (Application / Real-World Style)

A hiker leaves base camp and walks 3 miles northeast, then turns and walks 5 miles due east. Let east be the positive $x$-direction and north be the positive $y$-direction. Find the total displacement vector for the trip, then calculate the straight-line distance from base camp to the hiker's final position, rounded to the nearest tenth of a mile.

Worked Solution:

1. Northeast corresponds to a standard angle of $45^{\circ }$. The first displacement vector ${\vec{d}}_1$ has components: $d_{1x}=3\cos (45^{\circ })\approx 2.12$, $d_{1y}=3\sin (45^{\circ })\approx 2.12$, so ${\vec{d}}_1=⟨2.12,2.12⟩$.
2. The second displacement vector is 5 miles due east, so ${\vec{d}}_2=⟨5,0⟩$.
3. Total displacement $\vec{D}={\vec{d}}_1+{\vec{d}}_2=⟨2.12+5,2.12+0⟩=⟨7.12,2.12⟩$.
4. Straight-line distance is the magnitude of $\vec{D}$: $|\vec{D}|=\sqrt{(7.12{)}^2+(2.12{)}^2}\approx \sqrt{50.7+4.5}\approx \sqrt{55.2}\approx 7.4$ miles.

In context, the hiker is approximately 7.4 miles from base camp in a straight line after the two legs of the trip.

7. Quick Reference Cheatsheet

8. 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, which is a high-weight FRQ topic on the exam. Vectors also lay the groundwork for matrix transformations of the plane, where vectors represent points to be transformed by matrix multiplication. Beyond AP Precalculus, vectors are the core of multivariable calculus, linear algebra, and college physics, so mastering this topic gives a major head start for future STEM courses.

Parametric equations and planar motion Matrix transformations of the plane Matrices and systems of linear equations