17Calculus - Unit Vectors
Unit vectors are defined as vectors whose length is exactly one. This means you can use a unit vector to define direction and then assign a length to the vector to get a unique vector.
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The Standard Unit Vectors
There are three standard unit vectors listed in the table below and shown in the figure.
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The Three Standard Unit Vectors | |
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Vector |
Definition |
\(\hat{i}\) |
the unit vector in the direction of the x-axis |
\(\hat{j}\) |
the unit vector in the direction of the y-axis |
\(\hat{k}\) |
the unit vector in the direction of the z-axis |
Every vector in 3-dim space can be described as a linear combination of these three standard unit vectors. You will almost always see the same letters ('i', 'j' and 'k') used to indicate the standard unit vectors. However, you may see them written either as bold letters or with an 'arrow' instead of 'hat', i.e. \( \hat{i} = \vec{i} \).
Note - We recently ran across a book that used a little different notation. This book used \(\hat{x}\), \(\hat{y}\) and \(\hat{z}\). Also, some books use the 'hat' notation to always indicate unit vectors.
You now have two ways to describe a vector, \( \vec{v} = \langle v_1, v_2, v_3 \rangle \) or the same vector can be written \( \vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} \).
Okay, so let's watch a video explaining this in more detail. He goes through an example and explains this whole idea of the standard unit vectors as he works it.
Khan Academy - standard unit vectors [9mins-53secs]
video by Khan Academy |
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Finding a Unit Vector in a Specific Direction
If you have a vector, let's call it \(\vec{v}\), and you want to find a unit vector in the same direction as \(\vec{v}\), you just multiply the vector by the scalar \( 1/ \norm{\vec{v}} \). Said another way, you divide the vector by its length. This gives you the unit vector, \(\vec{u}\), in the same direction as vector \(\vec{v}\).
This is written as \(\displaystyle{ \vec{u} = \frac{\vec{v}}{\norm{\vec{v}}} }\).
You will often see a \(\vec{u}\) used to define a unit vector but not always. So make sure to check the context to determine exactly what is meant and keep track of unit vectors. They are important to vector calculus.
Okay, time for some practice problems. After that, your next topic is the dot product.
Practice
Find the unit vector in the direction of \(\langle 3,4 \rangle\).
Problem Statement |
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Find the unit vector in the direction of \(\langle 3,4 \rangle\).
Solution |
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1233 video
video by PatrickJMT |
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Find the unit vector in the direction of \(\vec{w}=\langle 1/2,1/8 \rangle\).
Problem Statement |
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Find the unit vector in the direction of \(\vec{w}=\langle 1/2,1/8 \rangle\).
Solution |
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1234 video
video by PatrickJMT |
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Find the unit vector in the direction of \(\vec{w}=\langle 0,5 \rangle\).
Problem Statement |
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Find the unit vector in the direction of \(\vec{w}=\langle 0,5 \rangle\).
Solution |
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1235 video
video by PatrickJMT |
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Find the sum of \( 3 \hat{i} + 5 \hat{j} \) and \( 2 \hat{i} - 7 \hat{j} \).
Problem Statement |
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Find the sum of the vectors \( 3 \hat{i} + 5 \hat{j} \) and \( 2 \hat{i} - 7 \hat{j} \).
Solution |
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1218 video
video by Krista King Math |
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Calculate \( \vec{a} + \vec{b} \) for \( \vec{a} = -3\hat{i} + 2\hat{j} \), \( \vec{b} = 2\hat{i} + 4\hat{j} \).
Problem Statement |
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Calculate \( \vec{a} + \vec{b} \) for \( \vec{a} = -3\hat{i} + 2\hat{j} \), \( \vec{b} = 2\hat{i} + 4\hat{j} \).
Solution |
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1236 video
video by Khan Academy |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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