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In a normed vector space E, if $c-a=t(b-a)$, with $t\geq 1$, then $d(a,c)=d(a,b)+d(b,c)$.

I see this problem like, a vector with extremes , a and c, then $c-a=t(b-a)$ is a separation of the segment $\widehat{ac}$. But I don't see the relation between $c-a=t(b-a)$ and $d(a,c)=d(a,b)+d(b,c)$

Ahna Akbar
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2 Answers2

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Note that $c=(1-t)a+tb$. Therefore, \begin{align*} d(a,b)+d(b,c)\underset{\phantom{t\geq1}}=&\,\|a-b\|+\|b-c\|=\|a-b\|+\|b-(1-t)a-tb\|\\ \underset{\phantom{t\geq1}}=&\,\|a-b\|+\|(1-t)(b-a)\|=\|a-b\|+|1-t|\|b-a\|\\ \underset{t\geq 1}{=}&\,\|a-b\|+(t-1)\|a-b\|=\color{blue}{t\|a-b\|}. \end{align*} Also, \begin{align*} d(a,c)=\|a-c\|=\|a-(1-t)a-tb\|=\|t(a-b)\|=|t|\|a-b\|=\color{blue}{t\|a-b\|}. \end{align*}

triple_sec
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  • very nice answer!!, I have more one question, If I have vector space with inner product, then $d(a,c)=d(a,b)+d(b,c)$ implies $c-a=t(b-a)$?? – Ahna Akbar Aug 16 '15 at 02:41
  • @AhnaAkbarperez Not necessarily. Take any three points $a,b,c$ in the space such that $a=b\neq c$. Then, $$d(a,c)=0+d(a,c)=d(a,a)+d(a,c)=d(a,b)+d(b,c),$$ but $c-a\neq0$, while $t(b-a)=0$ for any $t\in\mathbb R$. – triple_sec Aug 16 '15 at 03:10
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Hint: the condition $c -a = t(b -a) $ with $t \ge 1$ is telling you that $a$, $b$ and $c$ are collinear.

Rob Arthan
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