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(B) Since $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Let the

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$\vec{u}, \vec{v}$ are orthogonal vectors

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(B) Since $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Let the first term of this $A.P.$ be $A$ and the common difference be $D$.Then,$\frac{1}{a} = A + (p-1)D, \frac{1}{b} = A + (q-1)D, \frac{1}{c} = A + (r-1)D$.Subtracting these,we get $\frac{1}{a} - \frac{1}{b} = (p-q)D$,$\frac{1}{b} - \frac{1}{c} = (q-r)D$,and $\frac{1}{c} - \frac{1}{a} = (r-p)D$.Thus,$(q-r) = \frac{1/b - 1/c}{D} = \frac{c-b}{bcD}$,$(r-p) = \frac{a-c}{acD}$,and $(p-q) = \frac{b-a}{abD}$.Now,calculate the dot product $\vec{u} \cdot \vec{v} = (q-r)\frac{1}{a} + (r-p)\frac{1}{b} + (p-q)\frac{1}{c}$.Substituting the values: $\vec{u} \cdot \vec{v} = \frac{c-b}{bcD} \cdot \frac{1}{a} + \frac{a-c}{acD} \cdot \frac{1}{b} + \frac{b-a}{abD} \cdot \frac{1}{c}$.$\vec{u} \cdot \vec{v} = \frac{1}{abcD} (c-b + a-c + b-a) = \frac{1}{abcD} (0) = 0$.Since the dot product is $0$,the vectors $\vec{u}$ and $\vec{v}$ are orthogonal.

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then:

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