## 17Calculus - Vector Dot Product

Using Vectors

Applications

### Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

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Calculus 1 Practice

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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.

### Dot Product Study Hints

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 $$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$$. Another way to calculate the dot product is $$\vec{u} \cdot \vec{v} = \norm{\vec{u}} \norm{\vec{v}} \cos \theta$$ where $$\theta$$ is the angle between vectors $$\vec{u}$$ and $$\vec{v}$$.
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.

### PatrickJMT - overview of the dot product [1min-37secs]

video by PatrickJMT

It works the same for three dimensional vectors, shown in this video clip.

### PatrickJMT - dot product in three dimensions [23secs]

video by PatrickJMT

Properties of the Dot Product

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

$$k(\vec{u} \cdot \vec{v} ) = k\vec{u} \cdot \vec{v} = \vec{u} \cdot k\vec{v}$$

$$\vec{u} \cdot \vec{0} = 0$$

$$\vec{0}$$ 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.

Practice

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

Calculate the dot product of the two vectors $$\vec{a}=\langle 2,5\rangle$$, $$\vec{b}=\langle-3,1\rangle$$

Solution

### 1237 video

video by PatrickJMT

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$$\vec{u} = 2\vec{i} + \vec{j}-\vec{k}$$, $$\vec{v} = \vec{i} + 7\vec{j}$$

Problem Statement

Calculate the dot product of the two vectors $$\vec{u} = 2\vec{i} + \vec{j}-\vec{k}$$, $$\vec{v} = \vec{i} + 7\vec{j}$$

$$\vec{u} \cdot \vec{v} = 9$$

Problem Statement

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

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.

 $$\vec{u} \cdot \vec{v}$$ $$\left( 2\vec{i} + \vec{j} - \vec{k}\right) \cdot \left( \vec{i} + 7\vec{j} + 0\vec{k}\right)$$ $$2(1) + 1(7) - 1(0)$$ $$2 + 7 - 0 = 9$$

$$\vec{u} \cdot \vec{v} = 9$$

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$$\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{B} = \hat{i} + 3\hat{k}$$

Problem Statement

Calculate the dot product of the two vectors $$\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{B} = \hat{i} + 3\hat{k}$$

$$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + 3\hat{k}) = 14$$

Problem Statement

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

 $$\vec{A} \cdot \vec{B}$$ $$(2\hat{i}+3\hat{j}+4\hat{k}) \cdot (\hat{i}+3\hat{k})$$ $$2(1) + 3(0) + 4(3) = 2+0+12 = 14$$

$$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\hat{i} + 3\hat{k}) = 14$$

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$$\vec{u}=2\vec{i}+4\vec{j}-17\vec{k}$$, $$\vec{v}=\vec{i}+5\vec{j}+\vec{k}$$

Problem Statement

Calculate the dot product of the two vectors $$\vec{u}=2\vec{i}+4\vec{j}-17\vec{k}$$, $$\vec{v}=\vec{i}+5\vec{j}+\vec{k}$$

$$\vec{u}\cdot\vec{v}=5$$

Problem Statement

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

 $$\vec{u} \cdot \vec{v}$$ $$\left( 2\vec{i} + 4\vec{j} - 17\vec{k}\right) \cdot \left( \vec{i} + 5\vec{j} + \vec{k} \right)$$ $$2(1) + 4(5) - 17(1)$$ $$2 + 20 - 17 = 5$$

$$\vec{u}\cdot\vec{v}=5$$

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Applications of the Dot Product

Here is a list of a few applications where we can use the dot product to solve specific problems using the dot product. If you are first learning this material, you may want to study these applications in this order, since later ones depend on knowing the ones before them.

 1. Angle Between Two Vectors 2. Direction Cosines and Direction Angles 3. Vector Projections 4. Work

Here is an interesting video about the real world applications of the dot product.

### Zach Star - The real world applications of the dot product [12mins-48secs]

video by Zach Star

You CAN Ace Calculus

Wikipedia: Work (Physics)

Wikipedia - Dot Product

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

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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 Overview Properties of the Dot Product Practice Applications of the Dot Product

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