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

##### 17Calculus

Multi-Variable Calculus Learning Path

### Resources

basics of vectors

dot product

Wikipedia - Dot Product

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?

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 $$\vec{w} = 4\vec{i} + 1\vec{j} + 2\vec{k}$$.

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

### PatrickJMT - 1238 video 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 $$\vec{B} = \hat{i} + 3\hat{k}$$.

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

### PatrickJMT - 1245 video 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

### PatrickJMT - 1239 video 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

### Larson Calculus - 1807 video 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

### PatrickJMT - 1246 video 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 $$\langle3,-2\rangle$$.

$$\pi/2$$ 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.

### Larson Calculus - 1808 video solution

video by Larson Calculus

$$\pi/2$$ radians

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The curves $$\vec{r}_1(t) = \langle 5t+2, \ln(t), \sqrt{4-2t-t^2} \rangle$$ and $$\vec{r}_2(s) = \langle 7, \sqrt{2} s^{3/2}-4, e^{s^2-2s} \rangle$$ 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 $$\vec{r}_1(t) = \langle 5t+2, \ln(t), \sqrt{4-2t-t^2} \rangle$$ and $$\vec{r}_2(s) = \langle 7, \sqrt{2} s^{3/2}-4, e^{s^2-2s} \rangle$$ intersect at a point. Find $$\cos \theta$$ where $$\theta$$ is the angle of intersection (the angle between their respective tangent lines).

Solution

### Steve Butler - 4333 video solution

video by Steve Butler

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