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Vectors II. The Dot Product.

All those who seem stupid, they are, and also so are half of those who do not, Quevedo

Recall

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

Vectors II

Theorem. The dot product of two vectors is the product of their magnitudes multiplied by the cosine of the angle between them: $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$

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Proof.

Let’s start by observing that the dot product of a vector with itself, $\vec{A}·\vec{A}$, yields the square of the vector’s magnitude. Mathematically:

$\vec{A}·\vec{A} = \sum a_ia_i = a_1a_1 + a_2a_2 + a_3a_3 = ||\vec{A}||^2$.

This also corresponds to:

$||\vec{A}||^2·cos(0)$ because the cosine of 0 degrees (or radians) is 1.

The Law of cosine relates the lengths of the sides of a triangle to the cosine of one of its angle. For a triangle with sides a, b, and c, c2 = a2 + b2 -2abcos(θ) where θ is the angle opposite to the side c.

Using vectors, if $\vec{C} = \vec{A}-\vec{B}$ (Figure v), then: $||\vec{C}||^2 = ||\vec{A}||^2+||\vec{B}||^2-2||\vec{A}||·||\vec{B}||cos(θ)$ (i)

The square of the magnitude of the resultant vector $||\vec{C}||^2$ equals the sum of the squares of the magnitudes of the original vectors, minus twice the product of their magnitudes and the cosine of the angle between them.

Expanding the definition of the resultant vector $\vec{C}$ and simplifying, we find that$||\vec{C}||^2 =\vec{C}·\vec{C}=(\vec{A}-\vec{B})·(\vec{A}-\vec{B}) = \vec{A}·\vec{A}-\vec{A}·\vec{B}-\vec{B}·\vec{A}+\vec{B}·\vec{B}$ = [Since $\vec{A}·\vec{B} = \vec{B}·\vec{A}, \vec{A}·\vec{A}=||\vec{A}||^2$] $||\vec{A}||^2+||\vec{B}||^2-2\vec{A}·\vec{B}$ (ii)

Combining the equations (i) and (ii), we get $-2||\vec{A}||·||\vec{B}||cos(θ) = -2\vec{A}·\vec{B} $ ⇒[Simplifying we obtain] $2\vec{A}·\vec{B} = 2||\vec{A}||·||\vec{B}||cos(θ)⇒\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ)$ ∎

The dot product between two vectors, say $\vec{a}$ and $\vec{b}$, provides insight on the projection of one vector onto another.

To understand this, let’s replace $\vec{b}$ with the unit vectors that points in the same direction as $\vec{b}$ and call this unit vector $\vec{u}=\frac{\vec{b}}{||\vec{b}||}$.

First, verify that $\vec{u}$ is indeed a unit vector:

$||\vec{u}|| = \sqrt{\frac{b_1^2}{||\vec{b}||^2}+\frac{b_2^2}{||\vec{b}||^2}+\frac{b_3^2}{||\vec{b}||^2}} = \sqrt{\frac{b_1^2+b_2^2+b_3^2}{||\vec{b}||^2}} = \frac{\sqrt{b_1^2+b_2^2+b_3^2}}{||\vec{b}||} = \frac{||\vec{b}||}{||\vec{b}||} = 1$, i.e., $\vec{u}$ is an unit vector.

Vectors

The dot product of $\vec{a}$ with the unit vector $\vec{u}$, denoted as $\vec{a}·\vec{u}$, is the projection of $\vec{a}$ in the direction of $\vec{u}$, or the amount that $\vec{a}$ is pointing (or aligned) in the same direction as the unit vector $\vec{u}$. By forming a right triangle with the vector $\vec{a}$ and this shadow, you can calculate $cos(θ)=\frac{c_1}{||\vec{a}||} ⇒ c_1 = ||\vec{a}||·cos(θ)$.

$\vec{a}·\vec{u} =[\text{Previous theorem}] ||\vec{a}||·1·cos(θ) = c_1$ (Figure B). If $\vec{a}$ and $\vec{u}$ were to be perpendicular vectors, then there would be no shadow and that corresponds to the case when cos(θ) = cos(π2) = 0, $\vec{a}·\vec{u} = 0.$

Generalizing the previous result (where we are not dealing with a unit vector, but an arbitrary vector $\vec{b}$), $\vec{a}·\vec{b} =||\vec{a}||·||\vec{b}||·cos(θ)$ ⇒[Considering $\vec{a}·\vec{b} = \vec{a}·||\vec{b}||·\vec{u}$ and that the projection of $\vec{a}$ onto $\vec{b}$ is independent of the magnitude of $\vec{b}$, but rather depends solely on its direction] $|proj_{\vec{b}}\vec{a}| = ||\vec{a}||·cos(θ) = \frac{||\vec{a}||·||\vec{b}||·cos(θ)}{||\vec{b}||} = \frac{\vec{a}·\vec{b}}{||\vec{b}||},$ the projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$ (Figure C.).

Similarly, the projection of $\vec{b}$ onto $\vec{a}$ is $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}$.

The orthogonal projection of $\vec{a}$ with respect to $\vec{b}$ is $ort_{\vec{b}}\vec{a}=\vec{SR}=\vec{PR}-\vec{PS}=\vec{a}-proj_{\vec{b}}\vec{a}$. This represents the component of $\vec{a}$ that is orthogonal (perpendicular) to $\vec{b}$ (Figure 4).

Dot Product

  1. Projection of $\vec{v} = ⟨3, 4⟩$ onto $\vec{u} = ⟨1, 0⟩, proj_{\vec{u}}\vec{v} = \frac{\vec{v}·\vec{u}}{||\vec{u}||^2}\vec{u} = \frac{3⋅1+4⋅0}{1^0+0^2}⟨1, 0⟩ = \frac{3}{1}⟨1, 0⟩ = ⟨3, 0⟩$.
  2. The projection of $\vec{b}$ = ⟨2, -1, 3⟩ onto $\vec{a}$ = ⟨1, 1, 1⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{1⋅2+1⋅(−1)+1⋅3}{1^2+1^2+1^2}⟨1, 1, 1⟩=\frac{4}{3}⟨1, 1, 1⟩=⟨\frac{4}{3}, \frac{4}{3}, \frac{4}{3}⟩$.
  3. The projection of $\vec{b}$ = ⟨-1, 2, -3⟩ onto $\vec{a}$ = ⟨1, -1, 1⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{-6}{3}⟨1, -1, 1⟩=⟨-2, 2, -2⟩$.
  4. The projection of $\vec{b}$ = ⟨2, 1, -1⟩ onto $\vec{a}$ = ⟨1, 0, -2⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{4}{5}⟨1, 0, -2⟩=⟨\frac{4}{5}, 0, \frac{-8}{5}⟩$.
  5. The projection of $\vec{b}$ = ⟨3, 3, 4⟩ onto $\vec{a}$ = ⟨-1, -2, 2⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{3·(-1)+3·(-2)+2·4}{(-1)^2+(-2)^2+(-2)^2}⟨-1, -2, 2⟩=\frac{-1}{9}⟨-1, -2, 2⟩=⟨\frac{-1}{9}, \frac{2}{9}, \frac{-2}{9}⟩$.
  6. $ort_{\vec{a}}\vec{b}=\vec{b}-proj_{\vec{a}}\vec{b} = ⟨3, 3, 4⟩-⟨\frac{-1}{9}, \frac{2}{9}, \frac{-2}{9}⟩ = ⟨\frac{26}{9}, \frac{25}{9}, \frac{38}{9}⟩$.

Solved exercises

Let $\vec{w} = 2\vec{a}-3\vec{b} = 2(-3\vec{i}+4\vec{j})-3(\vec{i}+2\vec{j})=-9\vec{i}+2\vec{j}$

Let’s normalize $\vec{w}$ to find the unit vector $\vec{u}$. By construction, $\vec{u} || 2\vec{a}-3\vec{b}$ $\vec{u}=\frac{\vec{w}}{||\vec{w}||} = \frac{-9\vec{i}+2\vec{j}}{\sqrt{85}} = \frac{-9\sqrt{85}}{85}\vec{i}+\frac{2\sqrt{85}}{85}\vec{j}=\frac{\sqrt{85}}{85}(-9\vec{i}+2\vec{j})$

Scale $\vec{u}$ to have the magnitude $||\vec{v}||=3$:

$\vec{v}=||\vec{v}||·\vec{u} = 3·\vec{u}=3·\frac{\sqrt{85}}{85}(-9\vec{i}+2\vec{j})$

In three-dimensional space, any vector $\vec{v}$ can be represented as a linear combination of its components along the three coordinate axes (x, y, and z axes in Cartesian coordinate system). These angles (α, β, γ) are called direction angles and the cosines of these angles are called direction cosines, and their formulas are: $cos(α)=\frac{\vec{v}·\vec{i}}{||\vec{v}||} = \frac{v_1}{||\vec{v}||}, cos(β)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_2}{||\vec{v}||}, cos(γ)=\frac{\vec{v}·\vec{k}}{||\vec{v}||} = \frac{v_3}{||\vec{v}||}$ (Figure i).

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Calculate the magnitude of $||\vec{v}||$: $||\vec{v}|| = \sqrt{4 + 1 + 16} = \sqrt{21}$

Find the direction cosines:

$cos(α)=\frac{\vec{v}·\vec{i}}{||\vec{v}||} = \frac{v_1}{||\vec{v}||} = \frac{2}{\sqrt{21}}, cos(β)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_2}{||\vec{v}||} = \frac{1}{\sqrt{21}}, cos(γ)=\frac{\vec{v}·\vec{k}}{||\vec{v}||} = \frac{v_3}{||\vec{v}||}= \frac{-4}{\sqrt{21}}$

Calculate the direction angles:

$α = arccos(\frac{2}{\sqrt{21}}) = 1.119~ \text{radians} = 1.119·\frac{180}{π}=64.11° \text{(degrees),}~ β = arccos(\frac{1}{\sqrt{21}}) = 1.3531~ \text{radians} = 77.396° , γ = arccos(\frac{-4}{\sqrt{21}}) = 2.632~ \text{radians} = 150.794° \text{(degrees).}$

  1.119 radiant ----> x degrees
  π ----> 180 ⇒ x degrees = (1.119·180)/π

In three-dimensional space, any vector $\vec{v}$ can be represented as a linear combination of its components along the three coordinate axes (x, y, and z axes in Cartesian coordinate system). These angles (α, β, γ) are called direction angles and the cosines of these angles are called direction cosines, and their formulas are: $cos(α)=\frac{\vec{v}·\vec{i}}{||\vec{v}||} = \frac{v_1}{||\vec{v}||}, cos(β)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_2}{||\vec{v}||}, cos(γ)=\frac{\vec{v}·\vec{k}}{||\vec{v}||} = \frac{v_3}{||\vec{v}||}$ (Figure i), Here, v1, v2, and v3 are the components of the vector $\vec{v}$ along the x, y, and z axes, respectively.

Calculate the magnitude of $||\vec{v}||$: $||\vec{v}|| = \sqrt{9 + 25 + 16} = 5\sqrt{2}$

Find the direction angles:

$cos(α)=\frac{\vec{v}·\vec{i}}{||\vec{v}||} = \frac{v_1}{||\vec{v}||} = \frac{3}{5\sqrt{2}}⇒ α = arccos(\frac{3}{5\sqrt{2}})≈arccos(0.424)≈ 64.9°$

$cos(β)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_2}{||\vec{v}||} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} ⇒ α = arccos(\frac{1}{\sqrt{2}})≈arccos(0.707) ≈ 45°$

$cos(γ)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_3}{||\vec{v}||} = \frac{-4}{5\sqrt{2}} ⇒ α = arccos(\frac{-4}{\sqrt{2}})≈arccos(−0.566)≈ 124.4°$

Using the dot product formula, the cosine of the angle θ between the vectors is:

$cos(θ) = \frac{\vec{a}·\vec{b}}{||\vec{a}||·||\vec{b}||} = \frac{(3)(−4)+(−1)(0)}{\sqrt{3^2+(-1)^2}·\sqrt{(-4)^2}+0^2} = \frac{-12}{\sqrt{10}·\sqrt{16}}=\frac{-3}{\sqrt{10}}$

To find the angle, we take the arccosine of the result: $θ = arccos(\frac{-3}{\sqrt{10}}) ~\text{radians}≈ 2.82x\frac{180}{π} = 161.6°$ degrees.

The sign of the dot product $\vec{a}·\vec{b} = ||\vec{a}||·||\vec{b}||·cos(θ) = -12$ depends on the cosine of the angle between the vectors.

  1. If the angle between the vectors is acute (0 < θ < $\frac{π}{2}$), the dot product is positive.
  2. If the angle between the vectors is obtuse ($π < θ < \frac{π}{2}$), the dot product is negative. Since θ ≈ 161.6° and the dot product is -12 < 0, our angle θ is obtuse.
  3. If the angle between the vectors is exactly 90 degrees (orthogonal, θ is a right angle), the dot product is zero (Figure iii).

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Understanding work

Work is the measure of energy transfer that occurs when a force acts upon an object to cause its displacement. Mathematically, the work is defined as the product of the force vector $\vec{F}$ and the displacement caused by the force in its own direction.

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When the direction of the force is different from the direction of motion, the work is calculated as the component of the force that acts in the direction of the motion multiplied by the magnitude of the displacement.

The work W done by a force $\vec{F}$ causing a displacement vector $\vec{PQ}$ is given by $W = \vec{F}·\vec{PQ}$

This can be expanded using the magnitudes of the vectors (force and displacement) and the cosine of the angle θ between them: $W = \vec{F}·\vec{PQ}$ where $||\vec{PQ}|| = 50$ feets is the displacement magnitude, $||\vec{F}|| = 8$ lb is the force magnitude, and θ = 55° is the angle between force and displacement.

$W = \vec{F}·\vec{PQ} = ||\vec{F}||·||\vec{PQ}||cos(θ) = 8·50·cos(55°) = 8⋅50⋅0.5736 ≈229.44ft·lb$. So, the work done by the force is approximately 229.44 foot-pounds (ft·lb).

To solve for the forces acting on a 20-pound object sitting on a ramp at an angle of 45° from the horizontal, we need to resolve the weight of the object into two components:

  1. The force pulling the object down the ramp is the parallel component of the weight (this is the part of the gravitational force that makes the book slide down the ramp), $\vec{F_{||}}=W·sin(θ) = 20·sin(45)=20·\frac{\sqrt{2}}{2} = 10·\sqrt{2} ≈ 14.14 $pounds.
  2. The force pressing the object into the ramp is the perpendicular component of the weight. This is calculated using the cosine function: $\vec{F_{⊥}}=W·cos(θ) = 20·cos(45)=20·\frac{\sqrt{2}}{2} = 10·\sqrt{2}≈14.14 $pounds.

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When an object is on an inclined plane, gravity exerts a force straight down. The gravitational force acting on the cart is vertical and downwards, we can represent this force by the vector $\vec{F}=-200\vec{j}$ where $\vec{j}$ is the unit vector in the downward direction.

Gravity tries to pull the cart straight down, but because the ramp is tilted, this downward pull can be thought of as being split into two parts:

  1. One part tries to make the cart slide down the ramp.
  2. The other part presses the cart into the ramp. To keep the cart from sliding, we only need to counteract the part of the force pulling it down the ramp.

We define a unit vector $\vec{v}$ in the direction of the ramp, $\vec{v} = cos(30)\vec{i} + sen(30)\vec{j} = \frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j}$, where $\vec{i}$ and $\vec{j}$ are unit vectors in the horizontal and vertical directions, respectively.

To find the force required to prevent the cart from rolling down the ramp, we need to project the gravitation force $\vec{F}$ onto the unit vector representing (in the direction of) the ramp. This projection gives us the component of the gravitational force acting along the ramp $\vec{v}$. This projection, $w_1 = proj_{\vec{v}}\vec{F} = (\frac{\vec{F}·\vec{v}}{||\vec{v}||^2})·\vec{v} =$[Since $\vec{v}$ is a unit vector, $||\vec{v}|| = 1$] $(\vec{F}·\vec{v})·\vec{v} = (-200\vec{j})·(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})·\vec{v} = -100·\vec{v} = -100(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})$

The magnitude of this force (projection vector) is $||(-100\vec{j})(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})|| = 100(\frac{3}{4}+\frac{1}{4}) = 100·1 = 100$ ⇒ Therefore, a force of 100 pounds is required to keek the cart stationary, that is, to prevent the cart from rolling down the ramp.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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