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The dot product is one of two main ways we 'multiply' vectors (the other way is the cross product). We call this 'multiplication' a dot product since we write the dot product using a 'dot' between the vectors. This dot is usually a solid dot like \( \cdot \) instead of an open dot (or circle) like \( \circ \), which is usually used for the composite of two functions.
1. Make a list of all of the properties, identities and unique concepts about the dot product on one sheet of paper. Go through this website, your class notes and your textbook to make sure everything is included on your sheet. Post the sheet above your main study area in your dorm or apartment and refer to it as you are doing your homework. It is best to hand-write the list instead of typing. This will help you remember them as you do your assignments and prepare for exams. Go through and check the list carefully to make sure it is correct.
2. Notice the notation used in your textbook and on this site.
3. Make flashcards (handwritten is best but a system like quizlet will also work) to help prepare for exams. Study your cards on the bus, between classes or while waiting for class to start.
4. Check for additional general study hints on the study techniques page.
If we have vectors, \(\vec{u} = \langle u_1, u_2, u_3 \rangle \) and \(\vec{v} = \langle v_1, v_2, v_3 \rangle \), the dot product is
\( \vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\).
With dot products, we are required to write the dot between \(\vec{u}\) and \(\vec{v}\) to indicate the dot product. Writing the two vectors side by side with nothing in between like this \(\vec{u} \vec{v} \) makes no sense and is incorrect. In two dimensions we can think of \( u_3 = 0 \) and \( v_3 = 0 \) and the above equation holds. Notice the result of the dot product of two vectors is a scalar. This is why the dot product is sometimes called the scalar product.
Okay, so let's watch a video clip showing a quick overview of the dot product. This video is for two dimensional vectors.
video by PatrickJMT
It works the same for three dimensional vectors, shown in this video clip.
video by PatrickJMT
Properties of the Dot Product |
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If we let \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be vectors and \(k\) be a scalar, the following properties hold for the dot product.
\( \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \) |
commutative property | |
\( \vec{u} \cdot ( \vec{v} + \vec{w} ) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \) |
distributive property | |
\( k(\vec{u} \cdot \vec{v} ) = k\vec{u} \cdot \vec{v} = \vec{u} \cdot k\vec{v} \) |
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\(\vec{u} \cdot \vec{0} = 0\) |
\( \vec{0}\) is the zero vector | |
\(\vec{v} \cdot \vec{v} = \norm{\vec{v}} ^2 \) |
The double bars on \( \norm{\vec{v}} \) indicate the magnitude of the vector \(\vec{v}\), usually called the vector norm. See the Vector Representations panel on the main vectors page for detail on the vector norm. | |
Larson Calculus - properties proofs [2mins-34secs]video by Larson Calculus |
Before we go on to look at some applications of the dot product, let's work a few practice problems.
Instructions - - Unless otherwise instructed, calculate the dot product of each set of vectors giving your answers in exact form. For angles, give your answers in radians to 3 decimal places.
\(\vec{a}=\langle 2,5\rangle\), \(\vec{b}=\langle-3,1\rangle\)
Problem Statement |
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Calculate the dot product of the two vectors \(\vec{a}=\langle 2,5\rangle\), \(\vec{b}=\langle-3,1\rangle\)
Solution |
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video by PatrickJMT
close solution |
\( \vec{u} = 2\vec{i} + \vec{j}-\vec{k} \), \( \vec{v} = \vec{i} + 7\vec{j} \)
Problem Statement |
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Calculate the dot product of the two vectors \( \vec{u} = 2\vec{i} + \vec{j}-\vec{k} \), \( \vec{v} = \vec{i} + 7\vec{j} \)
Final Answer |
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\( \vec{u} \cdot \vec{v} = 9 \) |
Problem Statement |
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Calculate the dot product of the two vectors \( \vec{u} = 2\vec{i} + \vec{j}-\vec{k} \), \( \vec{v} = \vec{i} + 7\vec{j} \)
Solution |
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Even though the second vector looks like a two dimensional vector, we can assume that the missing coordinate is zero giving us 2 three dimensional vectors. Otherwise the question would not make sense.
\( \vec{u} \cdot \vec{v} \) |
\( \left( 2\vec{i} + \vec{j} - \vec{k}\right) \cdot \left( \vec{i} + 7\vec{j} + 0\vec{k}\right) \) |
\( 2(1) + 1(7) - 1(0) \) |
\(2 + 7 - 0 = 9 \) |
Final Answer |
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\( \vec{u} \cdot \vec{v} = 9 \) |
close solution |
\( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = \hat{i} + 3\hat{k} \)
Problem Statement |
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Calculate the dot product of the two vectors \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = \hat{i} + 3\hat{k} \)
Final Answer |
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\( (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + 3\hat{k}) = 14 \) |
Problem Statement |
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Calculate the dot product of the two vectors \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = \hat{i} + 3\hat{k} \)
Solution |
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\( \vec{A} \cdot \vec{B} \) |
\( (2\hat{i}+3\hat{j}+4\hat{k}) \cdot (\hat{i}+3\hat{k}) \) |
\( 2(1) + 3(0) + 4(3) = 2+0+12 = 14 \) |
Final Answer |
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\( (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + 3\hat{k}) = 14 \) |
close solution |
\(\vec{u}=2\vec{i}+4\vec{j}-17\vec{k}\), \(\vec{v}=\vec{i}+5\vec{j}+\vec{k}\)
Problem Statement |
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Calculate the dot product of the two vectors \(\vec{u}=2\vec{i}+4\vec{j}-17\vec{k}\), \(\vec{v}=\vec{i}+5\vec{j}+\vec{k}\)
Final Answer |
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\(\vec{u}\cdot\vec{v}=5\) |
Problem Statement |
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Calculate the dot product of the two vectors \(\vec{u}=2\vec{i}+4\vec{j}-17\vec{k}\), \(\vec{v}=\vec{i}+5\vec{j}+\vec{k}\)
Solution |
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\( \vec{u} \cdot \vec{v} \) |
\( \left( 2\vec{i} + 4\vec{j} - 17\vec{k}\right) \cdot \left( \vec{i} + 5\vec{j} + \vec{k} \right) \) |
\( 2(1) + 4(5) - 17(1) \) |
\( 2 + 20 - 17 = 5 \) |
Final Answer |
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\(\vec{u}\cdot\vec{v}=5\) |
close solution |
Applications of the Dot Product |
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Here are a few applications where we can use the dot product to solve specific problems. More applications can be found on the cross products page.
Angle Between Vectors |
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Using the Law of Cosines, we show that \(\displaystyle{ \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \norm{\vec{u}} \norm{\vec{v}} } }\) where \(\theta \) is the angle between the vectors \( \vec{u}\) and \( \vec{v} \).
We can use this equation as an alternate way to calculate the dot product as follows.
\( \vec{u} \cdot \vec{v} = \cos (\theta) \norm{\vec{u}} \norm{\vec{v}} \)
Theorem: Angle Between Two Vectors |
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If \(\theta\) is the angle between two nonzero vectors \(\vec{u}\) and \(\vec{v}\), then \[ \cos\theta = \frac{\vec{u} \cdot \vec{v}}{\norm{\vec{u}} \norm{\vec{v}} }\] |
Proof Overview - - We will use the triangle on the right and the Law of Cosines to prove this theorem.
The Law of Cosines tells us
\( \norm{\vec{a}}^2 = \norm{\vec{b}}^2 + \norm{\vec{c}}^2 - 2\norm{\vec{b}} \norm{\vec{c}} \cos(\theta)
\). In terms of these vectors, our goal equation is
\[ \cos(\theta) = \frac{\vec{b} \cdot \vec{c}}{\norm{\vec{b}} \norm{\vec{c}}}
\]
Since our goal equation does not have \( \norm{\vec{a}} \) in it, let's look at the triangle to see if we can find a relationship between the vector \(\vec{a}\) and the other two vectors.
From basic vector addition, we see from the triangle that \( \vec{c} + \vec{a} = \vec{b} \). Solving for vector \(\vec{a}\), we have \( \vec{a} = \vec{b} - \vec{c} \). We can take the norm and square both sides^{(1)} to get \( \norm{\vec{a}}^2 = \norm{\vec{b} - \vec{c}}^2 \). Let's look closer at the right side of this equation.
\(
\begin{array}{rcl}
\norm{\vec{b} - \vec{c}}^2 & = & \left( \vec{b} - \vec{c} \right) \cdot \left( \vec{b} - \vec{c} \right) \\
& = & \vec{b} \cdot \left( \vec{b} - \vec{c} \right) - \vec{c} \cdot \left( \vec{b} - \vec{c} \right) \\
& = & \vec{b} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} \\
& = & \norm{\vec{b}}^2 - 2\vec{b} \cdot \vec{c} + \norm{\vec{c}}^2
\end{array}
\)
Since this expression is equal to \( \norm{\vec{a}}^2 \), we can substitute this result for \( \norm{\vec{a}}^2 \) in the Law of Cosines equation above.
\(
\begin{array}{rcl}
\norm{\vec{b}}^2 - 2\vec{b} \cdot \vec{c} + \norm{\vec{c}}^2 & = & \norm{\vec{b}}^2 + \norm{\vec{c}}^2 - 2\norm{\vec{b}} \norm{\vec{c}} \cos(\theta) \\
- 2\vec{b} \cdot \vec{c} & = & - 2\norm{\vec{b}} \norm{\vec{c}} \cos(\theta) \\
\cos(\theta) & = & \frac{\vec{b} \cdot \vec{c} }{\norm{\vec{b}} \norm{\vec{c}}} \text{ [qed] }
\end{array}
\)
Notes - - -
(1) Be very careful when you square both sides of an equation. It is strictly true only when both sides are guaranteed to be positive. If one side could be negative, we have lost that information if we need to go back to it later. Since the norm is always positive, we can safely do it here with no fear of losing information.
Now, let's stop and think for a moment.
orthogonal |
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Perpendicular, orthogonal and normal all essentially mean the same thing - meeting at right angles (90 degrees, π/2 radians). But in mathematics we say that two vectors are orthogonal, two lines or planes are perpendicular and a vector is normal to a line or plane. |
What is the dot product when the vectors are orthogonal? |
video by Larson Calculus
Here are some practice problems involving angles between vectors and orthogonal vectors.
Calculate the angle between the vectors \( \vec{v} = 2\vec{i} + 3\vec{j} + 1\vec{k} \) and \( \vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k} \).
Problem Statement |
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Calculate the angle between the vectors \( \vec{v} = 2\vec{i} + 3\vec{j} + 1\vec{k} \) and \( \vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k} \).
Final Answer |
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The angle between the vectors \( \vec{v}=2\vec{i}+3\vec{j}+1\vec{k} \) and \(\vec{w}=4\vec{i}+1\vec{j}+2\vec{k}\) is \(\displaystyle{ \arccos\left(\frac{13}{7\sqrt{6}} \right) \approx 0.710}\) radians. |
Problem Statement |
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Calculate the angle between the vectors \( \vec{v} = 2\vec{i} + 3\vec{j} + 1\vec{k} \) and \( \vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k} \).
Solution |
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To calculate the angle between the two vectors, we use the equation \(\displaystyle{ \cos(\theta ) = \frac{\vec{v} \cdot \vec{w}}{\| \vec{v} \| \| \vec{w} \|} }\).
First, let's find the dot product.
\( \vec{v} \cdot \vec{w} = \left( 2\vec{i} + 3\vec{j} + 1\vec{k} \right) \cdot \left( 4\vec{i} + 1\vec{j} + 2\vec{k} \right) = 2(4) + 3(1) + 1(2) = 13\)
Now we need to find the norm (magnitude) of each of the vectors.
\( \| \vec{v} \| = \sqrt{ 2^2 + 3^2 + 1^2} = \sqrt{4+9+1} = \sqrt{14} \)
\( \| \vec{w} \| = \sqrt({ 4^2 + 1^2 + 2^2} = \sqrt{16+1+4} = \sqrt{21} \)
So now we have \(\displaystyle{ \cos(\theta) = \frac{13}{\sqrt{14}\sqrt{21}} = \frac{13}{\sqrt{2(7)}\sqrt{3(7)}} = \frac{13}{7\sqrt{6}}
}\)
Solving for \(\theta\) we get \(\displaystyle{ \theta = \arccos\left(\frac{13}{7\sqrt{6}} \right) \approx 0.710286
}\)
Final Answer |
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The angle between the vectors \( \vec{v}=2\vec{i}+3\vec{j}+1\vec{k} \) and \(\vec{w}=4\vec{i}+1\vec{j}+2\vec{k}\) is \(\displaystyle{ \arccos\left(\frac{13}{7\sqrt{6}} \right) \approx 0.710}\) radians. |
close solution |
Find the angle between the vectors \( \vec{a} = 6\hat{i} - 2\hat{j} - 3\hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \).
Problem Statement |
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Find the angle between the vectors \( \vec{a} = 6\hat{i} - 2\hat{j} - 3\hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \).
Solution |
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video by PatrickJMT
close solution |
Calculate the angle between \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \).
Problem Statement |
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Calculate the angle between \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \).
Final Answer |
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The angle between \(\vec{A}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{B}=\hat{i}+3\hat{k}\) is \(\displaystyle{\arccos\left( \frac{14}{\sqrt{290}} \right) \approx 0.606 }\) radians. |
Problem Statement |
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Calculate the angle between \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \).
Solution |
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From a previous practice problem, we know that that dot product of these vectors is \(14\). In order to find the angle between the vectors, we can use the formula \(\vec{A} \cdot \vec{B} = \|\vec{A}\| \|\vec{B}\|\cos(\theta)\) where \(\theta\) is the angle between the two vectors.
Let's find the magnitudes of the two vectors.
\(\|\vec{A}\| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}\)
\(\|\vec{B}\| = \sqrt{1^2+0^2+3^2} = \sqrt{1+9} = \sqrt{10}\)
So now we have
\(\displaystyle{ \cos(\theta) = \frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\| \|\vec{B}\|} }\) |
\(\displaystyle{ \cos(\theta) = \frac{14}{(\sqrt{29})(\sqrt{10})} }\) |
\(\displaystyle{ \theta = \arccos\left( \frac{14}{\sqrt{290}} \right) \approx 0.606 }\) |
Final Answer |
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The angle between \(\vec{A}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{B}=\hat{i}+3\hat{k}\) is \(\displaystyle{\arccos\left( \frac{14}{\sqrt{290}} \right) \approx 0.606 }\) radians. |
close solution |
Find the angle between the vectors \( \vec{a} = \langle 1,2,3 \rangle \) and \( \vec{b} = \langle -3,-1,4 \rangle \).
Problem Statement |
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Find the angle between the vectors \( \vec{a} = \langle 1,2,3 \rangle \) and \( \vec{b} = \langle -3,-1,4 \rangle \).
Solution |
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video by PatrickJMT
close solution |
Are the vectors \(\vec{a}=\langle2,4\rangle\) and \(\vec{b}=\langle4,-2\rangle\) orthogonal?
Problem Statement |
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Are the vectors \(\vec{a}=\langle2,4\rangle\) and \(\vec{b}=\langle4,-2\rangle\) orthogonal?
Solution |
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video by PatrickJMT
close solution |
Find the angle between the vectors \( \langle 5,2 \rangle \) and \( \langle 3,4 \rangle \).
Problem Statement |
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Find the angle between the vectors \( \langle 5,2 \rangle \) and \( \langle 3,4 \rangle \).
Solution |
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video by Larson Calculus
close solution |
Find the angle between the vectors \( \vec{a} = \langle 2,3,5 \rangle \) and \( \vec{b} = \langle 1,6,-4 \rangle \).
Problem Statement |
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Find the angle between the vectors \( \vec{a} = \langle 2,3,5 \rangle \) and \( \vec{b} = \langle 1,6,-4 \rangle \).
Solution |
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video by PatrickJMT
close solution |
Find the angle between the vectors \(\langle4,6\rangle\) and \(\langle3,-2\rangle\).
Problem Statement |
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Find the angle between the vectors \(\langle4,6\rangle\) and \(\langle3,-2\rangle\).
Final Answer |
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\(\pi/2\) radians |
Problem Statement |
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Find the angle between the vectors \(\langle4,6\rangle\) and \(\langle3,-2\rangle\).
Solution |
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In the video, he gives his answer in degrees. However, in calculus we usually use radians.
video by Larson Calculus
Final Answer |
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\(\pi/2\) radians |
close solution |
Direction Cosines and Direction Angles |
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Direction angles are the angles between a given vector \(\vec{v}\) and each coordinate axis (usually in three dimensions, so there are three of them). Basically, we use the equation for the angle between vectors to get the direction cosine equations and the direction angles.
For example, to find the direction cosine and the direction angle between a vector \(\vec{v}\) and the x-axis, we have
\(\displaystyle{ \cos(\alpha) = \frac{\vec{v} \cdot \hat{i}}{\norm{\vec{v}} \norm{\hat{i}}} }\)
Let's label the components of \(\vec{v}\) as
\(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k} \)
Since \(\hat{i}\) is the unit vector in the direction of the x-axis, we can write \(\hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}\) and \( \|\hat{i}\| = 1\).
\(\displaystyle{ \cos(\alpha) = \frac{\vec{v} \cdot \hat{i}}{\|\vec{v}\| \|\hat{i}\|} = \frac{v_1}{\|\vec{v}\|} }\)
Similar results can be obtained for the other two angles. Most textbooks and mathematicians use special greek letters for these angles as listed below.
Angle Description | Direction Cosine | Direction Angle | ||
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\(\alpha\) is the angle between \(\vec{v}\) and \(\hat{i}\) | \(\displaystyle{ \cos(\alpha) = \frac{v_1}{\|\vec{v}\|} }\) | \(\displaystyle{ \alpha = \arccos\left(\frac{v_1}{\|\vec{v}\|} \right) }\) | ||
\(\beta\) is the angle between \(\vec{v}\) and \(\hat{j}\) | \(\displaystyle{ \cos(\beta) = \frac{v_2}{\|\vec{v}\|} }\) | \(\displaystyle{ \beta = \arccos\left(\frac{v_2}{\|\vec{v}\|} \right) }\) | ||
\(\gamma\) is the angle between \(\vec{v}\) and \(\hat{k}\) | \(\displaystyle{ \cos(\gamma) = \frac{v_3}{\|\vec{v}\|} }\) | \(\displaystyle{ \gamma = \arccos\left(\frac{v_3}{\|\vec{v}\|} \right) }\) |
Study Hint: Since you already need to know the equation for the angle between two vectors, just remember what the direction cosines and direction angles are. You can derive the equations in the above table from that information. Additionally, if you just memorize the equations, you may not remember what they represent or where they come from. What are you going to do when your instructor asks you to define they mean and where they come from?
Find the direction angle of \( 3 \vhat{i} - 4\vhat{j} \).
Problem Statement |
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Find the direction angle of \( 3 \vhat{i} - 4\vhat{j} \).
Solution |
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video by Larson Calculus
close solution |
Projection and Vector Components |
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Projection
We know from basic trigonometry, the projection of vector \(\vec{A}\) onto vector \(\vec{B}\) (see figure to the right) is just \( \|\vec{A}\| \cos(\theta)\). However, what if we don't know the value of angle \(\theta\)? We can use the dot product to find the projection of one vector onto another. Let's derive the equation.
From the picture on the right, it is easy to see that the magnitude of the projection of vector \(\vec{A}\) onto vector \(\vec{B}\) is
\( \|proj_{\vec{B}} \vec{A} \| = \|\vec{A}\| \cos(\theta)\). Since we don't know the angle \(\theta\), we can use the angle between vectors formula to substitute for \(\cos(\theta)\) as follows.
\(\begin{array}{rcl}
\| proj_{\vec{B}} \vec{A}\| & = & \|\vec{A}\| \cos(\theta) \\
& = & \displaystyle{ \|\vec{A}\| \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|} } \\
& = & \displaystyle{ \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|} }
\end{array}\)
Notice that this result is just the magnitude of the projection of vector \(\vec{A}\) onto \(\vec{B}\). If we want to find the projection vector itself, we can multiply this result by the unit vector in the direction of \(\vec{B}\) and end up with
\(\displaystyle{
proj_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|} \frac{\vec{B}}{\|\vec{B}\|} = \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|^2} \vec{B}
}\)
This last equation is the preferred equation since calculating \(\|\vec{B}\|\) requires taking a square root, so squaring it removes that difficulty.
Here is a great video explaining vector projection in more detail.
video by Dr Chris Tisdell
Vector Component
Now that we have the projection vector, it is a simple matter to find the vector component of \(\vec{A}\) orthogonal to \(\vec{B}\) (the dashed line in the above picture) using basic addition of vectors. Let's call our unknown vector \(\vec{v}\). We know that \( proj_{\vec{B}} \vec{A} + \vec{v} = \vec{A}\). Solving for \(\vec{v}\), we have the vector component of \(\vec{A}\) that is orthogonal to \(\vec{B}\) which is
\( \vec{v} = \vec{A} - proj_{\vec{B}} \vec{A} \).
Study Note - - We highly recommend that you spend some going through this section carefully until you understand it well. When you get to lines and planes, you will need to understand and be able to use this concept.
Notation Note: You may find different notation for the projection vector than we used here. The notation \( proj_{\vec{B}} \vec{A} \) is fairly common but not used exclusively.
Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).
Problem Statement |
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Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).
Final Answer |
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\(\langle 36,13 \rangle\) |
Problem Statement |
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Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).
Solution |
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These videos are part of a longer video and this is just part of the example. You do not need to know the missing parts in order to extract the solution. The explanation is shown in three video clips.
video by Larson Calculus
video by Larson Calculus
video by Larson Calculus
Final Answer |
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\(\langle 36,13 \rangle\) |
close solution |
Calculating Work |
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One of the major applications of the dot product is to calculate work.
Work is defined as the magnitude of a force acting on an object times the distance the object moves. Force is a vector and the only part of the vector that contributes to the work is the part in the direction the object moves. So, if we define a vector \(\vec{d}\) that points in the direction that the object moves whose magnitude is the distance moved from point \(P\) to point \(Q\) (see the figure to the right), the work, \(W\), is
\( W = \| proj_{\vec{d}} \vec{F} \| \| \vec{d} \| \)
Now, the length of the projection of vector \(\vec{F}\) onto vector \(\vec{d}\) is
\( \| proj_{\vec{d}} \vec{F} \| = \|\vec{F}\| \cos(\theta) \)
So the work equation becomes \( W = \|\vec{F}\| \cos(\theta) \| \vec{d} \| = \vec{F} \cdot \vec{d} \).
Here we have shown there are two (equivalent) equations to calculate the work.
\( W = \| proj_{\vec{d}} \vec{F} \| \| \vec{d} \| \)
\( W = \vec{F} \cdot \vec{d} \)
Your next logical step is the vector cross product.