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Vectors
Rules of vector algebra

Rules of vector algebra

Here we start by defining some rules of vector arithmetic using two vectors, a\boldsymbol{a} and b\boldsymbol{b}. Note that we define a-\boldsymbol{a} to be the vector of the same magnitude as a\boldsymbol{a} but in the opposite direction.

Addition and subtraction

Given the vectors a=(a1,a2,a3)\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right) and b=(b1,b2,b3)\boldsymbol{b}=\left(b_{1}, b_{2}, b_{3}\right),

a±b=(a1±b1,a2±b2,a3±b3)\boldsymbol{a} \pm \boldsymbol{b}=\left(a_{1} \pm b_{1}, a_{2} \pm b_{2}, a_{3} \pm b_{3}\right)

The diagram below shows the geometric interpretation of the addition of two vectors; it is known as the parallelogram law and it shows that vector addition is commutative, i.e.

a+b=b+a\boldsymbol{a}+\boldsymbol{b}=\boldsymbol{b}+\boldsymbol{a}

The geometric interpretation of the subtraction of two vectors, ba\boldsymbol{b}-\boldsymbol{a} is given below. The subtraction can be thought of as the addition of the vectors b\boldsymbol{b} and a-\boldsymbol{a} since a-\boldsymbol{a} has the same magnitude as a\boldsymbol{a} but in the opposite direction.

Multiplication by a scalar

For a vector a\boldsymbol{a} and a nonzero scalar λ\lambda, the vector λa\lambda \boldsymbol{a} is the vector of magnitude λ×a|\lambda| \times|\boldsymbol{a}| and in the direction of a\boldsymbol{a} if λ>0\lambda>0 or the opposite direction if λ<0\lambda<0. For a=(a1,a2,a3)\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right), the scalar multiplication gives,

λa=(λa1,λa2,λa3).\lambda \boldsymbol{a}=\left(\lambda a_{1}, \lambda a_{2}, \lambda a_{3}\right) .

Using the two properties above, we can show what was stated in Definition 9.4: any 3D vector is a linear combination of the standard basis vectors in 3D. Starting from the vector a=(a1,a2,a3)\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right), we can use the addition rule to rewrite as

a=(a1,0,0)+(0,a2,0)+(0,0,a3)\boldsymbol{a}=\left(a_{1}, 0,0\right)+\left(0, a_{2}, 0\right)+\left(0,0, a_{3}\right)

and the scalar multiplication rule to express as,

a=a1(1,0,0)+a2(0,1,0)+a3(0,0,1)\boldsymbol{a}=a_{1}(1,0,0)+a_{2}(0,1,0)+a_{3}(0,0,1)

Figure 9.2: Three-dimensional Cartesian coordinate system with axes x,y,zx, y, z.

This shows that any vector can be written in terms of the standard basis vectors. Consider the Cartesian coordinates x,y,zx, y, z shown in Fig. 9.2. The unit vectors i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} are the standard basis vectors in Eq. (9.1). The vectors i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} are parallel to the x,y,zx, y, z axes, respectively.

The vector r=(a1,a2,a3)\boldsymbol{r}=\left(a_{1}, a_{2}, a_{3}\right) that starts at the point A=(0,0,0)A=(0,0,0) and ends at B=(a1,a2,a3)B=\left(a_{1}, a_{2}, a_{3}\right) is called a position vector. The vector v=(0,0,0)\boldsymbol{v}=(0,0,0) which has all its components equal to zero is called a zero vector, sometimes denoted by 0,0,0\mathbf{0}, \overrightarrow{0}, \underline{0}, etc.

The position vector with general coordinates (x,y,z)(x, y, z) is written as

r=xi+yj+zk\boldsymbol{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}

where x,y,zx, y, z are the components of r\boldsymbol{r}. The magnitude of r\boldsymbol{r} is given by

r=x2+y2+z2|\boldsymbol{r}|=\sqrt{x^{2}+y^{2}+z^{2}}

Example 9.1 Prove that the line joining a vertex of a parallelogram to the midpoint of the opposite side divides the diagonal in the ratio 1:2.

Solution Figure 9.3 shows a diagram of the parallelogram. We have

OP=μOB\overrightarrow{O P}=\mu \overrightarrow{O B}

for some scalar μ\mu and we want to show that μ=1/3\mu=1 / 3.

We define

DP=λDA\overrightarrow{D P}=\lambda \overrightarrow{D A}

for some scalar λ\lambda. Next, we consider

OD+DP=OP\overrightarrow{O D}+\overrightarrow{D P}=\overrightarrow{O P}

and we choose to express Eq. (9.2) in terms of a basis; we choose the position vectors OA\overrightarrow{O A} and OC\overrightarrow{O C} as the basis. By definition, we have

OD=12OC\overrightarrow{O D}=\frac{1}{2} \overrightarrow{O C}

it follows that in Eq. (9.2), we are left to express DP\overrightarrow{D P} and OP\overrightarrow{O P} in terms of OA\overrightarrow{O A} and OC\overrightarrow{O C}. We start with OP\overrightarrow{O P} :

OP=μOB=μ(OC+CB)=μ(OC+OA)\overrightarrow{O P}=\mu \overrightarrow{O B}=\mu(\overrightarrow{O C}+\overrightarrow{C B})=\mu(\overrightarrow{O C}+\overrightarrow{O A})

For DP\overrightarrow{D P} :

DP=λDA=λ(OAOD)=λ(OA12OC)\overrightarrow{D P}=\lambda \overrightarrow{D A}=\lambda(\overrightarrow{O A}-\overrightarrow{O D})=\lambda\left(\overrightarrow{O A}-\frac{1}{2} \overrightarrow{O C}\right)

Substituting Eqs. (9.3)-(9.5) in (9.2) and rearranging, yields

12OC=OC(μ+12λ)+OA(μλ)\frac{1}{2} \overrightarrow{O C}=\overrightarrow{O C}\left(\mu+\frac{1}{2} \lambda\right)+\overrightarrow{O A}(\mu-\lambda)

Comparing coefficients of OA\overrightarrow{O A} and OC\overrightarrow{O C} on both sides, we have μ=λ\mu=\lambda and μ=1/3\mu=1 / 3.

Figure 9.3: Diagram for question in Example 9.1.

Vector spaces

A vector is an element of a vector space. In broad terms, a vector space is a set of vectors together with rules for vector addition and scalar multiplication. Such operations must produce vectors in the space and satisfy certain conditions. Definition 9.5 gives said conditions.

A vector space VV is a set closed under finite vector addition and scalar multiplication. The scalars may come from any field F\mathbb{F}, e.g. on real F=R\mathbb{F}=\mathbb{R} or complex F=C\mathbb{F}=\mathbb{C} vector spaces. For all u,v\boldsymbol{u}, \boldsymbol{v}, and wV\boldsymbol{w} \in V and λ,μF\lambda, \mu \in \mathbb{F}, we have the following requirements:

(i) Commutativity

u+v=v+u\boldsymbol{u}+\boldsymbol{v}=\boldsymbol{v}+\boldsymbol{u}

(ii) Associativity of vector addition

(u+v)+w=u+(v+w)(\boldsymbol{u}+\boldsymbol{v})+\boldsymbol{w}=\boldsymbol{u}+(\boldsymbol{v}+\boldsymbol{w})

Additive identity

There exists the zero vector 0\mathbf{0} such that for all u\boldsymbol{u},

u+0=0+u=u.\boldsymbol{u}+\mathbf{0}=\mathbf{0}+\boldsymbol{u}=\boldsymbol{u} .

(iv) Existence of additive inverse

For any u\boldsymbol{u} there exists u-\boldsymbol{u}, such that,

u+(u)=0\boldsymbol{u}+(-\boldsymbol{u})=\mathbf{0}

(v) Associativity of scalar multiplication

μ(λu)=(μλ)u\mu(\lambda \boldsymbol{u})=(\mu \lambda) \boldsymbol{u}

(vi) Distributivity of scalar sums

(μ+λ)u=μu+λu.(\mu+\lambda) \boldsymbol{u}=\mu \boldsymbol{u}+\lambda \boldsymbol{u} .

(vii) Distributivity of vector sums

μ(u+v)=μu+μv.\mu(\boldsymbol{u}+\boldsymbol{v})=\mu \boldsymbol{u}+\mu \boldsymbol{v} .

(viii) Scalar multiplication identity

1u=u1 \boldsymbol{u}=\boldsymbol{u}

A fun exercise is to use the rules given in Definition 9.5 to prove that for all vV\boldsymbol{v} \in V,

0v=00 \boldsymbol{v}=\mathbf{0}