Objectives
- Understanding the definition of a vector, commonly used notation, and basic operations from an algebraic point of view.
Section 1.1 Algebraic View
Definition 1.1.1. Vectors in \(\R^2\).
A vector in \(\R^2\) is an ordered pair of real numbers \(\vec{u} = \left[\begin{array}{c} u_1\\u_2 \end{array}\right]\text{.}\)
Definition 1.1.2. Vectors in \(\R^3\).
A vector in \(\R^3\) is an ordered triple of real numbers, \(\vec{u} = \left[\begin{array}{c} u_1\\u_2\\u_3 \end{array}\right]\text{.}\)
Insight 1.1.3.
- We denote vectors by bold lower case, \(\mathbf{u} \text{,}\) or by an arrow over a character \(\vec{u}\text{.}\)
- When talking about vectors we will usually refer to column vectors and write them either as \(\left[\begin{array}{c} 1\\2 \end{array}\right]\) or as a transposed row vector \(\left[ 1, 2 \right]^{\mathsf{T}}\text{.}\)
Definition 1.1.4. Coordinates of a vector.
Given a vector \(\vec{u} = \left[ u_1, u_2 \right]^{\mathsf{T}}\text{,}\) the scalars \(u_1\) and \(u_2\) are called the coordinates or components of the vector.
Subsection 1.1.1 Basic Vector Operations in \(\R^2\)
Given \(\vec{u} = \left[ u_1, u_2\right]^{\T}\text{,}\) \(\vec{v} = \left[ v_1, v_2\right]^{\T}\text{,}\) and scalar \(c \text{,}\) vector addition, subtraction and scalar multiplication are defined as
\begin{equation*}
\vec{u} + \vec{v} = \left[\begin{array}{c} u_1+v_1\\ u_2+v_2 \end{array}\right], \hspace{1cm} \vec{u} - \vec{v} = \left[\begin{array}{c} u_1-v_1\\ u_2-v_2 \end{array}\right], \hspace{1cm} c \vec{u} = \left[\begin{array}{c} c u_1\\ c u_2 \end{array}\right].
\end{equation*}
Subsection 1.1.2 Basic Vector Operations in \(\R^3\)
Given two vectors \(\vec{u} = \left[ u_1, u_2, u_3\right]^{\T}\text{,}\) \(\vec{v} = \left[ v_1, v_2, v_3\right]^{\T}\text{,}\) and scalar \(c \text{,}\) we can define vector addition, subtraction and scalar multiplication as
\begin{equation*}
\vec{u} + \vec{v} = \left[\begin{array}{c} u_1+v_1\\ u_2+v_2\\ u_3+v_3 \end{array}\right], \hspace{1cm} \vec{u} - \vec{v} = \left[\begin{array}{c} u_1-v_1\\ u_2-v_2 \\u_3- v_3 \end{array}\right], \hspace{1cm} c \vec{u} = \left[\begin{array}{c} c u_1\\ c u_2\\ c u_3 \end{array}\right].
\end{equation*}
Subsection 1.1.3 Properties of vector addition and scalar multiplication
Let \(\vec{u}, \vec{v}, \vec{w}\) be vectors in either \(\R^2\) or \(\R^3\) and \(c, d\) scalars,
Properties of vector addition
-
Commutative property.\(\displaystyle \vec{u} + \vec{v} = \vec{v} + \vec{u} \)
-
Associative property.\(\displaystyle \vec{u} + \left( \vec{v} + \vec{w}\right) = \left( \vec{u} + \vec{v} \right) + \vec{w} \)
-
Zero vector, \(\vec{0}= \left[ 0, 0, 0 \right]^{\T} \).\(\displaystyle \vec{u} + \vec{0} = \vec{u}\)
-
Negative vector.\(\displaystyle \vec{u} + \left( -\vec{u}\right) = \vec{0}\)
Properties of scalar multiplication
-
Distributive property 1.\(\displaystyle c\,\left(\vec{u} + \vec{v}\right) = c\,\vec{u} + c\,\vec{v} \)
-
Distributive property 2.\(\displaystyle \left(c+d\right)\,\vec{u} = c\,\vec{u} + d\,\vec{u} \)
-
Distributive property 3.\(\displaystyle c\,\left(d\,\vec{u}\right) = \left(c\,d\right) \vec{u}\)
-
Scalar multiplication by 1.\(\displaystyle 1\,\vec{u} =\vec{u}\)
\(~\)
Checkpoint 1.1.5. Vector addition and scalar multiplication.
Find \(\vec{z} = {2} \vec{u} - {2} \vec{v} + {2} \vec{w} \text{,}\) where
\begin{equation*}
\vec{u} = \left[ \begin{array}{r} {-4} \\ {3} \\ {-4} \end{array}\right], \hspace{0.5cm}
\vec{v} = \left[ \begin{array}{r} {4}\\ {4} \\ {-1} \end{array}\right], \hspace{0.5cm} \text{and} \hspace{0.5cm}
\vec{w} = \left[ \begin{array}{r} {4}\\ {-3} \\ {2} \end{array}\right].
\end{equation*}
\(\displaystyle{}\)
\(\vec{z} = \left[ \right.\) , , \(\left. \right]^{\T}\)
Answer 1.
Answer 2.
Answer 3.
Solution.
\(-8\)
\(-8\)
\(-2\)
\begin{equation*}
\vec{z} = \left[ \begin{array}{r}
({2} \cdot {-4}) - ({2} \cdot {4}) + ({2} \cdot {4})\\
({2} \cdot {3}) - ({2} \cdot {4}) + ({2} \cdot {-3})\\
({2} \cdot {-4}) - ({2} \cdot {-1}) + ({2} \cdot {2})\\
\end{array} \right] =
\left[ \begin{array}{r}
({-8}) - ({8}) + ({8})\\
({6}) - ({8}) + ({-6})\\
({-8}) - ({-2}) + ({4})\\
\end{array} \right] =
\left[ \begin{array}{r}
{-8}\\ {-8} \\ {-2}
\end{array} \right]
\end{equation*}
\(~\)
Definition 1.1.6. Distance Between Two Vectors.
The distance between two vector is given by
\begin{equation*}
\text{distance } = \sqrt{\left(u_1 - v_1\right)^2 + \left(u_2 - v_2\right)^2 +\left(u_3 - v_3\right)^2}.
\end{equation*}
Example 1.1.7. Distance between two vectors.
Find the distance between the two vectors \(\vec{u} = \left[\begin{array}{c} 3\\-1\\4 \end{array}\right]\) and \(\vec{v} = \left[\begin{array}{c} 0\\2\\-2 \end{array}\right].\)
\begin{equation*}
\text{distance } = \sqrt{\left(3 - 0\right)^2 + \left(-1 - 2\right)^2 +\left(4 - (-2)\right)^2}
= \sqrt{9 + 9 +36} = \sqrt{54} = 7.3485.
\end{equation*}
Checkpoint 1.1.8. Distance between two vectors in \(\R^2\).
Find the distance between
\begin{equation*}
\vec{u} = \left[ \begin{array}{r} {-3} \\ {-1} \end{array}\right]
\hspace{0.5cm} \text{ and }\hspace{0.5cm}
\vec{v} = \left[ \begin{array}{r} {1}\\ {1} \end{array}\right].
\end{equation*}
\(\displaystyle{}\)
Distance = .
