## 17Calculus - Dot Product Application - Angle Between Two Vectors

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If we have two vectors, $$\vec{u}$$ and $$\vec{v}$$, the angle $$\theta$$ between them can be determined from this equation. $\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \norm{\vec{u}} \norm{\vec{v}} }$ (The derivation of this equation, using the Law of Cosines, is shown below.)

orthogonal

Perpendicular, orthogonal and normal all essentially mean the same thing - meeting at right angles (90 degrees, $$\pi/2$$ 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.

We can use this equation as an alternate way to calculate the dot product by solving for the dot product term. $\vec{u} \cdot \vec{v} = \norm{\vec{u}} \norm{\vec{v}} \cos (\theta)$ Now, let's stop and think for a moment.
What is the dot product when the vectors are orthogonal?

Okay, so let's look at how we get the equation of the angle between two vectors.

Theorem: Angle Between Two Vectors

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 shown here 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.

$$\begin{array}{rcl} \norm{\vec{b}}^2 - 2\vec{b} \cdot \vec{c} + \norm{\vec{c}}^2 & = & \norm{\vec{b}}^2 + \norm{\vec{c}}^2 - 2\norm{\vec{b}} \norm{\vec{c}} \cos(\theta) \\ - 2\vec{b} \cdot \vec{c} & = & - 2\norm{\vec{b}} \norm{\vec{c}} \cos(\theta) \\ \cos(\theta) & = & \displaystyle{ \frac{\vec{b} \cdot \vec{c} }{\norm{\vec{b}} \norm{\vec{c}}} } \text{ [qed](2) } \end{array}$$

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.
(2) The letters qed are placed at the end of proofs to indicate that we have proven what we set out to prove. It is from the Greek for quod erat demonstrandum [from wikipedia Q.E.D.].

Here is a video proof that will help you understand this better.

### Larson Calculus - proof of orthogonal vector [2mins-28secs]

video by Larson Calculus

Time for some practice problems involving angles between vectors and orthogonal vectors.

Practice

Unless otherwise instructed, find the angle between the given vectors. Give you answers in exact form.

$$\vec{v} = 2\vec{i} + 3\vec{j} + 1\vec{k}$$, $$\vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k}$$

Problem Statement

Calculate the angle between the vectors $$\vec{v} = 2\vec{i} + 3\vec{j} + 1\vec{k}$$ and $$\vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k}$$.

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

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

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 $$\displaystyle{ \theta = \arccos\left(\frac{13}{7\sqrt{6}} \right) \approx 0.710286 }$$

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.

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$$\vec{a} = 6\hat{i} - 2\hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} + \hat{k}$$.

Problem Statement

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

### 1238 video

video by PatrickJMT

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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 angle between $$\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$ and $$\vec{B} = \hat{i} + 3\hat{k}$$.

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

Calculate the angle between $$\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$ and $$\vec{B} = \hat{i} + 3\hat{k}$$.

Solution

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

 $$\displaystyle{ \cos(\theta) = \frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\| \|\vec{B}\|} }$$ $$\displaystyle{ \cos(\theta) = \frac{14}{(\sqrt{29})(\sqrt{10})} }$$ $$\displaystyle{ \theta = \arccos\left( \frac{14}{\sqrt{290}} \right) \approx 0.606 }$$

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.

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$$\vec{a} = \langle 1,2,3 \rangle$$, $$\vec{b} = \langle -3,-1,4 \rangle$$.

Problem Statement

Find the angle between the vectors $$\vec{a} = \langle 1,2,3 \rangle$$ and $$\vec{b} = \langle -3,-1,4 \rangle$$.

Solution

### 1245 video

video by PatrickJMT

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Are the vectors $$\vec{a}=\langle2,4\rangle$$ and $$\vec{b}=\langle4,-2\rangle$$ orthogonal?

Problem Statement

Are the vectors $$\vec{a}=\langle2,4\rangle$$ and $$\vec{b}=\langle4,-2\rangle$$ orthogonal?

Solution

### 1239 video

video by PatrickJMT

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$$\langle 5,2 \rangle$$, $$\langle 3,4 \rangle$$.

Problem Statement

Find the angle between the vectors $$\langle 5,2 \rangle$$ and $$\langle 3,4 \rangle$$.

Solution

### 1807 video

video by Larson Calculus

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$$\vec{a} = \langle 2,3,5 \rangle$$, $$\vec{b} = \langle 1,6,-4 \rangle$$.

Problem Statement

Find the angle between the vectors $$\vec{a} = \langle 2,3,5 \rangle$$ and $$\vec{b} = \langle 1,6,-4 \rangle$$.

Solution

### 1246 video

video by PatrickJMT

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$$\langle4,6\rangle$$, $$\langle3,-2\rangle$$.

Problem Statement

Find the angle between the vectors $$\langle4,6\rangle$$ and $$\langle3,-2\rangle$$.

$$\pi/2$$ radians

Problem Statement

Find the angle between the vectors $$\langle4,6\rangle$$ and $$\langle3,-2\rangle$$.

Solution

In the video, he gives his answer in degrees. However, in calculus we usually use radians.

### 1808 video

video by Larson Calculus

$$\pi/2$$ radians

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You CAN Ace Calculus

 basics of vectors dot product

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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 Theorem: Angle Between Vectors Practice

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