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Vector Algebra

Suppose that the displacements $ \stackrel{\displaystyle \rightarrow}{PQ}$ and $ \stackrel{\displaystyle \rightarrow}{QR}$ , shown in Figure A.2, represent the vectors $ {\bf a}$ and $ {\bf b}$ , respectively. 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

$\displaystyle {\bf c} = {\bf a} + {\bf b}.$ (A.1)

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

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

$\displaystyle {\bf c} = {\bf a} + {\bf b} = {\bf b} + {\bf a}.$ (A.3)

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

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

$\displaystyle {\bf a} + {\bf0} = {\bf a},$ (A.5)

where $ {\bf a}$ is a general vector. The negative of $ {\bf a}$ is defined as that vector which 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: that is,

$\displaystyle {\bf a} + (-{\bf a}) = {\bf0}.$ (A.6)

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

$\displaystyle {\bf c} = {\bf a} - {\bf b} = {\bf a} +(-{\bf b}).$ (A.7)

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

Figure A.3: Vector subtraction.
\begin{figure}
\epsfysize =1.75in
\centerline{\epsffile{AppendixA/figA.03.eps}}
\end{figure}

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

$\displaystyle n\,(m\,{\bf a})$ $\displaystyle = n\,m\,{\bf a} = m\,(n\,{\bf a}),$ (A.8)
$\displaystyle (n+m)\,{\bf a}$ $\displaystyle = n\,{\bf a} + m\,{\bf a},$ (A.9)
$\displaystyle n\,({\bf a} + {\bf b})$ $\displaystyle = n\,{\bf a} + n\,{\bf b}.$ (A.10)


next up previous
Next: Cartesian Components of a Up: Vectors and Vector Fields Previous: Scalars and Vectors
Richard Fitzpatrick 2016-03-31