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

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(B) Let the first term of the $GP$ be $u$ and the common ratio be $z$. Then, $T_p = u z^{p-1} = a$, $T_q = u z^{q-1} = b$, and $T_r = u z^{r-1} = c$.

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) Let the first term of the $GP$ be $u$ and the common ratio be $z$. Then, $T_p = u z^{p-1} = a$, $T_q = u z^{q-1} = b$, and $T_r = u z^{r-1} = c$. Taking the logarithm on both sides: $\log a = \log u + (p-1) \log z$ $\log b = \log u + (q-1) \log z$ $\log c = \log u + (r-1) \log z$ 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 $\vec{B} = (q-r) i + (r-p) j + (p-q) k$. The dot product $\vec{A} \cdot \vec{B} = 2 [(\log a)(q-r) + (\log b)(r-p) + (\log c)(p-q)]$. Substituting $\log a, \log b, \log c$ in terms of $p, q, r$: $(\log a)(q-r) + (\log b)(r-p) + (\log c)(p-q) = [\log u + (p-1)\log z](q-r) + [\log u + (q-1)\log z](r-p) + [\log u + (r-1)\log z](p-q)$. $= \log u (q-r+r-p+p-q) + \log z [(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$. $= \log u (0) + \log z [pq - pr - q + r + qr - qp - r + p + rp - rq - p + q] = 0$. Since the dot product is $0$, the vectors are perpendicular. Therefore, the angle $	heta = \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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