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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 on this separate page.)
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?
Think about it for a minute and then click here for the answer
Time for some practice problems involving angles between vectors and orthogonal vectors.
Practice
Unless otherwise instructed, find the angle between the given vectors. Give your 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 \( \vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k} \).
Final Answer 

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 \(\displaystyle{ \theta = \arccos\left(\frac{13}{7\sqrt{6}} \right) \approx 0.710286 }\)
Final Answer
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
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 \( \vec{B} = \hat{i} + 3\hat{k} \).
Final Answer 

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
\(\displaystyle{ \cos(\theta) = \frac{\vec{A}\cdot\vec{B}}{\\vec{A}\ \\vec{B}\} }\) 
\(\displaystyle{ \cos(\theta) = \frac{14}{(\sqrt{29})(\sqrt{10})} }\) 
\(\displaystyle{ \theta = \arccos\left( \frac{14}{\sqrt{290}} \right) \approx 0.606 }\) 
Final Answer
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
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
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
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
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 \(\langle3,2\rangle\).
Final Answer 

\(\pi/2\) 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.
video by Larson Calculus 

Final Answer
\(\pi/2\) radians
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The curves \( \vec{r}_1(t) = \langle 5t+2, \ln(t), \sqrt{42tt^2} \rangle \) and \( \vec{r}_2(s) = \langle 7, \sqrt{2} s^{3/2}4, e^{s^22s} \rangle \) intersect at a point. Find \(\cos \theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).
Problem Statement
The curves \( \vec{r}_1(t) = \langle 5t+2, \ln(t), \sqrt{42tt^2} \rangle \) and \( \vec{r}_2(s) = \langle 7, \sqrt{2} s^{3/2}4, e^{s^22s} \rangle \) intersect at a point. Find \(\cos \theta\) where \(\theta\) is the angle of intersection (the angle between their respective tangent lines).
Solution
video by Steve Butler 

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