3.2 Norm, Dot Product, and Distance in Rⁿ: Class Notes

Definition

If $𝐯 = (v_{1}, v_{2}, \dots, v_{n})$ is a vector in $R^{n}$ , then the norm of $𝐯$ (also called its length or magnitude) is denoted by $‖ 𝐯 ‖$ [by this author], and is defined by the formula $‖ 𝐯 ‖ = \sqrt{v_{1}^{2} + v_{2}^{2} + \dots + v_{n}^{2}}$ .

Theorem 3.2.1 Properties of the norm of a vector

If $𝐯$ is a vector in $R^{n}$ and $k$ is any scalar, then:

a) $‖ 𝐯 ‖ \geq 0$

b) $‖ 𝐯 ‖ = 0$ if and only if $𝐯 = 𝟎$

c) $‖ k 𝐯 ‖ = | k | ‖ 𝐯 ‖$

Proof of (c)

Let $\vec{v} = (v_{1}, v_{2}, \dots, v_{n})$ .

$\begin{matrix} k \vec{v} & = (k v_{1}, k v_{2}, \dots, k v_{n}) (definition of scalar multiplication) \\ ‖ k \vec{v} ‖ & = \sqrt{k^{2} v_{1}^{2} + k^{2} v_{2}^{2} + \dots + k^{2} v_{n}^{2}} (definition of norm) \\ = \sqrt{k^{2}} \sqrt{v_{1}^{2} + v_{2}^{2} + \dots + v_{n}^{2}} = | k | ‖ \vec{v} ‖ (definition of absolute value) \end{matrix}$

(The last step uses $\sqrt{k^{2}} = | k |$ .)

Definition

The norm of a unit vector is 1. We can obtain a unit vector from a nonzero vector $𝐯$ by multiplying by the reciprocal of its length. This process is called normalizing the vector.

$\frac{1}{‖ \vec{v} ‖} \vec{v} = \frac{\vec{v}}{‖ \vec{v} ‖} is a unit vector.$

#1

Find the norm of $𝐯$ and a unit vector that is oppositely directed to $𝐯$ .

a. $𝐯 = (2, 2, 2)$

b. $𝐯 = (1, 0, 2, 1, 3)$

Solution

a. $\vec{v} = (2, 2, 2) = 2 (1, 1, 1)$ , so

$‖ \vec{v} ‖ = 2 \sqrt{1^{2} + 1^{2} + 1^{2}} = 2 \sqrt{3} .$

Multiply by $- \frac{1}{‖ \vec{v} ‖}$ : the negative sign reverses the direction and dividing by the norm makes the length 1.

$- \frac{1}{2 \sqrt{3}} \vec{v} = (- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}})$

This has length 1 and is oppositely directed to $\vec{v}$ .

Answer: $‖ \vec{v} ‖ = 2 \sqrt{3}, (- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}})$

$‖ \vec{v} ‖ = \sqrt{1 + 0 + 4 + 1 + 9} = \sqrt{15}$

$- \frac{1}{\sqrt{15}} (1, 0, 2, 1, 3) = (- \frac{1}{\sqrt{15}}, 0, - \frac{2}{\sqrt{15}}, - \frac{1}{\sqrt{15}}, - \frac{3}{\sqrt{15}})$

Answer: $‖ \vec{v} ‖ = \sqrt{15}, (- \frac{1}{\sqrt{15}}, 0, - \frac{2}{\sqrt{15}}, - \frac{1}{\sqrt{15}}, - \frac{3}{\sqrt{15}})$

Definition

The standard unit vectors in $𝐑^{𝐧}$ are the standard basis vectors for $R^{n}$ , $𝐞_{1} = (1, 0, 0, \dots, 0)$ , $𝐞_{2} = (0, 1, 0, \dots, 0)$ , $\dots$ , $𝐞_{n} = (0, 0, 0, \dots, 1)$ .

Recall

In $R^{2}$ : ${\vec{e}}_{1} = (1, 0) = \hat{ı}$ , ${\vec{e}}_{2} = (0, 1) = \hat{ȷ}$ .

In $R^{3}$ : ${\vec{e}}_{1} = (1, 0, 0) = \hat{ı}$ , ${\vec{e}}_{2} = (0, 1, 0) = \hat{ȷ}$ , ${\vec{e}}_{3} = (0, 0, 1) = \hat{k}$ .

Every vector $\vec{v} = (v_{1}, v_{2}, \dots, v_{n})$ in $R^{n}$ can be expressed as a linear combination of the standard basis vectors:

$\vec{v} = v_{1} {\vec{e}}_{1} + v_{2} {\vec{e}}_{2} + \dots + v_{n} {\vec{e}}_{n} .$

Definition

If $𝐮 = (u_{1}, u_{2}, \dots, u_{n})$ and $𝐯 = (v_{1}, v_{2}, \dots, v_{n})$ are points in $R^{n}$ , then we denote the distance between $𝐮$ and $𝐯$ by $d (𝐮, 𝐯)$ and define it to be

$d (𝐮, 𝐯) = ‖ 𝐮 - 𝐯 ‖ = \sqrt{(u_{1} - v_{1})^{2} + (u_{2} - v_{2})^{2} + \dots + (u_{n} - v_{n})^{2}}$

#4

Evaluate the given expression with $𝐮 = (2, - 2, 3)$ , $𝐯 = (1, - 3, 4)$ , and $𝐰 = (3, 6, - 4)$ .

a. $‖ 𝐮 + 𝐯 + 𝐰 ‖$

b. $‖ 𝐮 - 𝐯 ‖$

c. $‖ 3 𝐯 ‖ - 3 ‖ 𝐯 ‖$

d. $‖ 𝐮 ‖ - ‖ 𝐯 ‖$

Solution

$\vec{u} + \vec{v} + \vec{w} = (2 + 1 + 3, - 2 - 3 + 6, 3 + 4 - 4) = (6, 1, 3),$

$‖ \vec{u} + \vec{v} + \vec{w} ‖ = \sqrt{36 + 1 + 9} = \sqrt{46} .$

Answer: $\sqrt{46}$

b. Note: this is the distance between $\vec{u}$ and $\vec{v}$ , $d (\vec{u}, \vec{v})$ .

$‖ \vec{u} - \vec{v} ‖ = \sqrt{(2 - 1)^{2} + (- 2 + 3)^{2} + (3 - 4)^{2}} = \sqrt{1^{2} + 1^{2} + (- 1)^{2}} = \sqrt{3}$

Answer: $\sqrt{3}$

c. By Theorem 3.2.1(c), $‖ 3 \vec{v} ‖ = 3 ‖ \vec{v} ‖$ , so

$‖ 3 \vec{v} ‖ - 3 ‖ \vec{v} ‖ = 3 ‖ \vec{v} ‖ - 3 ‖ \vec{v} ‖ = 0 .$

(Numerically, $‖ \vec{v} ‖ = \sqrt{1 + 9 + 16} = \sqrt{26}$ and $3 \sqrt{26} - 3 \sqrt{26} = 0$ .)

Answer: $0$

$‖ \vec{u} ‖ - ‖ \vec{v} ‖ = \sqrt{4 + 4 + 9} - \sqrt{1 + 9 + 16} = \sqrt{17} - \sqrt{26}$

Answer: $\sqrt{17} - \sqrt{26}$

Definition

Let $𝐮$ and $𝐯$ be nonzero vectors in $R^{n}$ positioned so that their initial points coincide. The angle between $𝐮$ and $𝐯$ is the angle $θ$ determined by $𝐮$ and $𝐯$ such that $0 \leq θ \leq π$ .

Two vectors u and v drawn from a common initial point, with the angle theta between them

$\vec{u}$ and $\vec{v}$ start at the same point; $θ$ is the angle between them.

Definition

If $𝐮$ and $𝐯$ are nonzero vectors in $R^{2}$ or $R^{3}$ , and if $θ$ is the angle between $𝐮$ and $𝐯$ , then the dot product or Euclidean inner product of $𝐮$ and $𝐯$ is denoted by $𝐮 \cdot 𝐯$ and is defined as $𝐮 \cdot 𝐯 = ‖ 𝐮 ‖ ‖ 𝐯 ‖ \cos θ$ . If $𝐮 = 𝟎$ or $𝐯 = 𝟎$ , then we define $𝐮 \cdot 𝐯$ to be 0.

Tip: For numbers, $3 (x)$ , $3 x$ , $3 \times x$ , and $3 \cdot x$ all mean the same product. In $\vec{u} \cdot \vec{v}$ the dot is required.

Definition

If $𝐮 = (u_{1}, u_{2}, \dots, u_{n})$ and $𝐯 = (v_{1}, v_{2}, \dots, v_{n})$ are vectors in $R^{n}$ , then the dot product or Euclidean inner product of $𝐮$ and $𝐯$ is denoted by $𝐮 \cdot 𝐯$ and is defined by $𝐮 \cdot 𝐯 = u_{1} v_{1} + u_{2} v_{2} + \dots + u_{n} v_{n}$ .

To connect the two definitions, let $\vec{u}$ and $\vec{v}$ share an initial point in $R^{3}$ , and let $Q$ and $P$ be their terminal points, respectively.

Triangle formed by u, v, and the vector PQ = u minus v joining their tips, with angle theta between u and v

$\vec{u}$ ends at $Q$ and $\vec{v}$ ends at $P$ , with angle $θ$ between them. The third side, $\vec{P Q} = \vec{u} - \vec{v}$ , has length $‖ \vec{u} - \vec{v} ‖$ . The other two sides have lengths $‖ \vec{u} ‖$ and $‖ \vec{v} ‖$ .

Recall: Law of Cosines

Triangle with vertices A, B, C and opposite sides a, b, c

Side $c$ is opposite angle $C$ , side $a$ opposite $A$ , and side $b$ opposite $B$ .

$c^{2} = a^{2} + b^{2} - 2 a b \cos C$

By the Law of Cosines,

${‖ \vec{u} - \vec{v} ‖}^{2} = {‖ \vec{u} ‖}^{2} + {‖ \vec{v} ‖}^{2} - 2 ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos θ .$

Write the left side and the squared norms in components:

$(u_{1} - v_{1})^{2} + (u_{2} - v_{2})^{2} + (u_{3} - v_{3})^{2} = u_{1}^{2} + u_{2}^{2} + u_{3}^{2} + v_{1}^{2} + v_{2}^{2} + v_{3}^{2} - 2 ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos θ .$

Each square on the left expands, e.g. $(u_{1} - v_{1})^{2} = u_{1}^{2} - 2 u_{1} v_{1} + v_{1}^{2}$ . The squared terms appear on both sides and cancel, leaving

$\begin{matrix} - 2 (u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}) & = - 2 ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos θ \\ ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos θ & = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3} = \vec{u} \cdot \vec{v} . \end{matrix}$

So the two definitions agree in $R^{3}$ . In $R^{2}$ the same argument gives just $u_{1} v_{1} + u_{2} v_{2}$ .

#10

Find $𝐮 \cdot 𝐯$ , $𝐮 \cdot 𝐮$ , and $𝐯 \cdot 𝐯$ .

a. $𝐮 = (1, 1, - 2, 3)$ , $𝐯 = (- 1, 0, 5, 1)$

b. $𝐮 = (2, - 1, 1, 0, - 2)$ , $𝐯 = (1, 2, 2, 2, 1)$

Solution

$\begin{matrix} \vec{u} \cdot \vec{v} & = 1 (- 1) + 1 (0) - 2 (5) + 3 (1) = - 8 \\ \vec{u} \cdot \vec{u} & = 1 (1) + 1 (1) - 2 (- 2) + 3 (3) = 15 \\ \vec{v} \cdot \vec{v} & = - 1 (- 1) + 0 (0) + 5 (5) + 1 (1) = 27 \end{matrix}$

Answer: $\vec{u} \cdot \vec{v} = - 8, \vec{u} \cdot \vec{u} = 15, \vec{v} \cdot \vec{v} = 27$

$\begin{matrix} \vec{u} \cdot \vec{v} & = 2 (1) - 1 (2) + 1 (2) + 0 (2) - 2 (1) = 0 \\ \vec{u} \cdot \vec{u} & = 2^{2} + (- 1)^{2} + 1^{2} + 0^{2} + (- 2)^{2} = 10 \\ \vec{v} \cdot \vec{v} & = 1^{2} + 2^{2} + 2^{2} + 2^{2} + 1^{2} = 14 \end{matrix}$

Notice $\vec{u} \cdot \vec{u} = {‖ \vec{u} ‖}^{2}$ .

Answer: $\vec{u} \cdot \vec{v} = 0, \vec{u} \cdot \vec{u} = 10, \vec{v} \cdot \vec{v} = 14$

#11

Find the Euclidean distance between $𝐮$ and $𝐯$ and the cosine of the angle between those vectors. State whether that angle is acute, obtuse, or $90^{\circ}$ .

a. $𝐮 = (3, 3, 3)$ , $𝐯 = (1, 0, 4)$

b. $𝐮 = (0, - 2, - 1, 1)$ , $𝐯 = (- 3, 2, 4, 4)$

Solution

From the definition, $\vec{u} \cdot \vec{v} = ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos θ$ , so $\cos θ = \frac{\vec{u} \cdot \vec{v}}{‖ \vec{u} ‖ ‖ \vec{v} ‖}$ .

$d (\vec{u}, \vec{v}) = \sqrt{(3 - 1)^{2} + (3 - 0)^{2} + (3 - 4)^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$

$‖ \vec{u} ‖ = 3 \sqrt{3}, ‖ \vec{v} ‖ = \sqrt{17}, \vec{u} \cdot \vec{v} = 3 (1) + 3 (0) + 3 (4) = 15$

$\cos θ = \frac{15}{3 \sqrt{3} \sqrt{17}} = \frac{5}{\sqrt{51}}$

Since $\cos θ > 0$ , $0 < θ < π / 2$ : the angle is acute.

Answer: $d = \sqrt{14}, \cos θ = \frac{5}{\sqrt{51}}, acute$

Tip: On the unit circle, $\cos θ > 0$ on the right half (angles between 0 and $π / 2$ ) and $\cos θ < 0$ on the left half (angles between $π / 2$ and $π$ ).

$d (\vec{u}, \vec{v}) = \sqrt{9 + 16 + 25 + 9} = \sqrt{59}$

$‖ \vec{u} ‖ = \sqrt{6}, ‖ \vec{v} ‖ = 3 \sqrt{5}, \vec{u} \cdot \vec{v} = - 4$

$\cos θ = \frac{- 4}{\sqrt{6} \cdot 3 \sqrt{5}} = - \frac{4}{3 \sqrt{30}} < 0 \Rightarrow π / 2 < θ < π$

Answer: $d = \sqrt{59}, \cos θ = - \frac{4}{3 \sqrt{30}}, obtuse$

Theorem 3.2.2 Properties of the dot product

If $𝐮$ , $𝐯$ , and $𝐰$ are vectors in $R^{n}$ and if $k$ is a scalar, then:

a) $𝐮 \cdot 𝐯 = 𝐯 \cdot 𝐮$ (symmetry property)

b) $𝐮 \cdot (𝐯 + 𝐰) = 𝐮 \cdot 𝐯 + 𝐮 \cdot 𝐰$ (distributive property)

c) $k (𝐮 \cdot 𝐯) = (k 𝐮) \cdot 𝐯$ (homogeneity property)

d) $𝐯 \cdot 𝐯 \geq 0$ and $𝐯 \cdot 𝐯 = 0$ if and only if $𝐯 = 𝟎$ (positivity property)

Proof of (b)

Let $\vec{u} = (u_{1}, \dots, u_{n})$ , $\vec{v} = (v_{1}, \dots, v_{n})$ , and $\vec{w} = (w_{1}, \dots, w_{n})$ .

$\begin{matrix} \vec{u} \cdot (\vec{v} + \vec{w}) & = (u_{1}, u_{2}, \dots, u_{n}) \cdot (v_{1} + w_{1}, v_{2} + w_{2}, \dots, v_{n} + w_{n}) \\ definition of vector addition \\ = u_{1} (v_{1} + w_{1}) + u_{2} (v_{2} + w_{2}) + \dots + u_{n} (v_{n} + w_{n}) \\ definition of dot product \\ = u_{1} v_{1} + u_{1} w_{1} + \dots + u_{n} v_{n} + u_{n} w_{n} \\ distributive property of real numbers \\ = u_{1} v_{1} + \dots + u_{n} v_{n} + u_{1} w_{1} + \dots + u_{n} w_{n} \\ commutative property of real numbers \\ = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \\ definition of dot product \end{matrix}$

Theorem 3.2.3 More properties of the dot product

If $𝐮$ , $𝐯$ , and $𝐰$ are vectors in $R^{n}$ and if $k$ is a scalar, then:

a) $𝟎 \cdot 𝐯 = 𝐯 \cdot 𝟎 = 0$

b) $(𝐮 + 𝐯) \cdot 𝐰 = 𝐮 \cdot 𝐰 + 𝐯 \cdot 𝐰$

c) $𝐮 \cdot (𝐯 - 𝐰) = 𝐮 \cdot 𝐯 - 𝐮 \cdot 𝐰$

d) $(𝐮 - 𝐯) \cdot 𝐰 = 𝐮 \cdot 𝐰 - 𝐯 \cdot 𝐰$

e) $k (𝐮 \cdot 𝐯) = 𝐮 \cdot (k 𝐯)$

Theorem 3.2.4 Cauchy-Schwarz Inequality

If $𝐮 = (u_{1}, u_{2}, \dots, u_{n})$ and $𝐯 = (v_{1}, v_{2}, \dots, v_{n})$ are vectors in $R^{n}$ , then $| 𝐮 \cdot 𝐯 | \leq ‖ 𝐮 ‖ ‖ 𝐯 ‖$ or in terms of components

$| u_{1} v_{1} + u_{2} v_{2} + \dots + u_{n} v_{n} | \leq {(u_{1}^{2} + u_{2}^{2} + \dots + u_{n}^{2})}^{1 / 2} {(v_{1}^{2} + v_{2}^{2} + \dots + v_{n}^{2})}^{1 / 2} .$

This is a necessary result for the cosine definition of the dot product in $R^{n}$ :

$\begin{matrix} | \vec{u} \cdot \vec{v} | \leq ‖ \vec{u} ‖ ‖ \vec{v} ‖ & \Rightarrow \frac{| \vec{u} \cdot \vec{v} |}{‖ \vec{u} ‖ ‖ \vec{v} ‖} \leq 1 \\ \Rightarrow - 1 \leq \frac{\vec{u} \cdot \vec{v}}{‖ \vec{u} ‖ ‖ \vec{v} ‖} \leq 1, \end{matrix}$

so $\cos θ = \frac{\vec{u} \cdot \vec{v}}{‖ \vec{u} ‖ ‖ \vec{v} ‖}$ always gives a value between $- 1$ and 1, and the angle $θ$ makes sense in $R^{n}$ .

Theorem 3.2.5 Triangle Inequalities

If $𝐮$ , $𝐯$ , and $𝐰$ are vectors in $R^{n}$ , then:

a) $‖ 𝐮 + 𝐯 ‖ \leq ‖ 𝐮 ‖ + ‖ 𝐯 ‖$ (triangle inequality for vectors)

b) $d (𝐮, 𝐯) \leq d (𝐮, 𝐰) + d (𝐰, 𝐯)$ (triangle inequality for distances)

Triangle formed by u, v placed tip to tail, and u + v

$\vec{u}$ and $\vec{v}$ placed tip to tail, with $\vec{u} + \vec{v}$ as the third side. A side of a triangle is never longer than the other two sides combined.

Proof of (a)

$\begin{matrix} {‖ \vec{u} + \vec{v} ‖}^{2} & = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = \vec{u} \cdot \vec{u} + 2 \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v} \\ \leq {‖ \vec{u} ‖}^{2} + 2 | \vec{u} \cdot \vec{v} | + {‖ \vec{v} ‖}^{2} \end{matrix}$

(by the definitions of norm and dot product and a property of absolute value: $\vec{u} \cdot \vec{v} \leq | \vec{u} \cdot \vec{v} |$ )

$\begin{matrix} \leq {‖ \vec{u} ‖}^{2} + 2 ‖ \vec{u} ‖ ‖ \vec{v} ‖ + {‖ \vec{v} ‖}^{2} \\ = {(‖ \vec{u} ‖ + ‖ \vec{v} ‖)}^{2} \end{matrix}$

(by the Cauchy-Schwarz inequality). Taking square roots yields $‖ \vec{u} + \vec{v} ‖ \leq ‖ \vec{u} ‖ + ‖ \vec{v} ‖$ .

Proof of (b)

Part (b) follows from part (a) and the definition of distance.

Note: An expression like $\vec{u} \cdot \vec{v} \cdot \vec{w}$ does not have meaning ( $\vec{u} \cdot \vec{v}$ is a number, and a number can’t be dotted with a vector). But we can consider matrix multiplication using the dot product: if $A$ is an $n \times n$ matrix and $\vec{u}$ and $\vec{v}$ are $n \times 1$ matrices, then

$\begin{matrix} A \vec{u} \cdot \vec{v} & = {\vec{v}}^{T} (A \vec{u}) = ({\vec{v}}^{T} A) \vec{u} = (A^{T} \vec{v})^{T} \vec{u} = \vec{u} \cdot A^{T} \vec{v} . \end{matrix}$

(Sizes: ${\vec{v}}^{T}$ is $1 \times n$ and $A \vec{u}$ is $n \times 1$ ; ${\vec{v}}^{T} A$ and $(A^{T} \vec{v})^{T}$ are $1 \times n$ ; $A^{T} \vec{v}$ is $n \times 1$ .)

$So A \vec{u} \cdot \vec{v} = \vec{u} \cdot A^{T} \vec{v} . Likewise, \vec{u} \cdot A \vec{v} = A^{T} \vec{u} \cdot \vec{v} .$

Last note: If we represent $A$ ( $m \times n$ ) by its rows and $B$ ( $n \times r$ ) by its columns,

$A = [\begin{matrix} {\vec{r}}_{1} \\ {\vec{r}}_{2} \\ ⋮ \\ {\vec{r}}_{m} \end{matrix}] and B = [\begin{matrix} {\vec{c}}_{1} & {\vec{c}}_{2} & \dots & {\vec{c}}_{r} \end{matrix}],$

then each entry of $A B$ is a dot product of a row of $A$ with a column of $B$ :

$A B = [\begin{matrix} {\vec{r}}_{1} \cdot {\vec{c}}_{1} & {\vec{r}}_{1} \cdot {\vec{c}}_{2} & \dots & {\vec{r}}_{1} \cdot {\vec{c}}_{r} \\ {\vec{r}}_{2} \cdot {\vec{c}}_{1} & {\vec{r}}_{2} \cdot {\vec{c}}_{2} & \dots & {\vec{r}}_{2} \cdot {\vec{c}}_{r} \\ ⋮ & ⋮ & ⋮ \\ {\vec{r}}_{m} \cdot {\vec{c}}_{1} & {\vec{r}}_{m} \cdot {\vec{c}}_{2} & \dots & {\vec{r}}_{m} \cdot {\vec{c}}_{r} \end{matrix}]$