Review Notes: Vectors

Introducitn Vectors

According to Merriam-Webster, the term vector stems from the past participle of Latin vehere "carry, convey". Mathematically, it denotes a tuple of numbers, e.g.

$$\left(1,2,3\right),\quad \left(-100.5,202\right),\quad \left(38,16,9,74,880\right).$$

The set of all vectors of length $n$, where $n$ is a positive integer, is called the $n$-dimensional Euclidean Space

$$\mathbb{R}^n:=\left\{\left(x_1, \cdots, x_n\right)\mid x_1\in\mathbb{R},\cdots,x_n\in\mathbb{R}\right\}.$$

In this course, we focus on $\mathbb{R}^2$ ($2$-vectors) and $\mathbb{R}^3$ ($3$-vectors). Each $2$-vector uniquely corresponds to a point on the plane; similarly, each $3$-vector uniquely corresponds to a point in the space.

Vectors

Figure 1. a vector is determined only by its direction and magnitude, not its starting point

We consider a vector as a direction plus a magnitude, often represented as an arrow. We do not care where the arrow starts. If two arrows with equal length are parallel to each other, they are considered as the same vector even if their starting points are different. Equivalenty speaking, we allow vectors to "freely parallel-translate".

Vector Operations

Given vectors $\vec{v}=\left(v_1,v_2,v_3\right)$, $\vec{w}=\left(w_1,w_2,w_3\right)$, we define their sum as

\begin{equation}\label{eq:sum} \vec{v} + \vec{w} := \left(v_1+w_1, v_2+w_2, v_3+w_3\right). \end{equation}

For any real number $c\in \mathbb{R}$, define the scalar multiplication of $c$ and $\vec{v}$ as

\begin{equation}\label{eq:scalar_mult} c \vec{v} = \left(cv_1, cv_2, cv_3\right). \end{equation}

With these two basic operations in mind, it is straightforward to define any linear combination of vectors, for instance

$$\vec{v}-\vec{w} = \vec{v} + \left(-1\right)\vec{w},$$

$$2\vec{u}-3\vec{v}+5\vec{w} = 2\vec{u}+\left(-3\right)\vec{v}+5\vec{u}.$$

Properties of the sum and the scalar multiplication includes the following:

$$\vec{v}+\vec{w} = \vec{w}+\vec{v},\quad \left(\vec{u}+\vec{v}\right)+\vec{w}=\vec{u}+\left(\vec{v}+\vec{w}\right),$$

$$c\left(\vec{v}+\vec{w}\right)=c\vec{v}+c\vec{w},\quad \left(c+d\right)\vec{x} = c\vec{x}+d\vec{x},\quad c \left(d\vec{x}\right)=\left(cd\right)\vec{x}.$$

All these formulae can be directly verified using \eqref{eq:sum} and \eqref{eq:scalar_mult}.

The dot product (a.k.a. scalar product or inner product) of two vectors $\vec{v}=\left(v_1,v_2,v_3\right)$, $\vec{w}=\left(w_1,w_2,w_3\right)$ is defined as

\begin{equation}\label{eq:dot_prod} \vec{v}\cdot\vec{w}=v_1w_1+v_2w_2+v_3w_3. \end{equation}

The name scalar follows from the fact that the result of the product is a scalar.

The following properties of the dot product can be verified directly via \eqref{eq:dot_prod}.

$$\vec{v}\cdot\vec{w} = \vec{w}\cdot\vec{v},\quad \vec{u}\cdot\left(\vec{v}+\vec{w}\right) = \vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w},\quad c\left(\vec{v}\cdot\vec{w}\right)=\left(c\vec{x}\right)\vec{w}.$$

When $\vec{v}=\vec{w}$, one has

$$\vec{v}\cdot\vec{v} = v_1^2+v_2^2+v_3^2,$$

which represents the squared distance from the origin to the point $\left(v_1,v_2,v_3\right)$. We thus define the magnitude of $\vec{v}$ as

$$\left|\vec{v}\right|=\sqrt{\vec{v}\cdot\vec{v}}.$$

Vector_Angle

The dot product has a geometric interpretation. Let $\vec{v}$, $\vec{w}$ be as in the figure on the right, and denote $\theta$ for the angle between $\vec{v}$ and $\vec{w}$. We have the following

Theorem 1. $\vec{v}\cdot\vec{w} = \left|\vec{v}\right|\left|\vec{w}\right|\cos\theta$.

Proof. Without loss of generality, assume $\vec{v}$, $\vec{w}$ are $3$-vectors. For the $2$-vector case, simply treat the last coordiantes of $\vec{v}$ and $\vec{w}$ as $0$.

The triangle on the right consists of three edges of length $\left|\vec{v}\right|$, $\left|\vec{w}\right|$, $\left|\vec{v}-\vec{w}\right|$. By the Law of Cosines, one has

$$\left|\vec{v}\right|^2+\left|\vec{w}\right|^2-\left|\vec{v}-\vec{w}\right|^2=2\left|\vec{v}\right|\left|\vec{w}\right|\cos\theta.$$

Note that

$$\left|\vec{v}-\vec{w}\right|^2=\left(\vec{v}-\vec{w}\right)\cdot\left(\vec{v}-\vec{w}\right)=\vec{v}\cdot\vec{v}+\vec{w}\cdot\vec{w}-2\vec{v}\cdot\vec{w}=\left|\vec{v}\right|^2+\left|\vec{w}\right|^2-2\vec{v}\cdot\vec{w},$$

thus the left hand side of the law of cosines can be written as

$$\left|\vec{v}\right|^2+\left|\vec{w}\right|^2-\left|\vec{v}-\vec{w}\right|^2=2\vec{v}\cdot\vec{w},$$

which yields

$$2\vec{v}\cdot\vec{w}=2\left|\vec{v}\right|\left|\vec{w}\right|\cos\theta,$$

and the conclusion follows immediately. $\blacksquare$

Theorem 1 builds the connection between dot product and the angle between two vectors. More specifically,

$$\vec{v}\cdot\vec{w}>0\quad \Leftrightarrow\quad \cos\theta >0\quad \Leftrightarrow\quad \theta \textrm{ is acute},$$

$$\vec{v}\cdot\vec{w}<0\quad \Leftrightarrow\quad \cos\theta <0\quad \Leftrightarrow\quad \theta \textrm{ is obtuse}.$$

If $\vec{v}\cdot\vec{w}=0$, $\vec{v}$ and $\vec{w}$ are said to be orthogonal or perpendicular to each other. When $\left|\vec{v}\right|\neq0$ and $\left|\vec{w}\right|\neq0$, this happens if and only if $\cos\theta = 0$.

The last important notion related to dot product is component. The component of a vector $\vec{v}$ along another vector $\vec{w}$ is defined as

$$\mathrm{comp}_{\vec{w}}\vec{v}=\frac{\vec{v}\cdot\vec{w}}{\left|\vec{w}\right|}.$$

Note that $\mathrm{comp}_{\vec{w}}\vec{v}$ can be understood as the dot product between $\vec{v}$ and a unit vector $\vec{w}\big/\left|\vec{w}\right|$. In particular, if we choose $\vec{w}$ to be one of the following three unit vectors

$$\vec{i}=\left(1,0,0\right),\quad \vec{j}=\left(0,1,0\right),\quad \vec{k}=\left(0,0,1\right),$$

then

$$\mathrm{comp}_{\vec{i}}\vec{v}=\vec{v}\cdot\vec{i}=v_1,$$

$$\mathrm{comp}_{\vec{j}}\vec{v}=\vec{v}\cdot\vec{j}=v_2,$$

$$\mathrm{comp}_{\vec{k}}\vec{v}=\vec{v}\cdot\vec{k}=v_3,$$

which are often referred to as the $x$-component, $y$-component, and $z$-component of $v$, respectively.

The cross product (a.k.a. vector product or outer product) of two $3$-d vectors $\vec{v}=\left(v_1,v_2,v_3\right)$, $\vec{w}=\left(w_1,w_2,w_3\right)$ is defined as

\begin{equation}\label{eq:cross_prod} \vec{v}\times\vec{w}=\left(v_2w_3-v_3w_2, -\left(v_1w_3-v_3w_1\right), v_1w_2-v_2w_1\right). \end{equation}

The name vector follows from the fact that the result of the product is a vector. Formally, $\vec{v}\times\vec{w}$ can be defined as the following determinant

$$\vec{v}\times\vec{w}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \notag \end{vmatrix}.$$

From either definitions, one has

$$\vec{v}\times\vec{w}=-\vec{w}\times\vec{v},$$

and in particular

$$\vec{v}\times\vec{v}=0.$$

By definition, it is straightforward to verify that $\vec{v}\times\vec{w}$ is orthogonal to both $\vec{v}$ and $\vec{w}$. For instance,

$$ \begin{align*} \vec{v}\cdot\left(\vec{v}\times\vec{w}\right)&=v_1\left(v_2w_3-v_3w_2\right)-v_2\left(v_1w_3-v_3w_1\right)+v_3\left(v_1w_2-v_2w_1\right)\\ &=v_1v_2w_3-v_1w_2v_3-v_1v_2w_3+w_1v_2v_3+v_1w_2v_3-w_1v_2v_3=0. \end{align*} $$

In other words, $\vec{v}\times\vec{w}$ is orthogonal to the plane spanned by $\vec{v}$ and $\vec{w}$, or equivalently, $\vec{v}\times\vec{w}$ is a normal vector of that plane.

We would like to determine the direction and magnitude of $\vec{v}\times\vec{w}$. The magnitude can be directly computated.

Theorem 2. $\left|\vec{v}\times\vec{w}\right|^2=\left|\vec{v}\right|^2\left|\vec{w}\right|^2-\left|\vec{v}\cdot\vec{w}\right|^2$.

Proof. Note that

$$ \begin{align*} \left|\vec{v}\times\vec{w}\right|^2&=\left(v_2w_3-v_3w_2\right)^2+\left(v_1w_3-v_3w_1\right)^2+\left(v_1w_2-v_2w_1\right)^2\\ &=v_2^2w_3^2+v_3^2v_2^2+v_1^2w_3^2+v_3^2w_1^2+v_1^2w_2^2+v_2^2w_1^2\\ &\qquad-2v_2w_3v_3w_2-2v_1w_3v_3w_1-2v_1w_2v_2w_1\\ &=\left(v_1^2+v_2^2+v_3^2\right)\left(w_1^2+w_2^2+w_3^2\right)-v_1^2w_1^2-v_2^2w_2^2-v_3^2w_3^2\\ &\qquad -2v_2w_2\cdot v_3w_3-2v_1w_1\cdot v_3w_3-2v_1w_1\cdot v_2w_2\\ &=\left(v_1^2+v_2^2+v_3^2\right)\left(w_1^2+w_2^2+w_3^2\right)-\left(v_1w_1+v_2w_2+v_3w_3\right)^2\\ &=\left|\vec{v}\right|^2\left|\vec{w}\right|^2-\left|\vec{v}\cdot\vec{w}\right|^2. \end{align*} $$

$\blacksquare$

Corollary 1. Let $\theta$ denote the angle between $\vec{v}$ and $\vec{w}$. Then

$$\left|\vec{v}\cdot\vec{w}\right|=\left|\vec{v}\right|\left|\vec{w}\right|\sin\theta.$$

Proof. Recall from Theorem 1 that $\vec{v}\cdot\vec{w}=\left|\vec{v}\right|\left|\vec{w}\right|\cos\theta$. Thus by Theorem 2

$$ \begin{align*} \left|\vec{v}\times\vec{w}\right|^2&=\left|\vec{v}\right|^2\left|\vec{w}\right|^2-\left|\vec{v}\cdot\vec{w}\right|^2=\left|\vec{v}\right|^2\left|\vec{w}\right|^2-\left|\vec{v}\right|^2\left|\vec{w}\right|^2\cos^2\theta\\ &=\left|\vec{v}\right|^2\left|\vec{w}\right|^2\left(1-\cos^2\theta\right)=\left|\vec{v}\right|^2\left|\vec{w}\right|^2\sin^2\theta. \end{align*} $$

The result then follows from taking square roots. $\blacksquare$

Corollary 2. If $\vec{v}$ and $\vec{w}$ are collinear, then $\vec{v}\times\vec{w}=0$.

Remark. Corollary 1 implies that $\left|\vec{v}\times\vec{w}\right|$ equals to the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$.

To facilitate describing the direction of $\vec{v}\times\vec{w}$, let us introduce the one last vector operations, the scalar triple product (a.k.a. mixed product or box product). For three $3$-d vectors $\vec{u}$, $\vec{v}$, $\vec{w}$ (the order matters!), define their scalar triple product as

$$\vec{u}\cdot\left(\vec{v}\times\vec{w}\right).$$

Theorem 3. The scalar triple product can be written in determinant form

$$ \begin{align*} \vec{u}\cdot\left(\vec{v}\times\vec{w}\right)=\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \notag \end{vmatrix} \end{align*} $$

Proof. Recall from the definition of the cross product that

$$\vec{v}\times\vec{w}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \notag \end{vmatrix}= \begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \notag \end{vmatrix}\vec{i} -\begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \notag \end{vmatrix}\vec{j} +\begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \notag \end{vmatrix}\vec{k},$$

thus

$$ \begin{align*} \vec{u}\cdot\left(\vec{v}\times\vec{w}\right)&=\left(u_1\vec{i}+u_2\vec{j}+u_3\vec{k}\right)\cdot\left(\begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \notag \end{vmatrix}\vec{i} -\begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \notag \end{vmatrix}\vec{j} +\begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \notag \end{vmatrix}\vec{k}\right)\\ &=u_1\begin{vmatrix} v_2 & v_3\\ w_2 & w_3 \notag \end{vmatrix} -u_2\begin{vmatrix} v_1 & v_3\\ w_1 & w_3 \notag \end{vmatrix} +u_3\begin{vmatrix} v_1 & v_2\\ w_1 & w_2 \notag \end{vmatrix}\\ &=\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \notag \end{vmatrix}. \end{align*} $$

$\blacksquare$

Corollary 3.

$$\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)=\vec{v}\cdot\left(\vec{w}\times\vec{u}\right)=\vec{w}\cdot\left(\vec{u}\times\vec{v}\right).$$

Proof. Use the properties of the determinant. $\blacksquare$

Definition. An ordered vector triple $\vec{u}, \vec{v}, \vec{w}$ is said to form a right-handed triple if

$$\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \notag \end{vmatrix}>0.$$

Theorem 4. If $\vec{v}\times\vec{w}\neq 0$, the ordered triple $\vec{v},\vec{w},\vec{v}\times\vec{w}$ forms a right-handed triple.

Proof. By Theorem 3, it suffices to show that

$$\vec{v}\cdot\left(\vec{w}\times\left(\vec{v}\times\vec{w}\right)\right)>0.$$

To see why this is true, use Corollary 3:

$$ \begin{align*} &\vec{v}\cdot\left(\vec{w}\times\left(\vec{v}\times\vec{w}\right)\right)=\left(\vec{v}\times\vec{w}\right)\cdot\left(\vec{v}\times\vec{w}\right)\\ &\qquad =\left|\vec{v}\times\vec{w}\right|^2>0, \end{align*} $$

where the last strict positivity follows from the assumption that $\vec{v}\times\vec{w}\neq 0$. $\blacksquare$

The fact that $\vec{v}, \vec{w}, \vec{v}\times\vec{w}$ form a right-handed triple suggests that we can determine the direction of $\vec{v}\times\vec{w}$ using the Right-hand Rule, as illustrated in the figure above. It can also be deduced directly from this geometric picture that

$$\vec{v}\times\vec{w}=-\vec{w}\times\vec{v}.$$

For this reason, sometimes people interprete $\vec{v}\times\vec{w}$ as the "signed area" of the parallelogram spanned by $\vec{v}$ and $\vec{w}$.

The last important point for vectors is the geometric interpretation of the scalar triple product. In fact, $\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)$ equals to the "signed volume" of the parallelepiped spanned by the three vectors $\vec{u},\vec{v},\vec{w}$, as explained in Figure 2.

Scalar Triple Product

Figure 2. scalar triple product equals to the "signed volume" of the parallelepiped

To understand this picture, note that

$$ \begin{align*} \vec{u}\cdot\left(\vec{v}\times\vec{w}\right)&=\left|\vec{u}\right|\left|\vec{v}\times\vec{w}\right|\cos\theta=\left|\vec{u}\right|\cos\theta\cdot\left|\vec{v}\times\vec{w}\right|\\ &=\left(\textrm{"signed height" of the parallelpiped}\right)\cdot\left(\textrm{based area of the parallelpiped}\right)\\ &=\textrm{"signed volume" of the parallelpiped}. \end{align*} $$

The sign of the "signed volume" only depends on the left- or right-handedness of the ordered triple $\vec{u},\vec{v},\vec{w}$. Note that the triples

$$\left(\vec{u},\vec{v},\vec{w}\right),\quad \left(\vec{v},\vec{w},\vec{u}\right),\quad \left(\vec{w},\vec{u},\vec{v}\right)$$

all have the same handedness, thus it is clear from this geometric interpretation that

$$\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)=\vec{v}\cdot\left(\vec{w}\times\vec{u}\right)=\vec{w}\cdot\left(\vec{u}\times\vec{v}\right).$$

Introducitn Vectors

Vector Operations

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