LINEAR ALGEBRA

Math 204 UL

Sir Calvin A. Gaye

2026-01-17

1 Euclidean Vector Spaces and Matrices

1.1 Vectors in \(\mathbb{R^2}\) and \(\mathbb{R^3}\)

Definition 1.1.1 | \(\mathbb{R^2}\)

  • We let \(\text{R}^{2}\) denote the set of all vectors of the form \(\begin{bmatrix} x_{1} \\ x_{2}\end{bmatrix}\), where \(x_{1}\) and \(x_{2}\) are real numbers called the components of the vector. Mathematically, we write \[R^{2} = \biggl\{ \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} | x_{1}, x_{2} \in \mathbb{R} \biggl\}\]

  • We say two vectors \(\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}\), \(\begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix} \in \mathbb{R}\) are equal if \(x_{1} = y_{1}\) and \(x_{2} = y_{2}\).

  • We write \[ \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix} \]

Definition 1.1.1 cont’d | \(\mathbb{R^2}\)

  • Although we are viewing the elements of \(\mathbb{R^{2}}\) as vectors, we can still interpret these geometrically as points.
Figure 1: Graphical representation of a vector

Definition 1.1.2 | Addition and Scalar Multiplication in \(\mathbb{R^2}\)

  • Let \(\overrightarrow{x} = \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}\), \(\overrightarrow{y} = \begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix} \in \mathbb{R^{2}}\). We define addition of vectors by
    \[\overrightarrow{x} + \overrightarrow{y} = \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} + \begin{bmatrix} y_{1} \\ y_{2} \end{bmatrix} = \begin{bmatrix} x_{1}+y_{1} \\ x_{2}+y_{2} \end{bmatrix}\]

  • We define scalar multiplication of \(\overrightarrow{x}\) by a factor of \(t \in \mathbb{R}\), called a scalar, by \[t\overrightarrow{x} = \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} tx_{1} \\ tx_{2} \end{bmatrix}\]

  • Remark It is important to note that \(\overrightarrow{x} - \overrightarrow{y}\) is to be interpreted as \(\overrightarrow{x} + (-1)\overrightarrow{y}\).

Example 1.1.1 | Solve and graph

  • Let \(\overrightarrow{x} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}\), \(\overrightarrow{y} = \begin{bmatrix} 5 \\ 1 \end{bmatrix} \in \mathbb{R^{2}}\). Calculate \(\overrightarrow{x} + \overrightarrow{y}\).

Solution 1.1.1 | Solve and graph

  • We have \(\overrightarrow{x} + \overrightarrow{y} = \begin{bmatrix} -2 + 5 \\ 3 + 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}\)

Example 1.1.2 | Finding vectors sum

  • Let \(\overrightarrow{u} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}\), \(\overrightarrow{v} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}\), \(\overrightarrow{w} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \in \mathbb{R^{2}}\).

  • Calculate

    1. \(\overrightarrow{u} + \overrightarrow{v}\)
    2. \(3\overrightarrow{w}\) and
    3. \(2\overrightarrow{v} + \overrightarrow{w}\)

Solution 1.1.2(a) | Finding vectors sum

  • We get \(\overrightarrow{u} + \overrightarrow{v} = \begin{bmatrix} 3 - 2 \\ 3 + 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}\)

Example 1.1.2(b) | Finding vectors sum

  • We get \(3\overrightarrow{w} = 3 \begin{bmatrix} 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ -3 \end{bmatrix}\) .

Solution 1.1.2(c) | Finding vectors sum

  • We have \[\begin{align} 2\overrightarrow{v} + \overrightarrow{w} \\ &= 2\begin{bmatrix} -2 \\ 3 \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \end{bmatrix} \\ &= \begin{bmatrix} -4 \\ 6 \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \end{bmatrix} \\ &= \begin{bmatrix} -4 \\ 5 \end{bmatrix} \end{align}\]

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Do It yourself 1.1.1 | Finding vectors sum

  • Let \(\overrightarrow{u} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\), \(\overrightarrow{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\), \(\overrightarrow{w} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \in \mathbb{R^{2}}\).

  • Calculate each of the following and illustrate with a sketch

    1. \(\overrightarrow{u} + \overrightarrow{w}\)
    2. \(-\overrightarrow{v}\) and
    3. \((\overrightarrow{u} + \overrightarrow{w}) - \overrightarrow{v}\)

Do It yourself 1.1.1(a) | Solution for finding vectors sum

  • We get \(\overrightarrow{u} + \overrightarrow{w} = \begin{bmatrix} 1 + 0 \\ -1 + 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)

Do It yourself 1.1.1(b) | Solution finding vectors sum

  • We get \(\overrightarrow{v} = (-1) \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ -1 \end{bmatrix}\) .

Do It yourself 1.1.1(c) | Solution Finding vectors sum

  • We have \((\overrightarrow{u} + \overrightarrow{w}) - \overrightarrow{v}\) \[\begin{align} &= \biggl(\begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}\biggl) + (-1) \begin{bmatrix} 2 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} -2 \\ -1 \end{bmatrix} \\ &= \begin{bmatrix} -1 \\ -1 \end{bmatrix} \end{align}\]

Definition 1.1.3 | Linear Combination

  • Let \(\overrightarrow{v}_{1}, \overrightarrow{v}_{2}, \dots, \overrightarrow{v}_{r} \in \mathbb{R^{2}}\) are vectors of the same size, and if \(c_{1}, c_{2}, \dots, c_{r}\) are scalars, then an expression of the form \(c_{1}\overrightarrow{v}_{1} + c_{2}\overrightarrow{v}_{2} + \dots + c_{r}\overrightarrow{v}_{r}\) is called a linear combination of \(\overrightarrow{v}_{1}, \overrightarrow{v}_{2}, \dots, \overrightarrow{v}_{r}\) with coefficients \(c_{1}, c_{2}, \dots, c_{r}\).

Theorem | 1.1.1

  • For all \(\overrightarrow{w}, \overrightarrow{x}, \overrightarrow{y} \in \mathbb{R^{2}}\) and \(s, t \in \mathbb{R}\) we have:

    • closed under addition: \(\overrightarrow{x} + \overrightarrow{y} \in \mathbb{R^{2}}\)

    • addition is commutative: \(\overrightarrow{x} + \overrightarrow{y} = \overrightarrow{y} + \overrightarrow{x}\)

Theorem | 1.1.1

  • addition is associative: \((\overrightarrow{x} + \overrightarrow{y}) + \overrightarrow{w} = \overrightarrow{x} + (\overrightarrow{y} + \overrightarrow{w})\)

  • zero vector: There exists vector \(\overrightarrow{0} \in \mathbb{R^{2}}\) such that \(\overrightarrow{z} + \overrightarrow{0} = \overrightarrow{z}\) for all \(\overrightarrow{z} \in \mathbb{R^{2}}\)

  • additive inverses: For each \(\overrightarrow{x} \in \mathbb{R^{2}}\) there exists a vector \(-\overrightarrow{x} \in \mathbb{R^{2}}\) such that \(\overrightarrow{x} + (-\overrightarrow{x}) = \overrightarrow{0}\)

  • closed under scalar multiplication: \(s\overrightarrow{x} \in \mathbb{R^{2}}\)

  • scalar multiplication is associative: \(s(t\overrightarrow{x}) =st(\overrightarrow{x})\)

  • a distributive law: \((s + t)\overrightarrow{x} =s\overrightarrow{x} + t\overrightarrow{x}\)

  • scalar multiplicative identity: \(1\overrightarrow{x} =\overrightarrow{x}\)

Definition 1.1.4 | The Vector Equation of a Line in \(\mathbb{R^2}\)

  • A line through the origin in \(\mathbb{R^{2}}\) is a set of the form \[\{t\overrightarrow{d}| t \in \mathbb{R}\}\] Often we do not use formal set notation but simply write a vector equation of the line: \[\overrightarrow{x} = t\overrightarrow{d}, \quad t \in \mathbb{R}\]

  • The non-zero vector \(\overrightarrow{d}\) is called a direction vector of the line.

  • Similarly, we define a line through \(\overrightarrow{p}\) with direction vector \(\overrightarrow{d} \neq \overrightarrow{0}\) to be the set

Definition 1.1.4 cont’d | The Vector Equation of a Line in \(\mathbb{R^2}\)

  • \[ \{\overrightarrow{p} + t\overrightarrow{d}| t \in \mathbb{R}\}\] which has vector equation \[\overrightarrow{x} = \overrightarrow{p} + t\overrightarrow{d}, \quad t \in \mathbb{R}\] This line is parallel to the line with equation \(\overrightarrow{x} = t\overrightarrow{d}, \quad t \in \mathbb{R}\) because of the parallelogram rule for addition.

  • Two lines are parallel if the direction vector of one line is a non-zero scalar multiple of the direction vector of the other line.

Definition 1.1.4 cont’d | The Vector Equation of a Line in \(\mathbb{R^2}\)

Figure 2: The line with vector equation \(\overrightarrow{x} = t\overrightarrow{d} + \overrightarrow{p}, \quad t \in \mathbb{R}.\)

Example 1.1.3 | Vector Equation of a Line

  • Write a vector equation of the line through \(P(1,2)\) parallel to the line with vector equation \(\overrightarrow{x} = t\begin{bmatrix} 2 \\ 3 \end{bmatrix}, \quad t \in \mathbb{R}\)

Solution 1.1.3 | Vector Equation of a Line

  • Since they are parallel, we can choose the same direction vector. Hence, a vector equation of the line is \(\overrightarrow{x} =\begin{bmatrix} 1 \\ 2 \end{bmatrix} + t\begin{bmatrix} 2 \\ 3 \end{bmatrix}, \quad t \in \mathbb{R}\)

Definitin 1.1.5 | Parametric Equations

  • Expanding a vector equation \(\overrightarrow{x} = \overrightarrow{p} + t\overrightarrow{d} \, t \in \mathbb{R}\) we get \[\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} p_{1} \\ p_{2} \end{bmatrix} + t\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix} = \begin{bmatrix} p_{1} + td_{1}\\ p_{2} + td_{2} \end{bmatrix}\] Comparing entries, we get parametric equations on the line: \[\begin{align*} \left\{ \begin {aligned} & x_{1} = p_{1} + td_{1} \\& & t \in \mathbb{R} \\ & x_{2} = p_{2} + td_{2} \quad \end{aligned} \right. \end{align*}\]

Definitin 1.1.6 | Scalar Equation

  • The familiar scalar equation of the line is obtained by eliminating the parameter \(t\). Provided that \(d_{1} \neq 0\) we solve the first equation for \(t\) to get \[\frac{x_{1} - p_{1}}{d_{1}} = t\] Substituting this into the second equation gives the scalar equation \[ x_{2} = p_{2} + \frac{d_{2}}{d_{1}}(x_{1} - p_{1}) \tag{1}\]

Example 1.1.4 | Vector, Parametric and Scalar Equations

  • Write a vector equation, scalar equation, and parametric equations of the line passing through the point \(P(3,4)\) with direction vector \(\begin{bmatrix} -5 \\ 1 \end{bmatrix}\).

Solution 1.1.4 | Vector, Parametric and Scalar Equations

  • A vector equation is \(\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} + t\begin{bmatrix} -5 \\ 1 \end{bmatrix}, t \in \mathbb{R}\).

  • So, the parametric equation are \(\begin{align*} \left\{ \begin {aligned} & x_{1} = 3 - 5t \\& & t \in \mathbb{R} \\ & x_{2} = 4 + t \end{aligned} \right. \end{align*}\)

Solution 1.1.4 cont’d | Vector, Parametric and Scalar Equations

  • Provided that \(d_{1} \neq 0\) we solve the first equation for \(t\) to get \[\frac{-x_{1} + 3}{5} = t\]

  • Substituting this into the second equation gives the scalar equation \[x_{2} = 4 - \frac{1}{5}(x_{1} - 3)\]

Directed Line Segments

  • We denote the directed line segment from point \(P\) to point \(Q\) by \(\overrightarrow{PQ}\) as in Figure 3.
Figure 3: The directed line segment \(\overrightarrow{PQ}\) from \(P\) to \(Q\).

Directed Line Segments

  • For arbitrary points \(Q, R, S,\) and \(T\) in \(\mathbb{R^{2}}\), we define \(\overrightarrow{QR}\) to be equivalent to \(\overrightarrow{ST}\) if they are both equivalent to the same \(\overrightarrow{OP}\) for some \(P\). That is, if \[\overrightarrow{r} - \overrightarrow{q} = \overrightarrow{p}\] and \[\overrightarrow{t} - \overrightarrow{s} = \overrightarrow{p}\] for the same \(\overrightarrow{p}\)

  • We have used one directed line segment \(\overrightarrow{OP}\) starting from the origin in our definition.

Example 1.1.5 | Directed Line Segment

  • For points \(Q(1, 3)\), \(R(6, -1)\), \(S(-2, 4)\), and \(T(3, 0)\), show that \(\overrightarrow{QR}\) is equivalent to \(\overrightarrow{ST}\) and sketch.

Solution 1.1.5 | Directed Line Segment

  • \(\overrightarrow{r} -\overrightarrow{q} = \begin{bmatrix} 6 \\ -1 \end{bmatrix} - \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 5 \\ -4 \end{bmatrix}\)

  • \(\overrightarrow{t} -\overrightarrow{s} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} - \begin{bmatrix} -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 5 \\ -4 \end{bmatrix}\)

  • Therefore \(\overrightarrow{QR} = \overrightarrow{r} -\overrightarrow{q} = \overrightarrow{ST} = \overrightarrow{t} -\overrightarrow{s}\).

Solution 1.1.5 cont’d | Directed Line Segment

Figure 4: The directed line segment \(\overrightarrow{QR} = \overrightarrow{ST}\) .

Example 1.1.6 | Vector Equation

  • Find a vector equation of the line through \(P(1, 2)\) and \(Q(3, -1)\).

Solution 1.1.6 | Vector Equation

  • \(\overrightarrow{PQ} = \overrightarrow{q} -\overrightarrow{p} = \begin{bmatrix} 3 \\ -1 \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}\)

  • Hence, a vector equation of the line with direction \(\overrightarrow{PQ}\) that passes through \(P(1,2)\) is \(\overrightarrow{x} = \overrightarrow{p} + t\overrightarrow{PQ} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} + t\begin{bmatrix} 2 \\ -3 \end{bmatrix}, \, t\in \mathbb{R}\)

Solution 1.1.6 | Vector Equation

Figure 5

Do It yourself 1.1.2 | Finding Vector Equation of a Line

  • Find a vector equation of the line through \(P(1, 1)\) and \(Q(-2, 2)\).

Do It yourself 2 | Solution 2

  • \(\overrightarrow{PQ} = \overrightarrow{q} -\overrightarrow{p} = \begin{bmatrix} -2 \\ 2 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \end{bmatrix}\)

  • Hence, a vector equation of the line with direction \(\overrightarrow{PQ}\) that passes through \(P(1,1)\) is \(\overrightarrow{x} = \overrightarrow{p} + t\overrightarrow{PQ} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} + t\begin{bmatrix} -3 \\ 1 \end{bmatrix}, \, t\in \mathbb{R}\)

Definition | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • We define \(\text{R}^{3}\) to be the set of all vectors of the form \(\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}\), with \(x_{1}\), \(x_{2}, x_{3} \in \mathbb{R}\).

  • Mathematically, we write \[ R^{3} = \biggl\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} | x_{1}, x_{2}, x_{3} \in \mathbb{R} \biggl\} \]

  • We say two vectors \(\overrightarrow{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}\), \(\overrightarrow{y} = \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix}\) are equal and write \(\overrightarrow{x} = \overrightarrow{y}\) if \(x_{i} = y_{i}\), for \(i = 1, 2, 3.\)

Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • Let \(\overrightarrow{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}\), \(\overrightarrow{y} = \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} \in \mathbb{R^{3}}\).

  • We define addition of vectors by \[ \overrightarrow{x} + \overrightarrow{y}= \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} + \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} x_{1} + y_{1} \\ x_{2} + y_{2}\\ x_{3} + y_{3} \end{bmatrix} \]

  • We define the scalar multiplication of \(\overrightarrow{x}\) by a scalar \(t \in \mathbb{R}\) by \[t\overrightarrow{x} = t\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} tx_{1} \\ tx_{2} \\ tx_{3} \end{bmatrix} \].

Example 1.1.7 | Vectors, Lines, and Planes in

  • Let \(\overrightarrow{u} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}\), \(\overrightarrow{v} = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}\) , \(\overrightarrow{w} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \in \mathbb{R^{3}}\) Calculate \(\overrightarrow{v} + \overrightarrow{u}\), \(-\overrightarrow{w}\), and \(-\overrightarrow{v} + 2\overrightarrow{w} - \overrightarrow{u}\).

Solution 1.1.7 | Vectors, Lines, and Planes in

  • \(\overrightarrow{u} + \overrightarrow{v} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\), \(-\overrightarrow{w} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}\), \(-\overrightarrow{v} + 2\overrightarrow{w} - \overrightarrow{u} = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}\),

Example 1.1.8 | Vectors, Lines, and Planes in

  • Find a vector equation and parametric equations of the line that passes through the points \(P(1, 5, -2)\) and \(Q(4,-1,3)\).

Solution 1.1.8 | Vectors, Lines, and Planes in

  • A direction vector is \(\overrightarrow{PQ} = \overrightarrow{q} -\overrightarrow{p} = \begin{bmatrix} 4 \\ -1 \\ 3 \end{bmatrix} - \begin{bmatrix} 1 \\ 5 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ -6 \\ 5 \end{bmatrix}\)

  • Hence, a vector equation of the line with direction \(\overrightarrow{PQ}\) that passes through \(P(1,5,-2)\) is \(\overrightarrow{x} = \overrightarrow{p} + t\overrightarrow{PQ} = \begin{bmatrix} 1 \\ 5 \\ -2 \end{bmatrix} + t\begin{bmatrix} 3 \\ -6\\ 5 \end{bmatrix}, \, t\in \mathbb{R}\)

Solution 1.1.8 cont’d | Vectors, Lines, and Planes in

  • \(\begin{align*} \left\{ \begin {aligned} & x_{1} = 1 + 3t \\ & x_{2} = 5 - 6t & t \in \mathbb{R} \\ & x_{3} = -2 + 5t \end{aligned} \right. \end{align*}\)

Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • Let \(\overrightarrow{u}\) and \(\overrightarrow{v}\) be vectos in \(\mathbb{R^{3}}\) that are not scalar multiples of each other.
  • This implies that the sets \(\{t\overrightarrow{u}|t \in \mathbb{R}\}\) and \(\{s\overrightarrow{v}|t \in \mathbb{R}\}\) are both lines in \(\mathbb{R^{3}}\) through the origin in different directions.
  • Thus, the set of all possible linear combinations of \(\overrightarrow{u}\) and \(\overrightarrow{v}\) forms a two-dimensional plane.

Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • That is, the set \[\{t\overrightarrow{u} + s\overrightarrow{v}|s,t \in \mathbb{R}\}\] is a plane through the origin in \(\mathbb{R^{3}}\) as we did with lines, we could say that \[\{\overrightarrow{p} + t\overrightarrow{u} + s\overrightarrow{v}|s,t \in \mathbb{R}\}\] is a plane through $ in \(\mathbb{R^{3}}\) and that \[\overrightarrow{x} =\overrightarrow{p} + t\overrightarrow{u} + s\overrightarrow{v}, \quad s,t \in \mathbb{R}\] is a vector equation for the plane.

Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • It is very important to note that if either \(\overrightarrow{u}\) or \(\overrightarrow{v}\) is a scalar multiple of the other, then the set \[\{ t\overrightarrow{u} + s\overrightarrow{v}|s,t \in \mathbb{R}\}\] would not be a plane.

Example 1.1.9 | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • Determine which of the following vectors are in the plane with vector equation \[\overrightarrow{x}= s\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}, \quad s,t \in \mathbb{R}\\ (a) \overrightarrow{p} = \begin{bmatrix} 5/2 \\ 1 \\ 3 \end{bmatrix} \quad (b) \overrightarrow{q} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\]

Solution 1.1.9(a) | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • The vector \(\overrightarrow{p}\) is in the plane if and only if there are scalars \(s, t \in \mathbb{R}\) such that \[\begin{bmatrix} 5/2 \\ 1 \\ 3 \end{bmatrix}= s\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\] Performing the linear combination on the right-hand side gives \[\begin{bmatrix} 5/2 \\ 1 \\ 3 \end{bmatrix}= \begin{bmatrix} s + t \\ 2t \\ s + 2t \end{bmatrix}\]

Solution 1.1.9(a) cont’d | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • For these vectors to be equal we must have \[s + t = 5/2, \quad 2t = 1, \quad s + 2t = 3\]

  • We find that the values \(t = 1/2\) and \(s = 2\) satisfies all three equations. Hence, we have found that \(\overrightarrow{p}\) is in the plane. In particular, \[\begin{bmatrix} 5/2 \\ 1 \\ 3 \end{bmatrix}= 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\]

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Solution 1.1.9(b) | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • We need to determine whether there exists \(s, t \in \mathbb{R}\) such that \[\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}= s\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\] Performing the linear combination on the right-hand side gives \[\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}= \begin{bmatrix} s + t \\ 2t \\ s + 2t \end{bmatrix}\]

Solution 1.1.9(c) cont’d | Vectors, Lines, and Planes in \(\mathbb{R^{3}}\)

  • For these vectors to be equal we must have \[s + t = 1, \quad 2t = 1, \quad s + 2t = 3\]

  • The middle equation shows us that we must have \(t = 1/2\) However, then we would \(s = 1/2\) for the equation and \(s = 0\) for the third equation. Therefore, there is no value of \(s\) and \(t\) that satisfies all three equations. Thus, \(\overrightarrow{q}\) is not a linear combination of \[ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \, \text{and} \, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\], so \(\overrightarrow{q}\) is not in the plane.

Do It Yourself 3 | Vectors, Lines, and Planes in

  • Let \(\overrightarrow{x} = \begin{bmatrix} 2 \\ 1 \\ -5 \end{bmatrix}, \overrightarrow{y} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \overrightarrow{z} = \begin{bmatrix} 3 \\ 0 \\ -1 \end{bmatrix} \in \mathbb{R^{3}}\).
  • Calculate \(2\overrightarrow{x}-\overrightarrow{y}+3\overrightarrow{z}\).

Do It Yourself 4 | Vectors, Lines, and Planes in

  • Find a vector equation and parametric equations of the line that passes through the points \(P(1,2,2)\) and \(Q(1, -2,3)\)

Do It Yourself 1.1.5 | Vectors, Lines, and Planes in

  • Consider the plane in \(\mathbb{R^{3}}\) with vector equation \[\overrightarrow{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad s, t \in \mathbb{R}\] Find two vectors \(\overrightarrow{u}, \overrightarrow{v} \in \mathbb{R^{3}}\) that are in plane, and find a vector \(\overrightarrow{w} \in \mathbb{R^{3}}\) that is not in the plane.

1.2 Spanning and Linear Independence in \(\mathbb{R^2}\) and \(\mathbb{R^3}\)

Definition 1.2.1 | Span in \(\mathbb{R^2}\)

  • Let \(B = \{\overrightarrow{v_{1}},\cdots, \overrightarrow{v_{k}}\}\) be a set of vectors in \(\mathbb{R^{2}}\) or a set of vectors in \(\mathbb{R^{3}}\). We define the span of B, denoted Span B, to be the set of all possible linear combinations of the vectors in B. Mathematically, \[\text{Span }B = \{c_{1}\overrightarrow{v_{1}}+\cdots +\overrightarrow{v_{k}}\} | c_{1}, \cdots, \in \mathbb{R}\}\]

  • A vector equation for Span B is \[\overrightarrow{x} = c_{1}\overrightarrow{v_{1}}+ \cdots + c_{k}\overrightarrow{v_{k}}, \quad c_{1}, \cdots c_{k} \in \mathbb{R} \]

  • If \(S = \text{Span }B\), then we say that B spans S, that B is a spanning set for S, and that S is spanned by B.

Example 1.2.1 | Span in \(\mathbb{R^2}\)

  • Describe the set spanned by \(\biggl\{\begin{bmatrix} 1 \\ 2 \end{bmatrix}\biggl\}\) geometrically.

Solution 1.2.1 | Span in \(\mathbb{R^2}\)

  • A vector equation for the spanned set is \[\overrightarrow{x} = s\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \quad s \in \mathbb{R}\]

  • Thus, the spanned set is a line in \(\mathbb{R^{2}}\) through the origin with direction vector \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\).

Example 1.2.2 | Span in \(\mathbb{R^2}\)

  • Is the vector \(\begin{bmatrix} 3 \\ 1 \end{bmatrix}\) in Span \(\biggl\{\begin{bmatrix} 1 \\ 2 \end{bmatrix},\begin{bmatrix} -1 \\ 1 \end{bmatrix}\biggl\}\)?

Solution 1.2.2 | Span in \(\mathbb{R^2}\)

  • Using the definition of span, the vector \(\begin{bmatrix} 3 \\ 1 \end{bmatrix}\) is in the spanned set if it can be written as a linear combination of the vectors in the spanning set. That is, we need to determine whether there exists \(c_{1}, c_{2} \in \mathbb{R}\) such that \(\begin{bmatrix} 3 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 1 \end{bmatrix}.\)

Solution 1.2.2 cont’d | Span in \(\mathbb{R^2}\)

  • Performing operations on vectors on the right-hand side gives \[\begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} c_{1} - c_{2} \\ 2c_{1} + c_{2} \end{bmatrix} \]

  • Since vectors are equal if and only if and only if their corresponding entries are equal, we get that this vector equation implies \[\begin{align*} 3 &= c_{1} - c_{2} \\ 1 &= 2c_{1} + c_{2} \end{align*} \]

Solution 1.2.2 cont’d | Span in \(\mathbb{R^2}\)

  • Adding the equation gives \(4 = 3c_{1}\) and so \(c_{1}=\frac{4}{3}\) and \(c_{2}=-\frac{5}{3}\). Hence, we have that \[\begin{bmatrix} 3 \\ 1 \end{bmatrix} = \frac{4}{3}\begin{bmatrix} 1 \\ 2 \end{bmatrix} - \frac{5}{3}\begin{bmatrix} -1 \\ 1 \end{bmatrix}\]

  • Thus, by definition, \[\begin{bmatrix} 3 \\ 1 \end{bmatrix} \in \text{Span} \biggl\{\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}\biggl\}\]

Example 1.2.3 | Span in \(\mathbb{R^2}\)

  • Let \(\overrightarrow{\textbf{i}}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(\overrightarrow{\textbf{j}}=\begin{bmatrix} 0 \\ 1 \end{bmatrix}\). Show that Span \(\{\overrightarrow{\textbf{i}}, \overrightarrow{\textbf{j}}\} = \mathbb{R^{2}}\).

Solution 1.2.3 | Span in \(\mathbb{R^2}\)

  • We need to determine whether there exists \(c_{1}, c_{2} \in \mathbb{R}\) such that \[\begin{bmatrix} x\\ y \end{bmatrix} = c_{1}\begin{bmatrix} 1\\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 0\\ 1 \end{bmatrix}\]

  • We observe that we can take \(c_{1} = x\) and \(c_{2} = y\). That is, we have

\[\begin{bmatrix} x\\ y \end{bmatrix} = x\begin{bmatrix} 1\\ 0 \end{bmatrix} + y\begin{bmatrix} 0\\ 1 \end{bmatrix}\] So, Span\(\{\overrightarrow{\textbf{i}},\overrightarrow{\textbf{j}}\}=\mathbb{R^{2}}.\)

Example 1.2.4 | Span in \(\mathbb{R^2}\)

  • Describe \(\text{Span} \biggl\{\begin{bmatrix} 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 6 \\ 4 \end{bmatrix}\biggl\}\) geometrically.

Solution 1.2.4 | Span in \(\mathbb{R^2}\)

  • Using the definition of span, a vector equation of the spanned set is \[\overrightarrow{x}=s\begin{bmatrix} 3 \\ 2 \end{bmatrix} + t\begin{bmatrix} 6 \\ 4 \end{bmatrix}, \quad s,t \in \mathbb{R}\]

  • Observe that we can rewrite this as \[\overrightarrow{x}=s\begin{bmatrix} 3 \\ 2 \end{bmatrix} + (2t)\begin{bmatrix} 3 \\ 2 \end{bmatrix}, \quad s,t \in \mathbb{R}\]

Solution 1.2.4 | Span in \(\mathbb{R^2}\)

  • \[\overrightarrow{x}= (s+2t)\begin{bmatrix} 3 \\ 2 \end{bmatrix}, \quad s,t \in \mathbb{R}\]

  • Since \(c=s+2t\) can take any real value, the spanned set is a line through the origin with direction vector \(\begin{bmatrix} 3 \\ 2 \end{bmatrix}\).

Example 1.2.5 | Span in \(\mathbb{R^2}\)

  • Describe \(\text{Span} \left\{ \begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix} \right\}\) geometrically.

Solution 1.2.5 | Span in \(\mathbb{R^3}\)

  • By definition, a vector equation of the spanned set is \[\overrightarrow{x}=c_{1}\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}+ c_{2}\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}, \\ c_{1},c_{2},c_{3} \in \mathbb{R}\]

  • We observe that \(\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}+ \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\).

Solution 1.2.5 | Span in \(\mathbb{R^3}\)

  • Hence, we can rewrite the vector equation as \[\vec{x}= c_{1}\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} + \\ c_{3}\left\{\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\right\},\\ c_{1},c_{2},c_{3} \in \mathbb{R}\]

    • \[ (c_{1}+c_{3})\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + (c_{2}+c_{3})\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix},\\ c_{1},c_{2},c_{3} \in \mathbb{R}\]

Solution 1.2.5 cont’d | Span in \(\mathbb{R^3}\)

  • Since \(\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix}\) and \(\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\) are not scalar multiples of each other, we cannot simplify the vector equation any more. Thus, the set is a plane with vector equation

    \[\overrightarrow{x}=s\begin{bmatrix} 3 \\ 1 \\ -3 \end{bmatrix} + t\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \quad s,t \in \mathbb{R}\]

Linear Independence and Bases | in \(\mathbb{R^2}\) and \(\mathbb{R^3}\) | Definition

  • Let \(B = \{\overrightarrow{v_{1}},\cdots,\overrightarrow{v_{k}}\}\) be a set of vectors in \(\mathbb{R^{2}}\) or a set in \(\mathbb{R^{3}}\). The set \(B\) is said to be linearly dependent if there exist real coefficients \(c_{1},\cdots,c_{k}\) not all zero such that \[c_{1}\overrightarrow{v_{1}} + \cdots + c_{k}\overrightarrow{v_{k}} = \overrightarrow{0}\]

  • The set \(B\) is said to be linearly independent if the only solution to \[c_{1}\overrightarrow{v_{1}} + \cdots + c_{k}\overrightarrow{v_{k}} = \overrightarrow{0}\] is \(c_{1} = c_{2}=\cdots = c_{k} = 0\) (called the trivial solution).

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Example 1.2.6 | Linearly independent

  • Determine whether the set \(B = \left\{\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \right\}\) in \(\mathbb{R^{3}}\) is linearly independent.

Solution 1.2.6 | Linearly independent

  • By definition, we need to find all solutions of the equation \[\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}=c_{1}\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}+ c_{3}\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\]

Solution 1.2.6 cont’d | Linearly independent

  • Comparing entries gives the system of equations \[c_{1}+c_{3}=0, \quad c_{2}+c_{3}=0, \quad c_{1}+c_{2}=0\]

  • Adding the first to the second and then subtracting the third gives \(2c_{3}=0\). Hence, \(c_{3}=0\) which then implies \(c_{1}=c_{2}=0\) from the first and second equations. Since \(c_{1}=c_{2}=c_{3}=0\) is the only solution, the set is linearly independent.

Example 1.2.7 | Linearly independent

  • Determine whether the set \(C = \left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} , \begin{bmatrix} 2 \\ 2 \end{bmatrix} \right\}\) in \(\mathbb{R^{3}}\) is linearly independent or linearly dependent.

Solution 1.2.7 | Linearly independent

  • We consider the equation \[\begin{bmatrix} 0 \\ 0 \end{bmatrix}=c_{1}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 0 \end{bmatrix}+ c_{3}\begin{bmatrix} 2 \\ 2 \end{bmatrix}\]

  • We observe that taking \(c_{1}=-2\), \(c_{2}=0\), and \(c_{3}=1\) satisfies the equation. Hence, by definition, the \(C\) is linearly dependent.

Theorem 1.2.2 | Linearly independent

  • Let \(B = \{\overrightarrow{v_{1}},\cdots,\overrightarrow{v_{k}} \}\) be a set of vectors in \(\mathbb{R^{2}}\) or a set of vectors in \(\mathbb{R^{3}}\). The set \(B\) is linearly dependent if and only if \(\overrightarrow{v_{i}} \in Span \{\overrightarrow{v_{1}}, \cdots, \overrightarrow{v}_{i-1}, \overrightarrow{v}_{i+1}, \cdots, \overrightarrow{v}_{k}\}\) for some \(i, 1\leq i \leq k\).

Definition 1.2.2 | Basis of \(\mathbb{R^{2}}\) and \(\mathbb{R^{3}}\)

  • Let \(B=\{\overrightarrow{v_{1}},\overrightarrow{v_{2}}\}\) be a set in \(\mathbb{R^{2}}\). If \(B\) is linearly independent and Span \(B=\mathbb{R^{2}}\), then the set \(B\) is called a basis of \(\mathbb{R^{2}}\).

  • Let \(B=\{\overrightarrow{v_{1}},\overrightarrow{v_{2}}, \overrightarrow{v_{3}}\}\) be a set in \(\mathbb{R^{3}}\). If \(B\) is linearly independent and Span \(B=\mathbb{R^{3}}\), then the set \(B\) is called a basis of \(\mathbb{R^{3}}\).

Remarks | Linearly independent

  1. The plural of basis is bases. As we will see, both \(\mathbb{R^{2}}\) and \(\mathbb{R^{3}}\) have infinitely many bases.

  2. Here we are relying on our geometric intuition to say that all bases of \(\mathbb{R^{2}}\) have exactly two vectors and all bases of \(\mathbb{R^{3}}\) have exactly three vectors. In Chapter 2, we will mathematically prove this assertion.

Example 1.2.8 | Linearly independent

  • Prove that the set \(B=\{\overrightarrow{e_{1}},\overrightarrow{e_{2}}, \overrightarrow{e_{3}}\}\) is a basis for \(\mathbb{R^{3}}\)

Solution 1.2.8 | Linearly independent

  • To show that \(B\) is a basis, we need to prove that it is linearly independent and spans \(\mathbb{R^{3}}\).

  • Linear independent consider: \[\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}=c_{1}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}+ c_{3}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}=\begin{bmatrix} c_{1} \\ c_{2} \\ c_{3} \end{bmatrix}\]

  • Comparing entries, we get that \(c_{1}=c_{2}=c_{3}=0\). Therefore, \(B\) is linearly independent.

Solution 1.2.8 | Linearly independent

  • Spanning: Let \(\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3}\end{bmatrix}\) be any vector in \(\mathbb{R^{3}}\). Observe that we have \[\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}=x_{1}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}+ x_{3}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]

  • Hence, Span\(B = \mathbb{R^{3}}\)

  • Since \(B\) is linearly independent spanning set for \(\mathbb{R^{3}}\), it is a basis for \(\mathbb{R^{3}}\).

Example 1.2.9 | Linearly independent

  • Prove that the set \(C = \left\{\begin{bmatrix} -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}\) is a basis for \(\mathbb{R^{2}}\).

Solution 1.2.9 | Linearly independent

  • We need to show that Span \(C=\mathbb{R^{2}}\) and that \(C\) is linearly independent.

  • Spanning: Let \(\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix} \in \mathbb{R^{2}}\) and consider \[\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}=c_{1}\begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}= \begin{bmatrix} -c_{1}+c_{2} \\ 2c_{1}+c_{2} \end{bmatrix}\]

  • Comparing entries, we get two equations in two unknowns

    \[\begin {aligned} & -c_{1} + c_{2} = x_{1} \\ & -c_{1} + c_{2} = x_{2} \\ \end{aligned}\]

  • Solving gives \(c_{1}=\frac{1}{3}(-x_{1}+x_{2})\) and \(c_{2}=\frac{1}{3}(2x_{1}+x_{2})\).

Solution 1.2.9 cont’d | Linearly independent

  • Hence, we have that

\[\frac{1}{3}(-x_{1}+x_{2})\begin{bmatrix} -1 \\ 2 \end{bmatrix} + \frac{1}{3}(2x_{1}+x_{2})\begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

  • Thus, Span\(C=\mathbb{R^{2}}\).

  • Linear Independence: Take \(x_{1}=x_{2}=0\) in equation above. Our general solution to that equation says that the only solution is \(c_{1}=\frac{1}{3}(-0+0)=0\) and \(c_{2}=\frac{1}{3}(2(0)+0)=0\). So \(C\) is also linearly independent. Therefore, \(C\) is a basis for \(\mathbb{R^{2}}\).

Definition 1.2.3 | Coordinates in \(\mathbb{R^{2}}\)

Figure 6: Geometric representation of the basis \(B\) in Example 1.2.9
  • Let \(B=\{\overrightarrow{v_{1}}, \overrightarrow{v_{2}}\}\) be a basis for \(\mathbb{R^{2}}\) and let \(\overrightarrow{x}\in \mathbb{R^{2}}\). If \(\overrightarrow{x}=c_{1}\overrightarrow{v_{1}}+c_{2}\overrightarrow{v_{2}}\), then the scalars \(c_{1}\) and \(c_{2}\) are called the coordinates of \(\overrightarrow{x}\) with respect to the basis \(B\).

Example 1.2.10 | Coordinates with respect to Basis

  • Find the coordinates of \(\overrightarrow{x}= \begin{bmatrix} 4 \\ 1 \end{bmatrix}\) with respect to the basis \(C=\left\{\begin{bmatrix} -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}\).

Solution 1.2.10 | Coordinates with respect to Basis

  • From our work in Example 1.2.9 we have that the coordinates are:

    • \(c_{1} =\frac{1}{3}(-4+1)=-1\), and
    • \(c_{2} =\frac{1}{3}(2(4)+1)=3\).

Solution 1.2.10 cont’d | Coordinates with respect to Basis

Figure 7

Example 1.2.11 | Coordinates with respect to Basis

  • Given that

\(B=\left\{\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \end{bmatrix}\right\}\) is a basis for \(\mathbb{R^{2}}\), find the coordinates of \(\overrightarrow{x}=\begin{bmatrix} 4 \\ 1 \end{bmatrix}\) with respect to \(B\).

Solution 1.2.11 | Coordinates with respect to Basis

  • We need to find \(c_{1},c_{2}\in \mathbb{R}\) such that \[\begin{bmatrix} 4 \\ 1 \end{bmatrix}=c_{1}\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 1 \end{bmatrix}= \begin{bmatrix} c_{1}+2c_{2} \\ 2c_{1}+c_{2} \end{bmatrix}\]

Solution 1.2.11 cont’d | Coordinates with respect to Basis

  • Comparing entries gives the system of two equations in two unknowns \[\begin {aligned} & c_{1} + 2c_{2} = 4 \\ & 2c_{1} + c_{2} = 1 \\ \end{aligned}\]

  • Solving, we find that the coordinates of \(\overrightarrow{x}\) with respect to the basis \(B\) are \(c_{1}=-2/3\) and \(c_{2}=7/3\).

1.3 Length and Angles in \(\mathbb{R^{n}}\)

In many physical applications, we are given measurements in terms of angles and magnitudes. We must convert this data into vectors so that we can apply the tools of linear algebra to solve problems. For example, we may need to find a vector representing the path(and speed) of a plane flying northwest at 1300km/h. To do this, we need to identify the length of a vector and the angle between two vectors. In this section, we see how we can calculate both of these quantities with the dot product operator.

1.3.1 Length of a Vector | Pythagorean Theorem

  • Recall the distance formula in the plane: the distance between two points \((x_{1},y_{1})\) and \((x_{2},y_{2})\) is \(d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}\) (see Figure 8).

  • This formula arises from the Pythagorean Theorem for right triangles.

  • The 2-vector between the points is \([a_1,a_2]\) where \(a_1 = x_2 -x_1\) and \(a_2 = y_2 - y_1\), so \(d = \sqrt{a_1^2 + a_2^2}\)

  • This formula motivates the following definition:

::::

1.3.1 Length of a Vector | Pythagorean Theorem

  • Figure 8. The line segment (and vector) connecting points \(A\) and \(B\), with length \(\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}=\sqrt{a_1^2 + a_2^2}\)

Definition 1.3.1 | Length of a Vector

  • The length (also known as the norm or magnitude) of a vector \(\textbf{a} = [a_1, a_2, \dots, a_n]\) in \(\mathbb{R^n}\) is \(||\textbf{a}||=\sqrt{a_1^2 + a_2^2 + \dots + a_n^2}\).

Example 1.3.1 | Length of a Vector

  • The length of the vector \(\textbf{a} = [4, -3, 0, 2]\) is given by what?

Solution 1.3.1 | Length of a Vector

  • \[\begin{align}||\textbf{a}|| &= \sqrt{4^2 + (-3)^2 + 0^2 + 2^2} \\ &= \sqrt{16 + 9 + 0 + 4} \\ &= \sqrt{29} \end{align}\]

Definition 1.3.2 | Unit Vector

  • Any vector of length 1 is called a unit vector.

Definition 1.3.2 | Unit Vector

  • In \(\mathbb{R^2}\), the vector \(\bigg[\dfrac{3}{5}, -\dfrac{4}{5}\bigg]\) is a unit vector, because \(\sqrt{\bigg(\dfrac{3}{5}\bigg)^2 + \bigg(-\dfrac{4}{5}\bigg)^2} = 1\).

  • Certain unit vectors are particularly useful: those with a single coordinate equal to 1 and all other coordinates equal to 0.

  • In \(\mathbb{R^3}\) they are represented by \(\textbf{i} = [1,0,0]\), \(\textbf{j} = [0,1,0]\), and \(\textbf{k} = [0,0,1]\). These vectors are called standard unit vectors.

Exercise 1.3.1 | Length of a Vector

  • Find a vector that represents a movement from the first (initial) point to the second (terminal) point. Then use this vector to find the distance between the given points.

    1. \((-4,3), (5,-1)\)
    2. \((8,-2,5), (-3,2,0)\)

Solution 1.3.1a | Length of a Vector

  • \[\begin{align}\textbf{v} &= \textbf{v}_2 - \textbf{v}_1 \\&=\begin{bmatrix} 5\\ -1\end{bmatrix} -\begin{bmatrix} -4\\ 3\end{bmatrix} = \begin{bmatrix} 9\\ -4\end{bmatrix}\end{align}\]

  • \[\begin{align}||\textbf{v}|| &= \sqrt{9^2 + (-4)^2} \\ &= \sqrt{81 + 16} \\ &= \sqrt{97} \end{align}\]

Solution 1.3.1b | Length of a Vector

  • \[\begin{align}\textbf{v} &= \textbf{v}_2 -\textbf{v}_1 \\&=\begin{bmatrix} -3\\ 2\\ 0\end{bmatrix} -\begin{bmatrix} 8\\ -2 \\5\end{bmatrix} = \begin{bmatrix} -11\\ 4\\ -5\end{bmatrix}\end{align}\]

  • \[\begin{align}||\textbf{v}|| &= \sqrt{(-11)^2 + (4)^2 + (-5)^2} \\ &= \sqrt{121 + 16 + 25} \\ &= \sqrt{162} \\ &= 9\sqrt{2} \end{align}\]

Theorem 1.3.1 | Unit Vector

  • If \(\textbf{x}\) is a nonzero vector in \(\mathbb{R^n}\), then \(\textbf{u} = \bigg(\frac{1}{||\textbf{x}||}\bigg)\textbf{x}\) is a unit vector in the same direction as \(\textbf{x}\).

Definition 1.3.1 | Unit Vector

  • The process of “dividing” a vector by its length to obtain a unit vector in the same direction is called normalizing the vector

  • Figure 9. Normalizing a vector \(\textbf{x}\) to obtain a unit vector \(\textbf{u}\) in the same direction (with \(||\textbf{x}||>1\))

Example 1.3.2 | Unit Vector

  • Consider the vector \(\textbf{x} = [2,3,-1,1]\) in \(\mathbb{R^4}\). Normalize \(\textbf{x}\).

Solution 1.3.2 | Unit Vector

  • The norm of \(||\textbf{x}|| = \sqrt{2^2 + 3^2 + (-1)^2 + 1^2} =\sqrt{15}\)

  • \[\begin{align} \textbf{u} &= \bigg(\frac{1}{||\textbf{x}||}\bigg)\textbf{x}\\ &= \bigg(\frac{1}{\sqrt{15}}\bigg)[2,3,-1,1] \\&= \bigg[\frac{2}{\sqrt{15}},\frac{3}{\sqrt{15}},\frac{-1}{\sqrt{15}},\frac{1}{\sqrt{15}} \bigg] \end{align}\]

Exercise 1.3.2 | Unit Vector

  • For \(\vec{u} = (1, 2, -1)\). Normalize \(\vec{u}\).

Solution 1.3.2 | Unit Vector

Solution 1.3.2 | Unit Vector

1.3.3 Physical Application of Addition and Scalar Multiplication

  • Addition and scalar multiplication of vectors are often used to solve problems in physics.

  • Recall the trigonometric fact that if \(\textbf{v}\) is a vector in \(\mathbb{R^2}\) forming an angle of \(\theta\) with the positive \(x-\)axis then \(\textbf{v}=[||\textbf{v}||cos\theta,||\textbf{v}||sin\theta]\), as in the Figure 10 below.

1.3.3 Physical Application of Addition and Scalar Multiplication

  • Figure 10. The vector \(\textbf{v}=[||\textbf{v}||cos\theta,||\textbf{v}||sin\theta]\) forming an angle of \(\theta\) with the positive \(x-\)axis

Example 1.3.3 | Resultant Velocity

  • Suppose a man swims 5 km/hr in calm water. If he is swimming toward the east in wide stream with a northwest current of 3 km/hr, what is his resultant velocity (net speed and direction)?

Solution 1.3.3 | Resultant Velocity

  • The velocities of the swimmer and current are shown as vectors in Figure 11, where we have, for convenience, placed the swimmer a the origin. Now, \(\textbf{v}_1 = [5,0]\) and \(\textbf{v}_2 = \text{[3 cos 135}^{\circ},\text{3 sin 135}^{\circ}] = [-3\sqrt{2}/2, 3\sqrt{2}/2]\).

Solution 1.3.3 | Resultant Velocity

  • Figure 11. Velocity \(\textbf{v}_1\) of swimmer, velocity \(\textbf{v}_2\) of the current, and resultant velocity \(\textbf{v}_1 + \textbf{v}_2\)

Solution 1.3.3 | Resultant Velocity

  • Thus, the total (resultant) velocity of the swimmer is the sum of these velocities, \(\textbf{v}_1 + \textbf{v}_2\), which is \([5-3\sqrt{2}/2, 3\sqrt{2}/2]\approx[2.88,2.12]\). Hence, each hour the swimmer is traveling about 2.9 km east and 2.1 km north. The resultant speed of the swimmer is \(||[5-3\sqrt{2}/2, 3\sqrt{2}/2]||\approx\) 3.58 km/hr and direction is \(\theta =40.4^{\circ}\) north of east.

Example 1.3.4 | Newton’s Second Law

  • Newton’s famous Second Law of Motion asserts that sum, \(\textbf{f}\), of the vector forces on an object is equal to the scalar multiple of the mass \(m\) of the object times the vector acceleration \(\textbf{a}\) of the object; that is, \(\textbf{f} = m\textbf{a}\). For example, suppose a mass of 5 kg (kilograms) in a three-dimensional coordinate system has two forces acting on it: a force \(\textbf{f}_1\) of 10 newtons in the direction of the vector \([-2,1,2]\) and a force \(\textbf{f}_2\) of 20 newtons in the direction of the vector \([6,3,-2]\). What is the acceleration of the object?

Solution 1.3.4 | Newton’s Second Law

  • To find vectors representing \(\textbf{f}_1\) and \(\textbf{f}_2\):

    1. First, we multiply the magnitude each vector by a unit vector in that vector’s direction. The magnitude \(\textbf{f}_1\) and \(\textbf{f}_2\) of are 10 and 20, respectively.

    2. Next, we normalize the direction vectors \([-2,1,2]\) and \([6,3,-2]\) to create unit vectors in those directions, obtaining \([-2,1,2]/||[-2,1,2]||\) and \([6,3,-2]/||[6,3,-2]||\), respectively.

Solution 1.3.4 | Newton’s Second Law

  • Therefore, \(\textbf{f}_1 = 10([-2,1,2]/||[-2,1,2]||)\), and \(\textbf{f}_2 = 20([6,3,-2]/||[6,3,-2]||)\)

  • Now, the net force on the object is \(\textbf{f} = \textbf{f}_1 + \textbf{f}_2\).

  • Thus, the net acceleration on the object is \[\begin{align}\textbf{a} &= \frac{1}{m}\textbf{f} = \frac{1}{m}(\textbf{f}_1 + \textbf{f}_2) \\&= \frac{1}{5}\bigg(10\bigg(\frac{[-2,1,2]}{||[-2,1,2]||}\bigg) + 20\bigg(\frac{[6,3,-2]}{||[6,3, -2]||}\bigg)\bigg)\end{align}\]

Solution 1.3.4 | Newton’s Second Law

  • which equals \(\frac{2}{3}[-2,1,2] + \frac{4}{7}[6,3,-2] = \bigg[\frac{44}{21},\frac{50}{21},\frac{4}{21} \bigg]\).

  • The length of \(\textbf{a}\) is approximately 3.18, so pulling out a factor of 3.18 from each coordinate, we can approximate \(\textbf{a}\) as \(3.18[0.66,0.75,0.06]\), where \([0.66,0.75,0.06]\) is a unit vector. Hence, the acceleration is about 3.18 m/sec\(^2\) in the direction \([0.66,0.75,0.06]\).

Exercise 1.3.3 | Equilibrium

  • If the sum of the forces on an object is \(\textbf{0}\), then the object is in equilibrium; there is no acceleration in any direction. For example, two forces, \(\textbf{a}\) and \(\textbf{b}\), are simultaneously applied along cables attached to a weight, as in Figure 12, to keep the weight in equilibrium by balancing the force of gravity (which is \(m\textbf{g}\), where \(m\) is the mass of the weight and \(\textbf{g}=[0,-g]\)(with units in m/sec\(^2\)) is the downward acceleration due to gravity). Solve for \(\textbf{a}\) and \(\textbf{b}\) in terms of \(m\) and \(g\)

Exercise 1.3.3 | Equilibrium

  • Figure 12. Forces in equilibrium

Definition 1.3.5 | Vectors in the same direction, opposite directions, and parallel

  • Two nonzero vectors \(\textbf{x}\) and \(\textbf{y}\) are in \(\mathbb{R^n}\) are in the same direction if and only if there is a positive real number c such that \(\textbf{y} = c\textbf{x}.\)

  • Two nonzero vectors \(\textbf{x}\) and \(\textbf{y}\) are in \(\mathbb{R^n}\) are in the opposite directions if and only if there is a negative real number c such that \(\textbf{y} = c\textbf{x}.\)

  • Two nonzero vectors are parallel if and only if they are in the same direction or in the opposite direction.

Example 1.3.5 | Vectors in the same direction, opposite directions, and parallel

  1. Which of the following pairs of vectors are parallel? Tell whether they are in the same direction or opposite directions.

    1. \([1,-3,2]\) and \([3,-9,6]\)
    2. \([-3,6,0,15]\) and \([4,-8,0,-20]\)

  • Figure 13. The parallel vectors \([1,−3,2]\) and \([3,−9,6]\)

Solution 1.3.5 | Vectors in the same direction, opposite directions, and parallel

  • Hence

    1. the vectors are in the same direction because \([3,-9,6]=3[1,-3,2]\) (or because \([1,-3,2]=\frac{1}{3}[3,-9,6]\))

    2. the vectors are in the opposite directions because \([4,-8,0,-20]=-\frac{4}{3}[-3,6,0,15]\)

Definition 1.3.5 | The Dot Product

  • Let \(\textbf{x} = [x_1,x_2,\dots,x_n]\) and \(\textbf{y} = [y_1,y_2,\dots,y_n]\) be two vectors in \(\mathbb{R^n}\). The dot (inner) product of \(\textbf{x}\) and \(\textbf{y}\) is given by \[\textbf{x}\cdot\textbf{y} = x_1y_1 + x_2y_2 + \cdots + x_ny_n = \sum_{k=1}^{n} x_ky_k.\]

Example 1.3.6 | The Dot Product

  • Calculate the dot product of the following pairs of vectors.

    1. \(\vec{x}=\begin{bmatrix} 1 \\ -2\\ 1 \end{bmatrix}\) and \(\vec{y}=\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}\) .

    2. \(\vec{u}=\begin{bmatrix} 1/2 \\ 0\\ 1/2 \end{bmatrix}\) and \(\vec{v}=\begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}\)

Solution 1.3.6 | The Dot Product

  1. We have \(\vec{x} \cdot \vec{y} = 1(2) + (-2)(1) + 1(3) = 3\)
  1. \(\vec{u} \cdot \vec{v} = \frac{1}{2}(1) + (0)(5) + \frac{1}{2}(-1) = 0\)

Theorem 1.3.1 | The Dot Product

  • If \(\textbf{x}, \textbf{y}\) and \(\textbf{z}\) are any vectors in \(\mathbb{R^n}\), and if c is any scalar, then

    1. \(\textbf{x}\cdot\textbf{y} = \textbf{y}\cdot\textbf{x}\hspace{5.4cm}\) Commutativity of Dot Product

    2. \(\textbf{x}\cdot\textbf{x} = ||\textbf{x}||^2 \ge 0 \hspace{4cm}\) Relationship Between Dot Product and Length

    3. \(\textbf{x}\cdot\textbf{x} = 0\) if and only if \(\textbf{x} = \textbf{0}\)

    4. \(c(\textbf{x} \cdot \textbf{y}) = (c\textbf{x})\cdot\textbf{y}=\textbf{x}\cdot (c\textbf{y}) \hspace{0.7cm}\) Relationship Between Scalar Multiplication and Dot Product

    5. \(\textbf{x} \cdot (\textbf{y} +\textbf{z}) = (\textbf{x}\cdot\textbf{y}) + (\textbf{x}\cdot \textbf{z}) \hspace{1cm}\) Distributive Laws of Dot Product

    6. \(\textbf{x} + (\textbf{y} \cdot \textbf{z}) = (\textbf{x}\cdot\textbf{z}) + (\textbf{y}\cdot \textbf{z}) \hspace{1cm}\) Over Addition

Example 1.3.7 | Proof of the Distributive Law

  • Prove the distributive law Part (5) above.

Example 1.3.7 | Proof of the Distributive Law

  • Let \(\textbf{x}=[x_1,x_2,\dots, x_n]\), \(\textbf{y}=[y_1,y_2,\dots, y_n]\), and \(\textbf{z}=[z_1,z_2,\dots, z_n]\). Then,

  • \[\begin{align} \textbf{x}\cdot (\textbf{y} +\textbf{z}) &= [x_1,x_2,\dots, x_n] \cdot ([y_1,y_2,\dots, y_n] + [z_1,z_2,\dots, z_n]) \\&= [x_1,x_2,\dots, x_n] \cdot [y_1 + z_1, y_2 + z_2, \dots , y_n + z_n] \\&= x_1(y_1 + z_1) + x_2(y_2 + z_2) + \cdots + x_n(y_n + z_n) \\&= (x_1y_1 + x_2y_2 + \cdots + x_ny_n) + (x_1z_1 + x_2z_2 + \cdots + x_nz_n). \end{align}\]

  • Also

  • \[\begin{align} (\textbf{x}\cdot \textbf{y}) + (\textbf{x}\cdot \textbf{z}) &= ([x_1,x_2,\dots, x_n] \cdot [y_1,y_2,\dots, y_n]) \\& + ([x_1,x_2,\dots, x_n] \cdot [z_1,z_2,\dots, z_n]) \\&= (x_1y_1 + x_2y_2 + \cdots + x_ny_n) + (x_1z_1 + x_2z_2 + \cdots + x_nz_n). \end{align}\]

  • Hence, \(\textbf{x}\cdot (\textbf{y} + \textbf{z}) = (\textbf{x} \cdot \textbf{y}) + (\textbf{x} \cdot \textbf{z})\)

Example 1.3.8 | Dot Product Distributive Law

  • Simplify the dot product of the expression \((5\textbf{x}-4\textbf{y})\cdot (-2\textbf{x}+3\textbf{y})\) using the distributive law.

Solution 1.3.8 | Dot Product Distributive Law

  • \[\begin{align} (5\textbf{x}-4\textbf{y})\cdot (-2\textbf{x}+3\textbf{y}) &= [(5\textbf{x}-4\textbf{y})\cdot (-2\textbf{x})] + [(5\textbf{x}-4\textbf{y})\cdot (3\textbf{y}) ] \\& = [(5\textbf{x})\cdot (-2\textbf{x})] + [(-4\textbf{y})\cdot (-2\textbf{x})] + [(5\textbf{x})\cdot(3\textbf{y})] + [(-4\textbf{y})\cdot(3\textbf{y})] \\&= -10(\textbf{x}\cdot\textbf{x}) + 8(\textbf{y}\cdot\textbf{x}) + 15(\textbf{x}\cdot\textbf{y}) - 12(\textbf{y}\cdot\textbf{y}) \\& = -10||\textbf{x}||^2 + 23(\textbf{x}\textbf{y}) -12||\textbf{y}||^2. \end{align}\]

1.3.6 | Inequalities Involving the Dot Product

  • The following lemma is used in the proof of Theorem 1.7 below.

  • A lemma is a theorem whose main purpose is to assist in the proof of a more powerful result.

Lemma 1.6 | Inequalities Involving the Dot Product

  • If \(\textbf{a}\) and \(\textbf{b}\) are unit vectors in \(\mathbb{R}^n\), then \[\begin{align} -1\le \textbf{a}\cdot\textbf{b}\le 1 \end{align}\].

Proof of Lemma 1.6 | Inequalities Involving the Dot Product

  • Proof. Notice that the term \(\textbf{a}\cdot \textbf{b}\) appears in the expression of \((\textbf{a}+\textbf{b})\cdot(\textbf{a}+\textbf{b})\), as well as in the expression of \((\textbf{a}-\textbf{b})\cdot(\textbf{a}-\textbf{b}).\) The first gives

    • \[\begin{align} & \hspace{0.7in} (\textbf{a}+\textbf{b})\cdot(\textbf{a}+\textbf{b}) = ||\textbf{a}+\textbf{b}||^2 \ge 0 \\& \implies (\textbf{a}\cdot\textbf{a})+(\textbf{b}\cdot\textbf{a})+(\textbf{a}\cdot\textbf{b})+(\textbf{b}\cdot\textbf{b}) \ge 0 \\&\implies ||\textbf{a}||^2 + 2(\textbf{a}\cdot\textbf{b}) + ||\textbf{b}||^2 \ge 0 \\& \implies 1 + 2 (\textbf{a}\cdot\textbf{b}) + 1 \ge 0 \\& \implies \textbf{a}\cdot\textbf{b} \ge -1. \end{align}\]

  • A similar argument beginning with \((\textbf{a}-\textbf{b})\cdot (\textbf{a}-\textbf{b}) = ||\textbf{a}-\textbf{b}||^2 \ge 0\) shows \(\textbf{a}\cdot\textbf{b}\le1\) (Do Exercise 8).

  • Hence \(-1 \le \textbf{a}\cdot \textbf{b}\le 1\).

Theorem 1.7 | Cauchy-Schwarz Inequality

  • If \(\textbf{x}\) and \(\textbf{y}\) are vectors in \(\mathbb{R}^n\), then

\[\begin{align}|\textbf{x}\cdot\textbf{y}| \le (||\textbf{x}||)(||\textbf{y}||) \end{align}\].

Proof Theorem 1.7 | Cauchy-Schwarz Inequality

  • Proof. If \(\textbf{x} = \textbf{0}\) or \(\textbf{y}=\textbf{0}\), the theorem is certainly true. Hence, we need only examine the case when both \(||\textbf{x}||\) and \(||\textbf{y}||\) are nonzero. We need to prove \(-(||\textbf{x}||)(||\textbf{y}||)\le\textbf{x}\cdot\textbf{y}\le(||\textbf{x}||)(||\textbf{y}||)\).

  • This statement is true if and only if \[\begin{align} -1 \le \frac{\textbf{x}\cdot\textbf{y}}{(||\textbf{x}||)(||\textbf{y}||)} \le 1 \end{align}\]

  • But \(\textbf{x}\cdot\textbf{y}/(||\textbf{x}||)(||\textbf{y}||)\) is equal to \((\textbf{x}/||\textbf{x}||)\cdot(\textbf{y}/||\textbf{y}||)\).

  • However, both \(\textbf{x}/||\textbf{x}||\) and \(\textbf{y}/||\textbf{y}||\) are unit vectors, so by Lemma 1.6, their dot product satisfies the required double inequality above.

Example 1.3.9 | Cauchy-Schwarz Inequality

  • Verify the Cauchy-Schwarz Inequality in this specific case where \(\textbf{x} =[0, 3, 4]\) and \(\textbf{y} = [-1, 1, 1]\)

Solution 1.3.9 | Cauchy-Schwarz Inequality

  • \(\textbf{x}\cdot\textbf{y} = 0 + 3 + 4 = 7\)

  • Also \(||\textbf{x}|| = \sqrt{0 + 9 + 16} = \sqrt{25}=5\) and \(||\textbf{y}|| = \sqrt{1 + 1 + 1} = \sqrt{3}\)

  • Then \(|\textbf{x}\cdot\textbf{y}| \le (||\textbf{x}||)(||\textbf{y}||)\), because \(|7| = 7 \le 5\sqrt{3}\approx 8.66\)

Theorem 1.8 | Minkowshi’s Inequallity (Triangle Inequality)

  • If \(\textbf{x}\) and \(\textbf{y}\) are vectors in \(\mathbb{R}^n\), then \[\begin{align} ||\textbf{x} + \textbf{y}||\le||\textbf{x}|| + ||\textbf{y}||. \end{align}\]

1.3.7 | The Angle Between Two Vectors

  • Consider the vectors \(\textbf{x}-\textbf{y}\) in Figure 14, which begins are the terminal point of \(\textbf{y}\) and ends at the terminal point of \(\textbf{x}.\) Because \(0\le\theta\le\pi\), it follows from the Law of Cosines (please derive this law from the given figure) that \[\begin{align} ||\textbf{x} - \textbf{y}||^2 = ||\textbf{x}||^2 + ||\textbf{y}||^2 -2(||\textbf{x}||)(||\textbf{y}||)cos\theta \end{align}\].

  • Figure 14. The angle \(\theta\) between two nonzero vectors \(\textbf{x}\) and\(\textbf{y}\) in \(\mathbb{R}^2\)

1.3.7 | The Angle Between Two Vectors

  • But, \[\begin{align} ||\textbf{x} - \textbf{y}||^2 &= (\textbf{x} - \textbf{y})\cdot (\textbf{x} - \textbf{y}) \\&= (\textbf{x}\cdot\textbf{x}) - 2(\textbf{x}\cdot\textbf{y}) + (\textbf{y}\cdot\textbf{y}) \\&= ||\textbf{x}||^2 - 2(\textbf{x}\cdot\textbf{y}) + ||\textbf{y}||^2. \end{align}\]

  • Equating these two expressions for \(||\textbf{x} - \textbf{y}||^2\), and then canceling like terms yields \[\begin{align} -2||\textbf{x}||||\textbf{y}||cos\theta = -2(\textbf{x}\cdot\textbf{y}) \end{align}\]

  • This implies \[\begin{align} ||\textbf{x}||||\textbf{y}||cos\theta = (\textbf{x}\cdot\textbf{y}) \end{align}\], and so

  • \[\begin{align} cos\theta =\frac{ (\textbf{x}\cdot\textbf{y})}{||\textbf{x}||||\textbf{y}||}. \end{align}\]

Example 1.3.10 | The Angle Between Two Vectors

  • Suppose \(\textbf{x} =[6,-4]\) and \(\textbf{y} = [-2, 3]\) and \(\theta\) is the angle between x and y in \(\mathbb{R}^2\). Find angle \(\theta\)

Solution 1.3.10 | The Angle Between Two Vectors

  • \[\begin{align} \text{cos } \theta &= \frac{\textbf{x} \cdot \textbf{y}}{(||\textbf{x}||)(||\textbf{y}||)} \\&= \frac{(6)(-2) + (-4)(3)}{\sqrt{6^2 + (-4)^2}\times \sqrt{(-2)^2 + 3^2}} \\&=-\frac{24}{\sqrt{4 \cdot 13}\sqrt{13}} = -\frac{24}{2\sqrt{13}\sqrt{13}} = -\frac{12}{13}\\ \therefore \theta &= cos^{-1}\bigg( -\frac{12}{13}\bigg) \approx 2.74 \text{ radians or 157.4}^{\circ} \end{align}\]

  • (Remember that \(0 \le \theta \le \pi\))

Definition 1.3.7 | The Angle Between Two Vectors

  • Let x and y be two nonzero vectors in \(\mathbb{R}^n\), for \(n \ge 2\). Then the angle between x and y is the unique angle from 0 to \(\pi\) radians whose cosine is \(\frac{(\textbf{x} \cdot \textbf{y})} {((||\textbf{x}||)(||\textbf{y}||))}\).

Do It Yourself 1.3.11 | The Angle Between Two Vectors

  • For \(\textbf{} = [-1, 4, 2, 0, -3]\) and \(\textbf{} = [2, 1, -4, -1, 0]\), find angle \(\theta\).

Solution DIY 1.3.11 | The Angle Between Two Vectors

  • \[\begin{align} \text{cos } \theta &= \frac{\textbf{x} \cdot \textbf{y}}{(||\textbf{x}||)(||\textbf{y}||)} \\&= \frac{(-1)(2) + (4)(1)+(2)(-4) + (0)(-1) + (-3)(0)}{\sqrt{30} \sqrt{22}} \\&=-\frac{6}{2\sqrt{165}} \approx -0.234 \\ \therefore \theta &= cos^{-1}(-0.234) \\ \theta & \approx 1.8 \text{ radians or 103.5}^{\circ} \end{align}\].

Theorem 1.9 | The Angle Between Two Vectors

  • Let x and y be nonzero vectors in \(\mathbb{R}^n\), and let \(\theta\) be the angle between x and y. Then,

    1. \(\textbf{x}\cdot \textbf{y} > 0\) if and only if \(0 \le \theta < \frac{\pi}{2}\) radians (0\(^{\circ}\) or acute).

    2. \(\textbf{x}\cdot \textbf{y} = 0\) if and only if \(\theta = \frac{\pi}{2}\) radians (90\(^{\circ}\)).

    3. \(\textbf{x}\cdot \textbf{y} < 0\) if and only if \(\frac{\pi}{2} \le \theta < \pi\) radians (180\(^{\circ}\) or obtuse).

Definition 1.3.8 | Special Cases: Orthogonal and Parallel Vectors

  • Two vectors x and y in \(\mathbb{R}^n\) are orthogonal if and only if \(\textbf{x}\cdot \textbf{y} = 0\).

    • Note that by Theorem 1.9, two nonzero vectors are orthogonal if and only if they are perpendicular to each other.

Example 1.3.12 | Orthogal and Parallel Vectors

  • Given that \(\textbf{x} = [2, -5]\) and \(\textbf{y} = [-10, -4]\). Are vectors \(\textbf{x}\) and \(\textbf{y}\) orthogonal? If so sketch.

Solution 1.3.12 | Orthogonal and Parallel Vectors

  • \(\textbf{x}\cdot \textbf{y} = (2\cdot -10) + (-4 \cdot -5) = -20 + 20 = 0\).

    • Yes, vectors \(\textbf{x}\) and \(\textbf{y}\) are orthogonal.

1.3.8 | Orthogonal and Parallel Vectors

  • In \(\mathbb{R}^3\), the set of vectors \(\textbf{{i, j, k}}\) is mutually orthogonal; that is, the dot product of any two different vectors from this set equals zero.

  • In general, in \(\mathbb{R}^n\) the standard unit vectors \(\textbf{e}_1 = [1,0,0,\dots,0]\), \(\textbf{e}_2 = [0,1,0,\dots,0]\), \(\dots \textbf{e}_n = [0,0,0,\dots,1]\) form a mutually orthogonal set of vectors.

Theorem 1.10 | Parallel Vectors

  • Let \(\textbf{x}\) and \(\textbf{y}\) be nonzero vectors in \(\mathbb{R}^n\). If \(\textbf{x}\cdot \textbf{y} =\pm ||x||\, ||y||\) (that is, \(cos\theta = \pm 1\), where \(\theta\) is the angle between \(\textbf{x}\) and \(\textbf{y}\)), then \(\textbf{x}\) and \(\textbf{y}\) are parallel (that is, \(\textbf{y} = c\textbf{x}\) for some \(c \ne 0\)).

Example 1.3.13 | Orthogal and Parallel Vectors

  • Given that \(\textbf{x} = [8, -20, 4]\) and \(\textbf{y} = [6, -15, 3]\). Are vectors \(\textbf{x}\) and \(\textbf{y}\) parallel?

Solution 1.3.13 | Orthogonal and Parallel Vectors

  • \[\begin{align} \text{cos } \theta = \frac{\textbf{x} \cdot \textbf{y}}{(||\textbf{x}||)(||\textbf{y}||)} = \frac{48 + 300 + 12}{\sqrt{480} \sqrt{270}} =\frac{360}{\sqrt{129600}} = 1 \end{align}\]

  • Thus, by Theorem 1.10 \(\textbf{x}\) and \(\textbf{y}\) are parallel.

  • (Note also that \(\textbf{x}\) and \(\textbf{y}\) are parallel by the definition of parallel vectors because \([8, -20, 4] = \frac{4}{3}[6, -15, 3]\))