Vectors and Vector Arithmetic for AP Calculus

Practice problems for these concepts can be found at:  Graphs of Functions and Derivatives Practice Problems for AP Calculus

Vectors

A vector represents a displacement of both magnitude and direction. The length, r, of the vector is its magnitude, and the angle, θ, it makes with the x -axis gives its direction. The vector can be resolved into a horizontal and a vertical component.

A unit vector is a vector of magnitude 1. If i = <1, 0> is the unit vector parallel to the positive x -axis, that is, a unit vector with direction angle θ =0, and j = <0, 1> is the unit vector parallel to the y -axis, with an angle , then any vector in the plane can be represented as xi +y j or simply as the ordered pair

Example 1

Find the magnitude and direction of the vector represented by .

Step 1: Calculate the magnitude

Step 2: The terminal point of the vector is in the fourth quadrant. Calculate –.464 radians. This angle falls in quadrant IV.

Example 2

Find the magnitude and direction of the vector represented by .

Step 1: Calculate the magnitude

Step 2: The terminal point of the vector is in the third quadrant. Calculate

Example 3

Find the magnitude and direction of the vector represented by

Step 1: Calculate the magnitude

Step 2: The terminal point of the vector is in the second quadrant. Calculate

Example 4

Find the ordered pair representation of a vector of magnitude 12 and direction x =12 cos

Vector Arithmetic

If C is a constant, r₁ = x₁, y₁ and r₂ = x₂, y₂, then:

Addition: r₁ +r₂ = x₁ +x₂, y₁ + y₂
Subtraction: r₁ – r₂ = x₁ – x₂, y₁ – y₂
Scalar Multiplication: Cr₁ = Cx₁, Cy₁ Note: ||Cr₁|| = ||C|| · ||r₁||
Dot Product: The dot product of two vectors is r₁ · r₂ =||r₁|| · ||r₂|| · cos θ or r₁ · r₂ =x₁x₂ + y₁ y₂.

Parallel and Perpendicular Vectors

If r₂ =Cr₁, then r₁ and r₂ are parallel.

If r₁ · r₂ =0, then r₁ and r₂ are perpendicular or orthogonal.

The angle between two vectors can be found by cos

Example 1

Given r₁ = 4, –7, = r₂ = –3, –2 and r₂ = –1, 5, find 3r₁ – 5r₂ +2r₃. 3r₁ – 5r₂ – 2r₃ =34, –7 – 5–3, –2+2–1, 5= 12, –21 – –15, –10+ –2, 10= 27, –11 – –2, 10 = 29, –21.

Example 2

Determine whether the vectors r₁ = 4, –7 and r₂ = –3, –2 are orthogonal. If the vectors are not orthogonal, approximate the angle between them.

Step 1. Find the dot product r₁ · r₂ = 4(–3)+(–7)(–2)=2. Since the dot product is not equal to zero, the vectors are not orthogonal.

Step 2. If θ is the angle between the vectors, then cos θ = The dot product is 2, ||r1 = ≈ 0.0688 and θ ≈ 1.5019 radians.

Practice problems for these concepts can be found at:  Graphs of Functions and Derivatives Practice Problems for AP Calculus