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4.8 Vectors

Lesson

Introduction

Learning objectives

  • 4.8.A Identify characteristics of a vector.
  • 4.8.B Determine sums and products involving vectors.
  • 4.8.C Determine a unit vector for a given vector.
  • 4.8.D Determine angle measures between vectors and magnitudes of vectors involved in vector addition.

Characteristics of vectors

A vector is a directed line segment that represents both a direction and a magnitude (or length). Vectors have various applications in physics, engineering, and other fields where quantities have both magnitude and direction.

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When a vector is placed in the plane, the point at the beginning of the line segment is called the tail, and the point at the end of the line segment is called the head.

Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a component.

A vector P_1P_2 with two components can be plotted in the xy-plane from P_1 = \left(x_1, y_1\right) to P_2 = \left(x_2, y_2\right). The vector is identified by a and b, where a = x_2 - x_1 and b = y_2 - y_1. The vector can be expressed as \left(a, b\right). A zero vector \left(0, 0\right) is the trivial case when P_1 = P_2.

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Consider vector V shown in the graph:

P_1=\left(-3,1\right) and P_2=\left(2,4\right).

Vector V is \left(5,3\right).

The magnitude of the vector is the square root of the sum of the squares of the components, which can be calculated using the Pythagorean theorem:

||\left(a,b\right)||=\sqrt{a^2+b^2}

The direction of the vector is parallel to the line segment from the origin to the point with coordinates \left(a,b\right).

For a vector represented geometrically in the plane, the components of the vector can be found using trigonometry. If the angle \theta is known, the components a and b can be calculated using the sine and cosine functions:

a=||\left(a,b\right)|| \cos \theta \\ b=||\left(a,b\right)|| \sin \theta

Examples

Example 1

A vector, A, is defined by the points P_1=\left(2,3\right) and P_2=\left(5,7\right).

a

Find vector A.

Worked Solution
Create a strategy

A vector P_1P_2 where P_1 = \left(x_1, y_1\right) to P_2 = \left(x_2, y_2\right) is identified by a and b, where a = x_2 - x_1 and b = y_2 - y_1.

Apply the idea

Find a and b;

a=x_2-x_1=5-2=3

b=y_2-y_1=7-3=4

Vector A is \left(3,4\right).

b

Determine the magnitude of vector A.

Worked Solution
Create a strategy

The magnitude of a vector is ||\left(a,b\right)||=\sqrt{a^2+b^2}.

Apply the idea
\displaystyle ||\left(a,b\right)||\displaystyle =\displaystyle \sqrt{a^2+b^2}Formula for magnitude
\displaystyle ||\left(3,4\right)||\displaystyle =\displaystyle \sqrt{3^2+4^2}Substitute a=4 and b=3
\displaystyle =\displaystyle \sqrt{9+16}Evaluate the squares
\displaystyle =\displaystyle \sqrt{25}Evaluate the addition
\displaystyle =\displaystyle 5Evaluate the square root

Example 2

Vector C is shown on the graph.

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a

Identify the head and tail of vector C.

Worked Solution
Create a strategy

The head is the starting point of the vector and the tail is at the opposite end of the segment.

Apply the idea

Tail: \left(-2,-2\right)

Head: \left(-4,8\right)

b

Determine the magnitude of vector C.

Worked Solution
Create a strategy

Find a and b where a=x_2-x_1 and b=y_2-y_1.Then use the formula ||\left(a,b\right)||=\sqrt{a^2+b^2}, to find the magnitude of the vector.

Apply the idea

Find a:

a=x_2-x_1=-4-\left(-2\right)=-2

Find b:

b=y_2-y_1=8-\left(-2\right)=10

Calculate the magnitude:

\displaystyle ||\left(a,b\right)||\displaystyle =\displaystyle \sqrt{a^2+b^2}Formula for magnitude
\displaystyle ||\left(-2,10\right)||\displaystyle =\displaystyle \sqrt{\left(-2\right)^2+10^2}Substitute a=-2 and b=10
\displaystyle =\displaystyle \sqrt{4+100}Evaluate the squares
\displaystyle =\displaystyle \sqrt{104}Evaluate the addition
\displaystyle =\displaystyle 2\sqrt{26}Simplify the square root

Example 3

Vector V forms an angle of 45 \degree with the positive x-axis and its magnitude is 10 units. Determine the components of vector V.

Worked Solution
Create a strategy

Find a and b using the formulas a=||\left(a,b\right)|| \cos \theta and b=||\left(a,b\right)|| \sin \theta.

Apply the idea

Find a:

\displaystyle a\displaystyle =\displaystyle ||\left(a,b\right)|| \cos \thetaFormula for a
\displaystyle =\displaystyle 10 \cos 45\degreeSubstitution
\displaystyle =\displaystyle 10 \left(\dfrac{\sqrt{2}}{2}\right)Evaluate \cos 45 \degree
\displaystyle =\displaystyle 5\sqrt{2}Simplify

Find b:

\displaystyle b\displaystyle =\displaystyle ||\left(a,b\right)|| \sin \thetaFormula for b
\displaystyle =\displaystyle 10 \sin 45\degreeSubstitution
\displaystyle =\displaystyle 10 \left(\dfrac{\sqrt{2}}{2}\right)Evaluate \sin 45 \degree
\displaystyle =\displaystyle 5\sqrt{2}Simplify

a=5\sqrt{2} and b=5\sqrt{2}.

Idea summary

A vector is a directed line segment characterized by its components, magnitude, and direction, with a tail at the beginning and a head at the end.

  • Components: a=x_2-x_1 and b=y_2-y_1
  • Magnitude: ||\left(a,b\right)||=\sqrt{a^2+b^2}
  • Components using trigonometry: a=||\left(a,b\right)|| \cos \theta and b=||\left(a,b\right)|| \sin \theta

Sums and products with vectors

When we multiply a vector by a scalar (constant), the result is a new vector. The process involves multiplying the scalar by each of the components of the original vector. The new vector will be parallel to the original vector and will have the same direction if the scalar is positive, or the opposite direction if the scalar is negative. The magnitude of the new vector will be the magnitude of the original vector multiplied by the absolute value of the scalar.

Given a vector v = \left(a, b\right) and a scalar k, the product k v is a new vector kv = \left(ka, kb\right).

To find the sum of two vectors, we add their corresponding components. The resulting vector will have the sum of the original vectors' components as its components.

For example, given vectors u = \left(a_1, b_1\right) and v = \left(a_2, b_2\right), the sum u + v is a new vector w = \left(a_1 + a_2, b_1 + b_2\right).

The dot product (also called the scalar product) of two vectors is a scalar value obtained by summing the products of their corresponding components.

Given vectors u = \left(a_1, b_1\right) and v = \left(a_2, b_2\right), the dot product u \cdot v is a scalar defined as u \cdot v = a_1 \cdot a_2 + b_1 \cdot b_2.

The dot product has several important properties, such as the geometric interpretation, which relates the dot product to the magnitudes of the vectors and the angle between them. It can also be used to determine whether two vectors are perpendicular or to find the projection of one vector onto another. We will explore these properties and applications in subsequent sections.

Examples

Example 4

Given the following vectors:

\langle\overrightarrow{v}\rangle=\langle2,3\rangle \\ \langle\overrightarrow{w}\rangle=\langle1,2\rangle \\ \langle\overrightarrow{q}\rangle=\langle4,6\rangle

a

Find 3 \overrightarrow{v}

Worked Solution
Create a strategy

Given a vector v = \left(a, b\right) and a scalar k, the product k \cdot v is a new vector kv = \left(ka, kb\right).

Apply the idea

\langle3\overrightarrow{v}\rangle=\langle3 \cdot 2,3 \cdot 3\rangle=\langle6,9\rangle

b

Find \overrightarrow{q}+\overrightarrow{w}

Worked Solution
Create a strategy

For vectors u = \left(a_1, b_1\right) and v = \left(a_2, b_2\right), the sum u + v is a new vector w = \left(a_1 + a_2, b_1 + b_2\right).

Apply the idea

\overrightarrow{q}+\overrightarrow{w}=\langle4,6\rangle+\langle1,2\rangle=\langle4+1,6+2\rangle=\langle5,8\rangle

c

Find \overrightarrow{v}\cdot \overrightarrow{w}

Worked Solution
Create a strategy

For vectors u = \left(a_1, b_1\right) and v = \left(a_2, b_2\right), the dot product u \cdot v is a scalar defined as u \cdot v = a_1 \cdot a_2 + b_1 \cdot b_2.

Apply the idea

\overrightarrow{v}\cdot \overrightarrow{w}=\langle2,3\rangle \cdot \langle1,2\rangle=2 \cdot 1 + 3 \cdot 2=2+6=8

Idea summary

Sums and products for vectors u=\left(a_1,b_1\right) and v=\left(a_2,b_2\right):

  • Scalar product: k u =\left(ka_1,kb_1\right)
  • Sum: u+v=\left(a_1+a_2,b_1+b_2\right)
  • Dot product: u \cdot v= a_1 \cdot a_2 + b_1 \cdot b_2

Unit vectors

A unit vector is a vector with a magnitude of 1. Unit vectors are used to represent the direction of a vector without considering its magnitude. In two-dimensional space \Reals^2, there are two standard unit vectors, i and j, which represent the x and y directions, respectively. The unit vector i has the components \left(1,0\right), while the unit vector j has the components \left(0,1\right).

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To find a unit vector in the same direction as a given nonzero vector, we divide the vector by its magnitude. This process is called normalization. The normalization of a vector scales its components so that the resulting vector has a magnitude of 1 while preserving its direction.

For example, given a vector v = \left(a, b\right) with magnitude \left|v\right|, the unit vector u in the same direction as v is obtained by dividing v by its magnitude:

u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle

By expressing a vector as a linear combination of the unit vectors i and j, we can represent any vector in \Reals^2. For a given vector \left(a,b\right), its equivalent representation in terms of i and j is:

\langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}

This representation is useful for performing operations on vectors, such as addition, subtraction, and scalar multiplication, as well as for finding the angle between two vectors or projecting one vector onto another.

Examples

Example 5

Find a unit vector in the direction of the vector V=\langle 1, \sqrt{3} \rangle and write it in terms of i and j.

Worked Solution
Create a strategy

Find the unit vector using u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle then write in the form \langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}.

Apply the idea

Find the magnitude of V:

\displaystyle ||V||\displaystyle =\displaystyle \sqrt{a^2+b^2}Formula for magnitude
\displaystyle =\displaystyle \sqrt{1^2+\sqrt{3}^2}Substitute a=1 and b=\sqrt{3}
\displaystyle =\displaystyle \sqrt{1+3}Evaluate the squares
\displaystyle =\displaystyle \sqrt{4}Add
\displaystyle =\displaystyle 2Evaluate the square root

Find the unit vector:

u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle=\langle \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \rangle

Write using i and j:

\langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}= \dfrac{1}{2} \overrightarrow{i} +\dfrac{\sqrt{3}}{2} \overrightarrow{j}

Example 6

Find a unit vector in the direction of the vector V=\langle -6,8 \rangle and write it in terms of i and j.

Worked Solution
Create a strategy

Find the unit vector using u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle then write in the form \langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}.

Apply the idea

Find the magnitude of V:

\displaystyle ||V||\displaystyle =\displaystyle \sqrt{a^2+b^2}Formula for magnitude
\displaystyle =\displaystyle \sqrt{\left(-6\right)^2+8^2}Substitute a=-6 and b=8
\displaystyle =\displaystyle \sqrt{36+64}Evaluate the squares
\displaystyle =\displaystyle \sqrt{100}Add
\displaystyle =\displaystyle 10Evaluate the square root

Find the unit vector:

u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle=\langle -\dfrac{6}{10}, \dfrac{8}{10} \rangle=\langle -\dfrac{3}{5}, \dfrac{4}{5} \rangle

Write using i and j:

\langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}= -\dfrac{3}{5} \overrightarrow{i} +\dfrac{4}{5} \overrightarrow{j}

Idea summary

The unit vector, u, in the coordinate plane is a vector of length 1.

The unit vector can be found using:

u=\dfrac{\overrightarrow{v}}{||v||}=\langle \dfrac{\overrightarrow{a}}{||v||}, \dfrac{\overrightarrow{b}}{||v||} \rangle

We often represent vectors in terms of the vectors i=\left(1,0\right) and j=\left(0,1\right) using the form:

\langle a,b \rangle=a \overrightarrow{i} +b \overrightarrow{j}

Measures of vectors

In this section, we will learn how to find the angle between two vectors and how to determine side lengths and angle measures of triangles formed by vector addition using the Law of Sines and Law of Cosines.

The dot product of two vectors can be used to find the angle between them. For two nonzero vectors v = \langle a_1, b_1\rangle and w = \langle a_2, b_2 \rangle, their dot product is defined as:

v \cdot w = a_1 \cdot a_2 + b_1 \cdot b_2

The dot product is also equal to the product of the magnitudes of the two vectors and the cosine of the angle \left( \theta \right) between them:

v \cdot w = \left|v\right| \cdot \left|w\right| \cdot \cos \left(\theta \right)

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Using this relationship, we can find the angle between the two vectors by rearranging the equation:

\theta = \arccos\left(\dfrac{v \cdot w }{\left|v\right| \cdot \left|w\right|}\right)

Vector addition can create triangles, and we can use the Law of Sines and Law of Cosines to find side lengths and angle measures of these triangles. Suppose we have two vectors u and v, and their sum is w:

w=u+v

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We can use the Law of Cosines to find the angle between the vectors:

\cos\left(\theta \right) = \dfrac{u \cdot v}{\left|u\right|\cdot \left|v\right|}

For side lengths, we can use the magnitudes of the vectors as the lengths of the triangle's sides. To find the other angles in the triangle, we can apply the Law of Sines:

\dfrac{\sin \left(\alpha \right)}{ \left|u\right|} = \dfrac{sin\left(\beta\right)}{\left|v\right|} = \dfrac{\sin \left(\gamma \right) }{ \left|w \right|}

Here, \alpha , \text{ } \beta, and \gamma are the angles opposite to the sides with lengths \left|u\right|, \text{ } \left|v\right|, and \left|w\right|, respectively.

By using the relationships between the angles and side lengths, we can determine the properties of triangles formed by vector addition, which is useful in various mathematical and real-world applications.

Examples

Example 7

Given vectors A = 4 \overrightarrow{i} + 3\overrightarrow{j} and B = -2 \overrightarrow{i} + 3\overrightarrow{j}, find the angle \theta between these two vectors.

Worked Solution
Create a strategy

To calculate the angle between the two vectors, we can use the formula: \cos\left(\theta \right) = \dfrac{A \cdot B}{\left|A\right|\cdot \left|B\right|}.

Apply the idea

First, let's calculate the dot product A \cdot B:

A \cdot B = 4 \cdot -2 + 3 \cdot 3=-8+9=1

Next, let's find the magnitudes of A and B:

||A||=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5

||B||=\sqrt{\left(-2\right)^2+3^2}=\sqrt{4+9}=\sqrt{13}=5

Now, let's substitute these values into the formula:

\cos \left( \theta \right)=\dfrac{1}{5 \sqrt{13}}

Apply the inverse cosine to solve for \theta:

\theta=\cos ^{-1} \left(\dfrac{1}{5 \sqrt{13}}\right)

Example 8

Given vectors A = 5\overrightarrow{i} + \overrightarrow{j} and B = 3\overrightarrow{i} - 4\overrightarrow{j}, find the resultant vector R and its magnitude.

Worked Solution
Create a strategy

The resultant vector R is found by vector addition.

Apply the idea

R=A+B=\left(5\overrightarrow{i} + \overrightarrow{j}\right)+ \left( 3\overrightarrow{i} - 4\overrightarrow{j} \right)=8\overrightarrow{i} - 3\overrightarrow{j}

The magnitude of R is calculated as:

||R|| = \sqrt{8² + \left(-3\right)²} = \sqrt{64 + 9} = \sqrt{73}

Idea summary
  • Angle between two vectors:

\theta = \arccos\left(\dfrac{v \cdot w }{\left|v\right| \cdot \left|w\right|}\right)

  • Side lengths of vectors:

\dfrac{\sin \left(\alpha \right)}{ \left|u\right|} = \dfrac{sin\left(\beta\right)}{\left|v\right|} = \dfrac{\sin \left(\gamma \right) }{ \left|w \right|}

Outcomes

4.8.A

Identify characteristics of a vector.

4.8.B

Determine sums and products involving vectors.

4.8.C

Determine a unit vector for a given vector.

4.8.D

Determine angle measures between vectors and magnitudes of vectors involved in vector addition.

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