Vectors - AP PreCalculus Study Guide

The Core Idea: Vectors

In mathematics and physics, some quantities can be described by a single number, like temperature or speed. These are called scalars. However, other quantities, such as force or velocity, require both a size (magnitude) and a direction. A vector is the mathematical object used to represent these quantities. It is conceptualized as a directed line segment, possessing both a specific length (its magnitude) and an orientation in the plane (its direction).

To work with vectors algebraically, we represent them by their components. These components describe the horizontal and vertical change from the vector's starting (initial) point to its ending (terminal) point. This component form allows us to apply arithmetic operations to vectors, such as addition, subtraction, and scaling, and to calculate properties like length and the angle between two vectors, transforming geometric ideas into concrete calculations.

Key Formulas & Operations

The following formulas are essential for performing calculations with vectors in the plane. Let $\vec{u}=⟨u_1,u_2⟩$ and $\vec{v}=⟨v_1,v_2⟩$ be two vectors, and let $k$ be a scalar (a real number).

Component Form

The component form of a vector with an initial point $P(x_1,y_1)$ and a terminal point $Q(x_2,y_2)$ is found by subtracting the initial coordinates from the terminal coordinates.

$$\overrightarrow{PQ}=⟨x_2-x_1,y_2-y_1⟩$$

Vector Addition and Subtraction

Vectors are added or subtracted by performing the operation on their corresponding components.

• Addition:$$\vec{u}+\vec{v}=⟨u_1+v_1,u_2+v_2⟩$$

• Subtraction:$$\vec{u}-\vec{v}=⟨u_1-v_1,u_2-v_2⟩$$

Scalar Multiplication

To multiply a vector by a scalar, multiply each component of the vector by the scalar.

$$k\vec{v}=k⟨v_1,v_2⟩=⟨k{v}_1,k{v}_2⟩$$

Magnitude (Norm)

The magnitude, or norm, of a vector is its length. It is calculated using the Pythagorean theorem on its components.

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

Dot Product

The dot product of two vectors is a scalar value, not a vector. It is found by multiplying the corresponding components and then summing the results.

$$\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2$$

Angle Between Two Vectors

The angle $\theta$ between two nonzero vectors $\vec{u}$ and $\vec{v}$ is found using the dot product and the magnitudes of the vectors.

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

To find the angle $\theta$, you must take the inverse cosine of the resulting value:

$$\theta =\arccos \left(\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}\right)$$

Understanding the Relationship Between Dot Product, Magnitude, and Angle

The formula for the angle between two vectors, $\cos (\theta )=\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$, is a central concept that connects three distinct vector calculations. It's crucial to understand how these pieces fit together.

1. The Dot Product ($\vec{u}\cdot \vec{v}$): The dot product itself is a simple calculation, but its significance lies in what it reveals about the relationship between two vectors. The result is a single number (a scalar). This scalar value is a measure of how much one vector "points in the same direction" as the other. A positive dot product indicates an acute angle between the vectors, a negative dot product indicates an obtuse angle, and a dot product of zero indicates the vectors are perpendicular (at a $90^{\circ }$ angle).

2. The Magnitudes ($|\vec{u}|$ and $|\vec{v}|$): The magnitudes are the lengths of the vectors. In the angle formula, the product of the magnitudes in the denominator serves as a normalization factor. It scales the dot product so that the final ratio, $\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$, will always be a value between -1 and 1, inclusive. This is necessary because the range of the cosine function is $[-1,1]$. Without dividing by the magnitudes, the dot product alone would not give us the cosine of the angle.

In essence, the formula isolates the directional relationship (the cosine of the angle) between two vectors by taking their dot product and dividing out the influence of their individual lengths.

Core Concepts & Rules

• Vector Definition: A vector is a directed line segment defined by a magnitude (length) and a direction.

• Component Form: A vector is represented algebraically by its components, $⟨v_x,v_y⟩$, which denote the horizontal and vertical change from its initial to its terminal point.

• Finding Components: To find the component form from an initial point $(x_1,y_1)$ to a terminal point $(x_2,y_2)$, calculate $⟨x_2-x_1,y_2-y_1⟩$.

• Vector Operations are Component-wise: Vector addition, subtraction, and scalar multiplication are all performed by applying the operation to the corresponding components of the vector(s).

• Magnitude is Length: The magnitude of a vector, denoted $|\vec{v}|$, is its length and is found using the formula $|\vec{v}|=\sqrt{v_1^2+v_2^2}$. The result is always a non-negative scalar.

• Dot Product is a Scalar: The dot product of two vectors, $\vec{u}\cdot \vec{v}$, results in a single numerical value (a scalar), calculated as $u_1{v}_1+u_2{v}_2$. It does not produce a new vector.

• Angle Calculation: The angle $\theta$ between two vectors is determined by the relationship between their dot product and their magnitudes. The formula $\cos (\theta )=\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$ is the key to finding this angle.

Step-by-Step Example 1: Basic Vector Operations

Let vector $\vec{a}=⟨-2,5⟩$ and vector $\vec{b}$ be defined by an initial point $P(1,3)$ and a terminal point $Q(4,-1)$.

Task: Find the following:

1. The component form of vector $\vec{b}$.

2. The vector $\vec{c}=2\vec{a}-\vec{b}$.

3. The magnitude of vector $\vec{a}$, denoted $|\vec{a}|$.

Step 1: Find the component form of vector $\vec{b}$

Use the component form formula $⟨x_2-x_1,y_2-y_1⟩$ with initial point $(1,3)$ and terminal point $(4,-1)$.

$$\vec{b}=⟨4-1,-1-3⟩$$

$$\vec{b}=⟨3,-4⟩$$

Step 2: Calculate the vector $\vec{c}=2\vec{a}-\vec{b}$

First, perform the scalar multiplication $2\vec{a}$.

$$2\vec{a}=2⟨-2,5⟩=⟨2(-2),2(5)⟩=⟨-4,10⟩$$

Next, subtract the components of $\vec{b}$ from the components of $2\vec{a}$.

$$\vec{c}=2\vec{a}-\vec{b}=⟨-4,10⟩-⟨3,-4⟩$$

$$\vec{c}=⟨-4-3,10-(-4)⟩$$

$$\vec{c}=⟨-7,14⟩$$

Step 3: Calculate the magnitude of vector $\vec{a}$

Use the magnitude formula $|\vec{v}|=\sqrt{v_1^2+v_2^2}$ with $\vec{a}=⟨-2,5⟩$.

$$|\vec{a}|=\sqrt{(-2{)}^2+5^2}$$

$$|\vec{a}|=\sqrt{4+25}$$

$$|\vec{a}|=\sqrt{29}$$

Step-by-Step Example 2: Exam-Style Application

Task: An airplane's displacement is represented by the vector $\vec{u}=⟨10,25⟩$. The wind's effect on the plane is represented by the vector $\vec{v}=⟨-3,-8⟩$. Find the angle $\theta$ between the airplane's intended displacement and the wind vector, to the nearest tenth of a degree.

Step 1: Calculate the dot product $\vec{u}\cdot \vec{v}$

Use the dot product formula $\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2$.

$$\vec{u}\cdot \vec{v}=(10)(-3)+(25)(-8)$$

$$\vec{u}\cdot \vec{v}=-30+(-200)$$

$$\vec{u}\cdot \vec{v}=-230$$

Step 2: Calculate the magnitude of each vector, $|\vec{u}|$ and $|\vec{v}|$

Calculate the magnitude of $\vec{u}=⟨10,25⟩$.

$$|\vec{u}|=\sqrt{10^2+25^2}=\sqrt{100+625}=\sqrt{725}$$

Calculate the magnitude of $\vec{v}=⟨-3,-8⟩$.

$$|\vec{v}|=\sqrt{(-3{)}^2+(-8{)}^2}=\sqrt{9+64}=\sqrt{73}$$

Step 3: Substitute the values into the angle formula

Use the formula $\cos (\theta )=\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$.

$$\cos (\theta )=\frac{-230}{\sqrt{725}\cdot \sqrt{73}}$$

Step 4: Solve for $\theta$ using the inverse cosine

Now, calculate the value of the fraction and apply the inverse cosine function to find the angle.

$$\cos (\theta )\approx \frac{-230}{(26.9258)(8.544)}\approx \frac{-230}{230.09}\approx -0.9996$$

$$\theta =\arccos (-0.9996)$$

$$\theta \approx 178.4^{\circ }$$

The angle between the airplane's displacement vector and the wind vector is approximately $178.4^{\circ }$.

Using Your Calculator

For vector problems, a graphing calculator is primarily used for accurate numerical computations, especially when finding magnitudes and the final angle. The core vector operations (component form, addition, dot product) are typically straightforward to compute by hand.

Let's focus on the final, most critical calculator step: finding the angle from the cosine value, as in Example 2.

Problem: Find $\theta$ if $\cos (\theta )=\frac{-230}{\sqrt{725}\sqrt{73}}$.

TI-84 Style Steps:

1. Set the Mode: First, determine if the question asks for the angle in degrees or radians. Press the [MODE] key. Navigate down to the RADIAN DEGREE line and select DEGREE. Press [2nd][MODE](QUIT) to return to the home screen. 2. **Enter the Expression:** Use the inverse cosine function, $cos⁻¹(. This is accessed by pressing [2nd]` $[COS]$.

2. Input the Fraction: Carefully enter the fraction from the formula. Use parentheses to ensure the entire denominator is calculated before the division occurs.

   • Press `[2nd][COS]to get $cos⁻¹(.

   • Enter the numerator: $-230$.

   • Press the division key: $÷$.

   • Open a parenthesis for the denominator: $($.

   • Enter the first magnitude: [2nd]$[x^2]$ (for √) $725$$)$.

   • Press the multiplication key: $\times$.

   • Enter the second magnitude: [2nd]$[x^2]$$73$$)$.

   • Close the parenthesis for the denominator: $)$.

   • Close the parenthesis for the $co{s}^{-1}$ function: $)$.

   Your screen should look like this:

   $co{s}^{-1}(-230/(\surd (725)\ast \surd (73)))$

3. Calculate: Press `[ENTER]. The calculator will display the result, which should be approximately $178.4.

Checking Your Work: You can also use the calculator to compute the intermediate values (dot product, magnitudes) to reduce the chance of arithmetic errors in your handwritten work.

AP Exam Quick Hit

Common Question Types

• Component Form and Magnitude from Points: You will be given the initial and terminal points of a vector and asked to find its component form and/or its magnitude.

  • Example: "Vector $\vec{w}$ has an initial point at $(-5,2)$ and a terminal point at $(-1,-1)$. What is the component form of $\vec{w}$?"

• Linear Combinations of Vectors: You will be given two or more vectors in component form and asked to find a new vector that is the result of scalar multiplication and addition/subtraction.

  • Example: "Given $\vec{u}=⟨8,-2⟩$ and $\vec{v}=⟨1,3⟩$, find the vector $3\vec{v}-\vec{u}$."

• Angle Between Two Vectors: You will be given two vectors (either in component form or defined by points) and asked to calculate the angle between them. This is a comprehensive question that tests multiple skills.

  • Example: "Calculate the measure of the angle $\theta$ between the vectors $\vec{a}=⟨4,1⟩$ and $\vec{b}=⟨-2,3⟩$."

Common Mistakes

• Incorrect Component Form Calculation: Reversing the subtraction order for component form, calculating $⟨x_1-x_2,y_1-y_2⟩$ instead of the correct $⟨x_2-x_1,y_2-y_1⟩$. Always remember: Terminal minus Initial.

• Sign Errors with Magnitude: When squaring a negative component for the magnitude calculation, students may incorrectly write $(-4{)}^2=-16$ instead of the correct $16$. Remember that squaring any non-zero real number results in a positive value.

• Confusing Dot Product and Vector Multiplication: The dot product results in a scalar (a single number). A common mistake is to write the result as a new vector, e.g., for $⟨2,3⟩\cdot ⟨4,5⟩$, writing the answer as $⟨8,15⟩$ instead of the correct scalar value $2(4)+3(5)=8+15=23$.

• Stopping Before the Final Step: When finding the angle between vectors, a frequent error is to calculate the value of $\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$ and report this decimal as the final answer, forgetting to take the inverse cosine ($arccos$ or $co{s}^{-1}$) to find the actual angle $\theta$.

• Calculator in the Wrong Mode: Calculating the angle in radians when the question specifically asks for degrees, or vice-versa. Always check the mode on your calculator before performing inverse trigonometric calculations.