If p-th,q-th,and r-th terms of a geometric pr | vedclass

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(B) Let the first term of the $GP$ be $u$ and the common ratio be $z$. Then,$T_p = u z^{p-1} = a \Rightarrow \log u + (p-1) \log z = \log a$ $(i)$ $T_

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$\frac{\pi}{2}$

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(B) Let the first term of the $GP$ be $u$ and the common ratio be $z$. Then,$T_p = u z^{p-1} = a \Rightarrow \log u + (p-1) \log z = \log a$ $(i)$ $T_q = u z^{q-1} = b \Rightarrow \log u + (q-1) \log z = \log b$ (ii) $T_r = u z^{r-1} = c \Rightarrow \log u + (r-1) \log z = \log c$ (iii) Let $\vec{A} = (\log a^2) i + (\log b^2) j + (\log c^2) k = 2(\log a) i + 2(\log b) j + 2(\log c) k$ Let $\vec{B} = (q-r) i + (r-p) j + (p-q) k$ From $(i)$,(ii),and (iii),we have: $q-r = \frac{\log b - \log a}{\log z} - \frac{\log c - \log b}{\log z} = \frac{\log b - \log c}{\log z}$ (This is not quite right,let's use the property: $\log b - \log c = (q-r) \log z$). Actually,$\log b - \log c = (q-r) \log z$ and $\log c - \log a = (r-p) \log z$ and $\log a - \log b = (p-q) \log z$. The dot product $\vec{A} \cdot \vec{B} = 2(\log a)(q-r) + 2(\log b)(r-p) + 2(\log c)(p-q)$. Substituting the differences: $= \frac{2}{\log z} [(\log a)(\log b - \log c) + (\log b)(\log c - \log a) + (\log c)(\log a - \log b)]$. $= \frac{2}{\log z} [\log a \log b - \log a \log c + \log b \log c - \log b \log a + \log c \log a - \log c \log b] = 0$. Since the dot product is $0$,the angle $	heta$ between the vectors is $\frac{\pi}{2}$.

If $p$-th,$q$-th,and $r$-th terms of a geometric progression are the positive numbers $a, b,$ and $c$ respectively,then the angle between the vectors $(\log a^2) i + (\log b^2) j + (\log c^2) k$ and $(q-r) i + (r-p) j + (p-q) k$ is

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