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Next: Cartesian Components of a Up: Vector Algebra and Vector Previous: Scalars and Vectors

Vector Algebra

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.97. 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
\begin{displaymath}
{\bf c} = {\bf a} + {\bf b}.
\end{displaymath} (1261)

This defines vector addition. By completing the parallelogram $PQRS$, we can also see that
\begin{displaymath}
\stackrel{\displaystyle \rightarrow}{PR} \,= \, \stackrel{\...
...le \rightarrow}{PS}+\stackrel{\displaystyle \rightarrow}{SR}.
\end{displaymath} (1262)

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 above equation is equivalent to
\begin{displaymath}
{\bf c} = {\bf a} + {\bf b} = {\bf b} + {\bf a}.
\end{displaymath} (1263)

We conclude that the addition of vectors is commutative. It can also be shown that the associative law holds: i.e.,
\begin{displaymath}
{\bf a} +
({\bf b} + {\bf c}) = ({\bf a} + {\bf b}) + {\bf c}.
\end{displaymath} (1264)

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

\begin{displaymath}
{\bf a} + {\bf0} = {\bf a},
\end{displaymath} (1265)

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: i.e.,
\begin{displaymath}
{\bf a} + (-{\bf a}) = {\bf0}.
\end{displaymath} (1266)

We can also define the difference of two vectors, ${\bf a}$ and ${\bf b}$, as
\begin{displaymath}
{\bf c} = {\bf a} - {\bf b} = {\bf a} +(-{\bf b}).
\end{displaymath} (1267)

This definition of vector subtraction is illustrated in Figure A.98.

Figure A.98: Vector subtraction.
\begin{figure}
\epsfysize =1.75in
\centerline{\epsffile{AppendixA/figA.02a.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, since $n\,{\bf a} = \vert n\vert\,(-{\bf a})$, 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

$\displaystyle n\,(m\,{\bf a})$ $\textstyle =$ $\displaystyle n\,m\,{\bf a} = m\,(n\,{\bf a}),$ (1268)
$\displaystyle (n+m)\,{\bf a}$ $\textstyle =$ $\displaystyle n\,{\bf a} + m\,{\bf a},$ (1269)
$\displaystyle n\,({\bf a} + {\bf b})$ $\textstyle =$ $\displaystyle n\,{\bf a} + n\,{\bf b}.$ (1270)


next up previous
Next: Cartesian Components of a Up: Vector Algebra and Vector Previous: Scalars and Vectors
Richard Fitzpatrick 2011-03-31