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If $\Large\frac{1}{a},\frac{1}{b},\frac{1}{c}$are the $\large p^{th},q^{th}and\;r^{th}\;$ terms of an AP and $ \bar{u}=(q-r)\bar{i}+(r-p)\bar{j}+(p-q)\bar{k}\;and\;\bar{v}=\large\frac{1}{a}\bar{i}+\frac{1}{b}\bar{j}+\frac{1}{c}\bar{k}$ then prove that $\bar{u}\; and\;\bar{v}$ are orthogonal vectors.

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  • $n^{th}$ term, $t_n$ of an A.P. having ' A' as first term and 'd' as common difference is $A+(n-1)d$
  • If two vectors $\overrightarrow u\:and\:\overrightarrow v$ are orthogonal then $\overrightarrow u.\overrightarrow v=0$
Given $t_p=\frac{1}{a},\:\:t_q=\frac{1}{b}\:\:and\:\:t_r=\frac{1}{c}$ of an A.P.
$\Rightarrow \:A+(p-1)d=\frac{1}{a}$, $A+(q-1)d=\frac{1}{b}$ and
$\frac{1}{a}-\frac{1}{b}=(A+(p-1)d)-(A+(q-1)d)$ $=(p-q)d$
Similarly $\frac{1}{b}-\frac{1}{c}=(q-r)d$ and
$\Rightarrow \:p-q=\frac{b-a}{abd}$,
$q-r=\frac{c-b}{bcd}$ and
Given $\overrightarrow u=(q-r)\hat i+(r-p)\hat j+(p-q)\hat k$
$=\frac{c-b}{bcd}\hat i+\frac{a-c}{acd}\hat j+\frac{b-a}{abd}\hat k$
Also given $\overrightarrow v=\frac{1}{a}\hat i+\frac{1}{b}\hat j+\frac{1}{c}\hat k$
$\overrightarrow u.\overrightarrow v=(\frac{c-b}{bcd}\hat i+\frac{a-c}{acd}\hat j+\frac{b-a}{abd}\hat k).(\frac{1}{a}\hat i+\frac{1}{b}\hat j+\frac{1}{c}\hat k)$
$\Rightarrow \overrightarrow u.\overrightarrow v=0$
Hence $\overrightarrow u\:\:and\:\:\overrightarrow v$ are orthogonal.
Hence proved.
answered Apr 20, 2013 by rvidyagovindarajan_1

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