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'parallelogram principle'
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| Title of object: |
parallelogram principle |
| Canonical Name: |
ParallelogramPrinciple |
| Type: |
Topic |
| Created on: |
2008-02-06 16:39:48 |
| Modified on: |
2008-02-06 16:39:48 |
| Classification: |
msc:53A45 |
| Defines: |
sum of vectors, sum |
| Synonyms: |
parallelogram principle=addition of vectors
parallelogram principle=vector addition |
Preamble:
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Content:
\PMlinkescapeword{right}
A starting \PMlinkescapetext{point} for learning vectors is to think that they are {\em directed line segments}. Thus a vector $\vec{u}$ has a direction and a length (magnitude) and nothing else. Therefore, if two vectors $\vec{u}$ and $\vec{v}$ have a same direction and a same length, one can keep them identical (and denote\, $\vec{u} = \vec{v}$). So the location of a certain vector in the plane (or in the space) is insignificant; in fact one may also think that this vector consists of all possible directed line segments having a common direction and a common length.
However, a vector $\vec{u}$ as an infinite set of directed segments is quite uncomfortable to handle, and one can choose from all possible representants of $\vec{u}$ one individual directed segment $\overrightarrow{AB}$, i.e. a line segment directed from a certain point $A$ (the {\em initial point}) to another certain point $B$ (the {\em terminal point}). Although $\overrightarrow{AB}$ is only a representant of $\vec{u}$, one may write\, $\overrightarrow{AB} = \vec{u}$\, or\, $\vec{u} = \overrightarrow{AB}$.
For describing a vector $\vec{u}$, it's convenient to know its position in the coordinate system of the plane (or the space); there one can say e.g. how great a displacement $\vec{u}$ means from left to right (i.e. in the direction of $x$-axis) and how great from below upwards (i.e. in the direction of the $y$-axis); those displacements may be expressed with two numbers.\, One may for example write
\begin{align}
\vec{u} = \left(\!\begin{array}{c} +5\\-1 \end{array}\!\right)\!,
\end{align}
where the first (upper) number $+5$ tells that the vector leads 5 length-units to the right and the second (lower) number $-1$ that it leads 1 length-unit downwards.
Then to the {\em addition of vectors}!\, Since the vector may be interpreted as a \PMlinkescapetext{combination} of a horizontal displacement and a vertical displacement, it's meaningful that by the addition of two vectors the horizontal displacements are summed and likewise the vertical displacements.\, Accordingly, if we have
\begin{align}
\vec{v} = \left(\!\begin{array}{c} +1\\-3 \end{array}\!\right)\!,
\end{align}
then the {\em sum} of the vectors (1) and (2) is
$$\vec{u}+\vec{v} =
\left(\!\begin{array}{c} +5\\-1 \end{array}\!\right)+\left(\!\begin{array}{c} +1\\-3 \end{array}\!\right) =
\left(\!\begin{array}{c} 5+1\\-1-3 \end{array}\!\right) = \left(\!\begin{array}{c} +6\\-4 \end{array}\!\right)$$ |
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