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Pythagorean theoremColumn vector

Vector notation

Here you will learn about the magnitude of a vector, including what the magnitude of a vector is and how to calculate it.

Students will first learn about the magnitude of a vector as part of the number system in high school.

The **magnitude of a vector** is the length of a vector. The magnitude of the vector \textbf{a} is written as \lvert \textbf{a} \rvert.

Vector a has two components. There is a horizontal component and a vertical component.

a=\langle x, y\rangle^vector in component form where x represents the horizontal component and y represents the vertical component.

Below is the magnitude of a vector formula. Notice how it incorporates the pythagorean theorem (or the distance formula).

|\textbf{a}|=\sqrt{x^2+y^2}If a vector has a magnitude of 1, it is a unit vector.

For example,

a=\langle 3,4\ranglex=3 (horizontal component)

y=4 (vertical component)

Using the initial point and the terminal point of the vector, the components of the vector make a right triangle. The length of the vector is the hypotenuse of the right triangle.

The magnitude of vector a is \sqrt{4^2+3^2}=\sqrt{25}=5.

Note: That the answer is the absolute value of the square root of the sum of the vector components, since you are solving for the magnitude not the direction of a vector.

The length of the vector is 5.

Note: This page covers only 2 -dimensional vectors.

How does this relate to high school math?

**The Number System (HSN.VM.A.1)**

Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g.,, |*v*|, ||*v*||,*v**v*).

Use this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEIn order to calculate the magnitude of a vector:

**Identify the components of the vector.****Calculate the distance.**

Find the magnitude of vector b, rounding to the nearest tenth:

b=\langle 5,2\rangle**Identify the components of the vector.**

The horizontal component is x=5

The vertical component is y=2

2**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{5^2+2^2}=\sqrt{29}The magnitude of the vector is \sqrt{29}=5.3851…=5.4.

Note: Calculating the magnitude of the vector does not address the direction of the vector.

Find the magnitude of vector v, rounding to the nearest tenth:

v=\langle-1,3\rangle**Identify the components of the vector.**

The horizontal component is x=-1

The vertical component is y=3

**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{(-1)^2+3^2}=\sqrt{10}

The magnitude of the vector is \sqrt{10}=3.1622…=3.2.

Find the magnitude of vector a, rounding to the nearest tenth:

a=\langle-4,5\rangle**Identify the components of the vector.**

The horizontal component is x=-4

The vertical component is y=5

**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{(-4)^2+5^2}=\sqrt{41}

The magnitude of the vector is \sqrt{41}=6.4031…=6.4.

Find the magnitude of vector b, rounding to the nearest tenth:

b=\langle-2,-3\rangle**Identify the components of the vector.**

The horizontal component is x=-2

The vertical component is y=-3

**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{13}

The magnitude of the vector is \sqrt{13}=3.6055…=3.6.

Find the magnitude of vector c, rounding to the nearest tenth:

c=\langle-4,-6\rangle**Identify the components of the vector.**

The horizontal component is x=-4

The vertical component is y=-6

**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}

The magnitude of the vector is \sqrt{52}=7.2111…=7.2.

Find the magnitude of vector v, rounding to the nearest tenth:

v=\langle 2,-7\rangle**Identify the components of the vector.**

The horizontal component is x=2

The vertical component is y=-7

**Calculate the distance.**

Use the distance formula:

\sqrt{x^2+y^2}=\sqrt{2^2+(-7)^2}=\sqrt{53}

The magnitude of the vector is \sqrt{53}=7.2801…=7.3.

- Introduce vectors with worksheets that include their position clearly on the x -axis and y -axis. This helps students make connections to previous learning about cartesian coordinates, right triangles and distance.

- Have students discover the component form of vectors by having them use the distance formula or online platforms such as Desmos.

**Confusing vectors and scalars**

Scalar quantities only involve magnitude, not direction. Vectors involve both.

**Confusing vectors and matrices**

A vector is a list of numbers that can be in row or column form. It is a special form of a matrix. Anything that has more than 1 row and/or column is not a vector, it is a matrix.

**Making mistakes when squaring negative values**

A negative number squared gives a positive answer. If you need to square a negative number using a calculator, it is a good idea to use brackets to make sure that you get the correct answer.

- Vector multiplication
- Component form of a vector
- Subtracting vectors
- Adding vectors
- Vector practice problems

1. Calculate the magnitude of the given vector. Round your answer to the nearest tenth.

\langle 3,8\rangle

8.6

4.8

4.9

8.5

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{3^2+8^2}=\sqrt{73}=8.544…=8.5

2. Calculate the magnitude of the vector. Round your answer to the nearest tenth.

\langle 4,7\rangle

8.0

5.3

8.1

5.2

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{4^2+7^2}=\sqrt{65}=8.062…=8.1

3. Calculate the magnitude of the vector. Round your answer to the nearest tenth.

\langle 1,-3\rangle

3.2

3.1

1.7

1.8

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}=3.162…=3.2

4. Calculate the magnitude of the vector. Round your answer to the nearest tenth.

\langle 2,-4\rangle

4.4

5.1

5.2

4.5

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{2^2+(-4)^2}=\sqrt{20}=4.472…=4.5

5. Calculate the magnitude of the vector. Round your answer to the nearest tenth.

\langle-3,2\rangle

3.7

3.6

2.4

2.5

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{(-3)^2+2^2}=\sqrt{13}=3.605…=3.6

6. Calculate the magnitude of the vector. Round your answer to the nearest tenth.

\langle -1,-4\rangle

4.2

5.1

4.1

5.2

The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{(-1)^2+(-4)^2}=\sqrt{17}=4.123…=4.1

A vector defined in a 3 -dimensional space. For example,

The component form shows the change in x and y that occur in the vector between endpoints. To write a vector in component form, find the change in x and then the change in y.

Is the answer to a vector addition problem, or the sum of at least two vectors.

Yes, the magnitude of a vector can be used to calculate the dot product of two vectors. It is also possible to calculate the cross product of two vectors.**Step-by-step guide:** Vector multiplication

- Lines
- Angles
- Trig Graphs
- Quadratic equations
- Functions in algebra

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