<?xml version="1.0" encoding="UTF-8"?>

<record version="3" id="10249">
 <title>difference of vectors</title>
 <name>DifferenceOfVectors</name>
 <created>2008-02-09 13:14:03</created>
 <modified>2008-08-26 06:28:17</modified>
 <type>Definition</type>
<parent id="10244">parallelogram principle</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="53A45"/>
 </classification>
 <defines>
	<concept>difference vector</concept>
	<concept>opposite vector</concept>
 </defines>
 <synonyms>
	<synonym concept="difference of vectors" alias="vector difference"/>
	<synonym concept="difference of vectors" alias="vector subtraction"/>
 </synonyms>
 <related>
	<object name="CommonPointOfTriangleMedians"/>
	<object name="Difference2"/>
	<object name="ProvingThalesTheoremWithVectors"/>
	<object name="VectorValuedFunction2"/>
 </related>
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\usepackage{amssymb}
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\newtheorem*{thmplain}{Theorem}
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 <content>Let $\vec{a}$ and $\vec{b}$ be two vectors in the plane (or in a vector space).\, The {\em difference vector} or {\em difference} \,$\vec{a}\!-\!\vec{b}$\, of $\vec{a}$ and $\vec{b}$ is a vector $\vec{d}$\, such that
                 $$\vec{b}+\vec{d} = \vec{a}.$$
Thus we have
\begin{align}
\vec{b}+(\vec{a}\!-\!\vec{b}) = \vec{a}.
\end{align}

According to the procedure of forming the sum of vectors by setting the addends one after the other, the equation (1) tallies with the picture below; when the minuend and the subtrahend emanate from a common initial point, their difference vector can be directed from the terminal point of the subtrahend to the terminal point of the minuend. 

\begin{center}
\begin{pspicture}(0,-1)(4,2.5)
\psdot[linecolor=black](0,0)
\psline[arrows=-&gt;,arrowsize=5pt,linecolor=black](0,0)(-1.5,2)
\psline[arrows=-&gt;,arrowsize=5pt,linecolor=red](-1.5,2)(4,1)
\psline[arrows=-&gt;,arrowsize=5pt,linecolor=blue](0,0)(4,1)
\rput[a](2.1,0.2){$\vec{a}$}
\rput[a](-1,0.7){$\vec{b}$}
\rput[a](1.6,1.76){$\vec{a}\!-\!\vec{b}$}
\end{pspicture}
\end{center}


\textbf{Remark.}\, It is easily seen that the difference $\vec{a}\!-\!\vec{b}$ is same as the sum vector
$$\vec{a}\!+\!(-\vec{b})$$
where $-\vec{b}$ is the {\em opposite vector} of $\vec{b}$:\, it may be represented by the directed line segment from the terminal point of $\vec{b}$ to the initial point of $\vec{b}$.

</content>
</record>
