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We are able to use vectors to mathematically represent physical quantities that have both magnitude and direction. These quantities can be added in such a way that takes into account both direction and magnitude. Position, displacement, velocity, acceleration, force, momentum and torque are just some of the physical quantities that can be represented using vectors.
Properties of a Vector
A vector, of notation ${\vec A}$, is a quantity that has both a direction and a magnitude, ${|\vec A|}$ . Graphically, a vector is depicted as an arrow with the length equivalent to the magnitude and an arrow illustrating the direction.
Vector Addition
Given two vectors, say ${\vec A}$ and ${\vec B}$, we can define a resultant vector ${\vec C}$ using vector addition so that:
\begin{equation*}
{\vec C=\vec A+\vec B}
\end{equation*}
We can illustrate the addition of two vectors by connecting the tails of the ${\vec A}$ and ${\vec B}$ vectors, then drawing the resultant vector ${\vec C}$ in one of two ways depending on the direction of each vector. If both the vectors go in the same general direction, then to draw the resultant vector ${\vec C}$, we start at the arrow end of ${\vec A}$ and draw towards the arrow end of ${\vec B}$ so that:
If on the other hand, vectors ${\vec A}$ and ${\vec B}$ go in different directions, then to determine the resultant vector ${\vec C}$, we need to first establish a dotted line perpendicular to each vector. The point where the dotted lines cross is the end point of our resultant vector.
When discussing Vector addition, there are four unique properties that must be noted:
Commutivity: When adding vectors, it does not matter what order you put them in, such that:
\begin{equation*}
{\vec A+\vec B=\vec B+\vec A}
\end{equation*}
Associativity: If you are adding three vectors, it does not matter which two vectors you start with in your analysis, such that:
\begin{equation*}
{(\vec A+\vec B)+\vec C=\vec A+(\vec B+\vec C)}
\end{equation*}
Identify Element: For vector addition, there is always a unique vector, ${\vec O}$, known as the identity element. This element creates an instance where for all ${\vec A}$:
\begin{equation*}
{\vec A+\vec O=\vec O+\vec A=\vec A}
\end{equation*}
Inverse element: For vector addition, for all ${\vec A}$, there is always a unique inverse vector, such that:
\begin{equation*}
{(-1)\vec A=-\vec A}
\end{equation*}
so:
\begin{equation*}
{\vec A+(-\vec A)=\vec O}
\end{equation*}
Both vectors have the same magnitude, $|\vec A|=|-\vec A|$, but face in opposite directions:
Scalar Multiplication of Vectors
All vectors can be multiplied by a scalar, which can be any real number. Say c is the scalar that is set to multiply against a given vector ${\vec A}$, such that:
\begin{equation*}
{c\vec A}
\end{equation*}
The magnitude of this new vector is the scalar multiplied by the magnitude of the original vector so that:
\begin{equation*}
{c|A|=|A|c}
\end{equation*}
The direction of this vector is dependent on the sign of the scalar. A negative sign will flip the direction of the original vector, where as a positive sign will maintain the original direction.
Some common properties to note with scalars:
Associativity: If you are given two scalars and one vector, it doesn’t matter what order you complete the multiplication, such that:
\begin{equation*}{b(cA)=(\normalsubformula{\text{bc}})A}\end{equation*} or any combination thereof
Distribution: The distribution law can be applied to the multiplication of vectors, such that:
\begin{equation*}
{(b+c)\VEC A=b\VEC A+c\VEC A}
\end{equation*}
Identify Element: For vector multiplication, the unique identity element is always the number 1, such that:
\begin{equation*}
{1\VEC A=\VEC A}
\end{equation*}
Application of Vectors
We can use vectors to represent physical quantities such as displacement, velocity, acceleration, force, impulse, momentum, torque, and angular momentum, to name a few. It must be stressed, though, that the physical formality of the vector we define is very important. In saying this, we can’t add a force vector to a momentum vector, or an acceleration vector to a velocity vector.
A vector can be any point, P, in space which has a unique magnitude and direction. Any vector with the same magnitude and direction as another are equal despite their location in space.
Any vector in the xy coordinate system (z coordinate left out for simplicity sake) can be decomposed in to component vectors, such that:
\begin{equation*}
{\vec A ={\vec A}_{{x}}+{\vec A}_{{y}}}
\end{equation*}
All points in space can be defined by a set of three unit vectors. These vectors, given a scalar variable ${\hat i$, ${\hat j$, and ${\hat k$ relating to the x, y, and z coordinates respectively, each have a magnitude such that ${|\hat{{i}}|=\text{1, }|\hat{{j}}|=\text{1, and }|\hat{{k}}|=1}$.
With the unit vectors defined for a given point, we can now define further the x and y components of a vector. Recall from above that:
\begin{equation*}
{\vec A={\vec A}_{{x}}+{\vec A}_{{y}}}
\end{equation*}
Expanding the x component:
\begin{equation*}
{{\vec A}_{{x}}=A_{{x}}\hat i}
\end{equation*}
Where ${A_{{x}}}$ is the x component of the ${\vec A}$ vector, not the magnitude, and can be positive, negative, or zero.
Doing the same for the x and z component, we get the fundamental form of a vector as:
\begin{equation*}
{\vec A=(A_{{x}},A_{{y}},A_{{z}})=(A_{{x}}\hat{{i}}+A_{{y}}\hat{{j}}+A_{{z}}\hat{{k}})}
\end{equation*}
The magnitude of a vector can be found such that:
\begin{equation*}
{A=\sqrt{A_{{x}^{{2}}}+A_{{y}^{{2}}}+A_{{z}^{{2}}}}}
\end{equation*}
The angle (direction) counterclockwise from the x axis can be determined from:
\begin{equation*}
{A_{{x}}=A\text{cos}\Theta }
\end{equation*}
\begin{equation*}
{A_{{y}}=A\text{sin}\Theta }
\end{equation*}
\begin{equation*}
{\vec A=(A\text{cos}\Theta \hat{{i}}+A\text{sin}\Theta \hat{{j}})}
\end{equation*}
\begin{equation*}
{\Theta =\text{tan}^{{-1}}(\frac{A_{{y}}}{A_{{x}}})}
\end{equation*}
Vector Addition
As discussed above, given two vectors, say ${\vec A}$ and ${\vec B}$, we can define a resultant vector ${\vec C}$ through vector addition:
\begin{equation*}
{\vec C=\vec A+\vec B}
\end{equation*}
Expanding, say ${\vec A}$ and ${\vec B}$ are both vectors in the x-y coordinate system with a certain angle ${\Theta _{{\vec A}}}$ and ${\Theta _{{\vec B}}}$. We can define the vectors as:
\begin{equation*}
{\vec A=(A\text{cos}\Theta _{{\vec A}}\hat{{i}}+A\text{sin}\Theta _{{\vec A}}\hat{{j}})}
\end{equation*}
and
\begin{equation*}
{\vec B=(B\text{cos}\Theta _{{\vec B}}\hat{{i}}+B\text{sin}\Theta _{{\vec B}}\hat{{j}})}
\end{equation*}
Then
\begin{equation*}
\begin{matrix}{C=(A\text{cos}\Theta _{{\vec A}}\hat{{i}}+A\text{sin}\Theta _{{\vec A}}\hat{{j}})+(B\text{cos}\Theta_{{\vec B}}\hat{{i}}+B\text{sin}\Theta _{{\vec B}}\hat{{j}})}\hfill\null \\=C(\text{cos}\Theta _{{\vec C}}+\text{sin}\Theta_{{\vec C}})\hfill\null \end{matrix}\hfill
\end{equation*}
Dot Product
When given two vectors, we can transform them in to a single point of magnitude, known as a scalar, using the vector operation known as the Dot Product.
The dot product of vectors ${\vec A}$ and ${\vec B}$ with angle ${\Theta }$ between them is given by:
\begin{equation*}
{\vec A\cdot {\vec B}=\normalsubformula{\text{AB}}\text{cos}\Theta }
\end{equation*}
Where A and B are the magnitudes of vectors ${\vec A}$ and ${\vec B}$.
There are certain properties and relationships that we must note regarding the Dot Product, they are as follows:
\begin{equation*}
{c\vec A\cdot {\vec B}=c(\vec A\cdot {\vec B})}
\end{equation*}
\begin{equation*}
{(\vec A+B)\cdot {\vec C}=\vec AC\cdot {\vec B}C}
\end{equation*}
\begin{equation*}
{\vec A\cdot {\vec B}=\vec B\cdot {\vec A}}
\end{equation*}
Using the above relationships, we can develop an algebraic expression for the Dot Product in terms of the vector components.
Generally given, for two vectors ${\vec A}$ and ${\vec B}$, the dot product can be written as:
\begin{equation*}
{\vec A\cdot {\vec B}= A_{{x}}B_{{x}}+A_{{y}}B_{{y}}+A_{{z}}B_{{z}}}
\end{equation*}
Example
Determine the angle between:
${\vec A=\langle 1,-4,3\rangle }$ And ${\vec B=\langle 2,5,0\rangle }$
To get ${\Theta }$ we need to use the general equation:
\begin{equation*}
{\vec A\cdot {\vec B}=\normalsubformula{\text{AB}}\text{cos}\Theta }
\end{equation*}
Solving for all the unknowns:
\begin{equation*}
{\vec A\cdot {\vec B}=1(2)+-4(5)+3(0)=\text{18}}
\end{equation*}
\begin{equation*}
{A=\sqrt{1^{{2}}+(-4)^{{2}}+3^{{2}}=\sqrt{\text{26}}}}
\end{equation*}
${B=\sqrt{2^{{2}}+(5)^{{2}}+0^{{2}}=\sqrt{\text{29}}}}$
Plugging in to solve for ${\Theta }$:
\begin{equation*}
{\text{18}=\sqrt{\text{26}}(\sqrt{\text{29})}\text{cos}\Theta }
\end{equation*}
\begin{equation*}
{\Theta =\text{cos}^{{-1}}(\frac{\text{18}}{\sqrt{\text{26}}(\sqrt{\text{29})}})=\text{143}{}^{\circ}}
\end{equation*}
Cross Product
When given two vectors, we can combine them to create a new vector using an operation known as the Cross Product.
The cross product of vectors ${\vec A}$ and ${\vec B}$ with angle ${\Theta }$ between them is given by:
\begin{equation*}
{|\vec Ax\vec B|=\normalsubformula{\text{AB}}\text{sin}\Theta }
\end{equation*}
Where A and B are the magnitudes of vectors ${\vec A}$ and ${\vec B}$.
The following properties and relationships should be noted regarding the Cross Product:
\begin{equation*}
{\vec Ax\vec B=-{\vec B}x\vec A}
\end{equation*}
\begin{equation*}
{(\vec A+\vec B)xC=\vec AxC+\vec BxC}
\end{equation*}
\begin{equation*}
{c\vec Ax\vec B=c(\vec Ax\vec B)}
\end{equation*}
Using the above relationships, we can develop an algebraic expression for the Cross Product in terms of the vector components.
Generally given, for two vectors ${\vec A}$ and ${\vec B}$, the cross product can be written as:
\begin{equation*}
{\vec Ax\vec B=\langle (A_{{y}}B_{{z}}-A_{{z}}B_{{y}}),(A_{{x}}B_{{z}}-A_{{z}}B_{{x}}),(A_{{x}}B_{{y}}-A_{{y}}B_{{x}})\rangle }
\end{equation*}
Example
Compute the Cross Product of:
${\vec A=\langle 1,-4,3\rangle }$ And ${\vec B=\langle 2,5,0\rangle }$
\begin{equation*}
{\vec Ax\vec B=\left[\begin{matrix}i&j&k\\1&-4&3\\2&5&0\end{matrix}\right]}
\end{equation*}
\begin{equation*}
{\vec Ax\vec B=\hat{{i}}|\begin{matrix}-4&3\\5&0\end{matrix}|-\hat{{j}}|\begin{matrix}1&3\\2&0\end{matrix}|+\hat{{k}}|\begin{matrix}1&-4\\2&5\end{matrix}|}
\end{equation*}
\begin{equation*}
{\vec Ax\vec B=\hat{{i}}(-4-\text{15})-\hat{{j}}(0-6)+\hat{{k}}(5-(-8))}
\end{equation*}
\begin{equation*}
{\vec Ax\vec B=-\text{19 \{}\hat{\normalsubformula{{i}}}}+6\hat{{j}}-3\hat{{k}}
\end{equation*}
