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Irony is wasted on the stupid, Oscar Wilde.

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

The point A from where the vector starts is called its initial point, and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as $|\vec{AB}|$, or $||\vec{AB}||$. The arrow indicates the direction of the vector.

Typically, the standard basis vectors are often denoted $\vec{i},\vec{j},\vec{k}$ thus $\vec{A} = a_1\vec{i}+a_2\vec{j}+a_3\vec{k}$ = < a1, a2, a3 > (Figure i).  

Definition. The magnitude or length of the vector $\vec{A}$ is given by $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, 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 $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

Definition. Any vector with magnitude of 1, i.e., $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2} = 1$, is called a unit vector.

Vector Arithmetic

Subtraction is very similar $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a1-b1), (a2+b2), (a3+b3) >, e.g., $\vec{A}-\vec{B} = ⟨2,−1,3⟩ - ⟨−3,4,1⟩ = ⟨5,-5,2⟩$

It may be represented graphically by placing the tail of the arrow b at the head of the arrow a, and then drawing an arrow from the tail of a to the head of b. The new arrow drawn represents the vector a + b and this is called the parallelogram rule because a and b form the sides of a parallelogram and a+b is one of the diagonals.

To subtract $\vec{B}$ from $\vec{A}$, place the tails of $\vec{A}$ and $\vec{B}$ at the same point, and then draw an arrow from the head of $\vec{B}$ to the head of $\vec{A}$.

Intuitively, multiplying a vector by a scalar r stretches it out by a factor of r. Geometrically, this can be visualized as placing r copies of the vector in a line where the endpoint of one vector is the initial point of the next vector. If r is negative, then the vector changes direction or flips around (by an angle of 180°).

The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is defined as follows: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3,$ e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$

Properties: $\vec{u}·(\vec{v}+\vec{w})=\vec{u}·\vec{v}+\vec{u}·\vec{w}, (c\vec{v})·\vec{w} = \vec{v}·(c\vec{w})=c·(\vec{v}·\vec{w}), \vec{v}·\vec{w}=\vec{w}·\vec{v}, \vec{u}·\vec{0} = 0, \vec{u}·\vec{u}=||\vec{u}||^2, if~ \vec{u}·\vec{u} = 0 ⇒ \vec{u} = \vec{0.}$

Theorem. $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$


Proof. Notice that $\vec{A}·\vec{A} = \sum a_ib_i = a_1a_1 + a_2a_2 + a_3a_3 = ||\vec{A}||^2 = ||\vec{A}||^2·cos(0).$

The Law of cosine states: $||\vec{C}||^2 = ||\vec{A}||^2+||\vec{B}||^2-2||\vec{A}||·||\vec{B}||cos(θ)$ (i)

$||\vec{C}||^2 =\vec{C}·\vec{C}=$[Figure v]$(\vec{A}-\vec{B})·(\vec{A}-\vec{B}) = \vec{A}·\vec{A}-\vec{A}·\vec{B}-\vec{B}·\vec{A}+\vec{B}·\vec{B} = ||\vec{A}||^2+||\vec{B}||^2-2\vec{A}·\vec{B}$ (ii)

Combining (i) and (ii), $-2||\vec{A}||·||\vec{B}||cos(θ) = -2\vec{A}·\vec{B} ⇒ 2\vec{A}·\vec{B} = 2||\vec{A}||·||\vec{B}||cos(θ)$ ∎

The dot product between two vectors, say $\vec{a}$ and $\vec{b}$, is based on the projection of one vector onto another. First, let’s replace $\vec{b}$ with the unit vectors that points in the same direction as $\vec{b}$ and call this vector $\vec{u}=\frac{\vec{b}}{||\vec{b}||}$. The reader should realize that $||\vec{u}|| = 1$, i.e., $\vec{u}$ is an unit vector

The dot product of $\vec{a}$ with an unit vector $\vec{u}$, denoted $\vec{a}·\vec{u}$, is defined to be the projection of a in the direction of u, or the amount that a is pointing in the same direction as unit vector u. By forming a right triangle with a and this shadow, you can calculate (by forming a right triangle with the vector $\vec{a}$ and its shadow $cos(θ)=\frac{c_1}{||\vec{a}||} ⇒ c_1 = ||\vec{a}||·cos(θ)$ ) $\vec{a}·\vec{u} = ||\vec{a}||·1·cos(θ) = c_1$ (Figure B). If $\vec{a}$ and $\vec{u}$ were perpendicular, there would be no zero and that corresponds to the case when cos(θ) = cos(π2) = 0, $\vec{a}·\vec{u} = 0.$

Generally, $\vec{a}·\vec{b} =||\vec{a}||·||\vec{b}||·cos(θ) ⇒ |proj_{\vec{b}}\vec{a}| = \frac{||\vec{a}||·||\vec{b}||·cos(θ)}{||\vec{b}||} = ||\vec{a}||·cos(θ), proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$ (Figure C.).

Similarly, $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}$, e.g., the projection of $\vec{b}$ = ⟨2, 1, -1⟩ onto $\vec{a}$ = ⟨1, 0, -2⟩, $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{4}{5}⟨1, 0, 2⟩=⟨\frac{4}{5}, 0, \frac{-8}{5}⟩$.

Solved exercises

Let $\vec{w} = 2\vec{a}-3\vec{b} = 2(-3\vec{i}+4\vec{j})-3(\vec{i}+2\vec{j})=-9\vec{i}+2\vec{j}; \vec{u}=\frac{\vec{w}}{||\vec{w}||} = \frac{-9\vec{i}+2\vec{j}}{\sqrt{85}} = \frac{-9\sqrt{85}}{85}\vec{i}+\frac{2\sqrt{85}}{85}\vec{j}=\frac{\sqrt{85}}{85}(-9\vec{i}+2\vec{j})$

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

$\vec{PQ}·\vec{PR} = ||\vec{PQ}||·||\vec{PQ}||cos(θ) ⇒ cos(θ) = \frac{\vec{PQ}·\vec{PR}}{||\vec{PQ}||·||\vec{PQ}||} =$ [$\vec{PQ}$ represent the displacement from point P to point Q, then its components can be found by subtracting the components of P from the components of Q = ⟨xq −xp, yq −yp, zq −zp⟩. Another way is this $\vec{OP}+\vec{PQ} = \vec{OQ} ⇒ \vec{PQ} = \vec{OQ} - \vec{OP}$] $\frac{⟨-1,1,0⟩·⟨-1, 0, 2⟩}{\sqrt{(-1)^2+1^2+0^2}\sqrt{(-1)^2+0^2+2^2}}=\frac{1+0+0}{\sqrt{2}\sqrt{5}} = \frac{1}{\sqrt{10}}$ ⇒ θ = cos-1($\frac{1}{\sqrt{10}}$) ≈ 71.5°.

The sign of the dot product $\vec{a}·\vec{b}$ depends on the cosine of the angle between the vectors.

  1. If the angle between the vectors is acute (θ less than 90 degrees), the dot product is positive.
  2. If the angle between the vectors is obtuse (θ greater than 90 degrees), the dot product is negative.
  3. If the angle between the vectors is exactly 90 degrees (orthogonal), the dot product is zero.

Let say P is a point defined by the given equation.

We can define two vectors, namely $\vec{OP} = ⟨x, y, x⟩, \vec{A} = ⟨1, 2, 3⟩$. where $\vec{OP}·\vec{A} = 0 ↭ \vec{OP}⟂\vec{A}$. It is a plane through the origin O and perpendicular to $\vec{A}$ (Figure A).

AreaTriangle = $\frac{1}{2}·||\vec{A}||·||\vec{B}||·sin(θ)$

Let $\vec{A’} = \vec{A}$ rotated 90° ⇒[$θ’=\frac{π}{2}-θ$ ↭ θ and θ’ are complementary angles] cos(θ’) = sin(θ).

$||\vec{A}||·||\vec{B}||·sin(θ) = ||\vec{A’}||·||\vec{B}||·cos(θ’) = \vec{A’}·\vec{B} =$[🚀]

Let’s rotate $\vec{A} = ⟨a_1, a_2⟩$ 90° counterclockwise (Figure i) ⇒ $\vec{A’} = ⟨-a_2, a_1⟩$. If we rotate $\vec{A} = ⟨a_1, a_2⟩$ 90° clockwise ⇒ $\vec{A’} = ⟨a_2, -a_1⟩$


$\vec{A’}·\vec{B} =$[🚀] $⟨-a_2, a_1⟩·⟨b_1, b_2⟩ = a_1b_2-a_2b_1 =det(\vec{A}, \vec{B}) = [\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}]$ and its measure ± the area of the parallelogram with sides $\vec{A}~and~ \vec{B}$ (Figure ii).

AreaTriangle = $\frac{1}{2}·|\vec{A}||·||\vec{B}||·sin(θ)| = \frac{1}{2}·|det(\vec{A}, \vec{B})| = \frac{1}{2}·|[\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}]|$


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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