Vector Algebra

Figure A.2: Vector addition.

Suppose that the displacements $\stackrel{\displaystyle \rightarrow}{PQ}$ and $\stackrel{\displaystyle \rightarrow}{QR}$ represent the vectors ${\bf a}$ and ${\bf b}$, respectively. See Figure A.2. It can be seen that the result of combining these two displacements is to give the net displacement $\stackrel{\displaystyle \rightarrow}{PR}$. Hence, if $\stackrel{\displaystyle \rightarrow}{PR}$ represents the vector ${\bf c}$ then we can write

$${\bf c} = {\bf a} + {\bf b}.$$ (1.1)

This defines vector addition. By completing the parallelogram $PQRS$, we can also see that

$$\stackrel{\displaystyle \rightarrow}{PR} \,= \, \stackrel{\displa... ...ackrel{\displaystyle \rightarrow}{PS}+\stackrel{\displaystyle \rightarrow}{SR}.$$ (1.2)

However, $\stackrel{\displaystyle \rightarrow}{PS}$ has the same length and direction as $\stackrel{\displaystyle \rightarrow}{QR}$, and, thus, represents the same vector, ${\bf b}$. Likewise, $\stackrel{\displaystyle \rightarrow}{PQ}$ and $\stackrel{\displaystyle \rightarrow}{SR}$ both represent the vector ${\bf a}$. Thus, the previous equation is equivalent to

$${\bf c} = {\bf a} + {\bf b} = {\bf b} + {\bf a}.$$ (1.3)

We conclude that the addition of vectors is commutative. It can also be shown that the associative law holds; that is,

$${\bf a} + ({\bf b} + {\bf c}) = ({\bf a} + {\bf b}) + {\bf c}.$$ (1.4)

The null vector, ${\bf0}$, is represented by a displacement of zero length and arbitrary direction. Because the result of combining such a displacement with a finite length displacement is the same as the latter displacement by itself, it follows that

$${\bf a} + {\bf0} = {\bf a},$$ (1.5)

where ${\bf a}$ is a general vector. The negative of ${\bf a}$ is defined as that vector that has the same magnitude, but acts in the opposite direction, and is denoted $-{\bf a}$. The sum of ${\bf a}$ and $-{\bf a}$ is thus the null vector; in other words,

$${\bf a} + (-{\bf a}) = {\bf0}.$$ (1.6)

We can also define the difference of two vectors, ${\bf a}$ and ${\bf b}$, as

$${\bf c} = {\bf a} - {\bf b} = {\bf a} +(-{\bf b}).$$ (1.7)

This definition of vector subtraction is illustrated in Figure A.3.

Figure A.3: Vector subtraction.

If $n>0$ is a scalar then the expression $n\,{\bf a}$ denotes a vector whose direction is the same as ${\bf a}$, and whose magnitude is $n$ times that of ${\bf a}$. (This definition becomes obvious when $n$ is an integer.) If $n$ is negative then, because $n\,{\bf a} = \vert n\vert\,(-{\bf a})$, it follows that $n\,{\bf a}$ is a vector whose magnitude is $\vert n\vert$ times that of ${\bf a}$, and whose direction is opposite to ${\bf a}$. These definitions imply that if $n$ and $m$ are two scalars then

$$n\,(m\,{\bf a}) = n\,m\,{\bf a} = m\,(n\,{\bf a}),$$ (1.8)

$$(n+m)\,{\bf a} = n\,{\bf a} + m\,{\bf a},$$ (1.9)

$$n\,({\bf a} + {\bf b}) = n\,{\bf a} + n\,{\bf b}.$$ (1.10)