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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?
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DOWNLOAD FREEIn order to calculate the magnitude of a vector:
Find the magnitude of vector b, rounding to the nearest tenth:
b=\langle 5,2\rangleThe horizontal component is x=5
The vertical component is y=2
2Calculate 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}=5.3851β¦=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\rangleIdentify 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}=3.1622β¦=3.2.
Find the magnitude of vector a, rounding to the nearest tenth:
a=\langle-4,5\rangleIdentify 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}=6.4031β¦=6.4.
Find the magnitude of vector b, rounding to the nearest tenth:
b=\langle-2,-3\rangleIdentify 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}=3.6055β¦=3.6.
Find the magnitude of vector c, rounding to the nearest tenth:
c=\langle-4,-6\rangleIdentify 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}=7.2111β¦=7.2.
Find the magnitude of vector v, rounding to the nearest tenth:
v=\langle 2,-7\rangleIdentify 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}=7.2801β¦=7.3.
1. Calculate the magnitude of the given vector. Round your answer to the nearest tenth.
\langle 3,8\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{3^2+8^2}=\sqrt{73}=8.544β¦=8.5
2. Calculate the magnitude of the vector. Round your answer to the nearest tenth.
\langle 4,7\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{4^2+7^2}=\sqrt{65}=8.062β¦=8.1
3. Calculate the magnitude of the vector. Round your answer to the nearest tenth.
\langle 1,-3\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}=3.162β¦=3.2
4. Calculate the magnitude of the vector. Round your answer to the nearest tenth.
\langle 2,-4\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{2^2+(-4)^2}=\sqrt{20}=4.472β¦=4.5
5. Calculate the magnitude of the vector. Round your answer to the nearest tenth.
\langle-3,2\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{(-3)^2+2^2}=\sqrt{13}=3.605β¦=3.6
6. Calculate the magnitude of the vector. Round your answer to the nearest tenth.
\langle -1,-4\rangle
The magnitude of the vector is \sqrt{x^2+y^2}=\sqrt{(-1)^2+(-4)^2}=\sqrt{17}=4.123β¦=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.
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