If $p$ th, $q$ th, $r$ th terms of a geometric progression are the positive numbers $a, b$ and $c$…
If $p$ th, $q$ th, $r$ th terms of a geometric progression are the positive numbers $a, b$ and $c$ respectively, then the angle between the vectors $\left(\log a^2\right) \mathbf{i}+\left(\log b^2\right) \mathbf{j}+\left(\log c^2\right) \mathbf{k} \quad$ and $(q-r) \mathbf{i}+(r-p) \mathbf{j}+(p-q) \mathbf{k}$ is
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\sin ^{-1} \frac{1}{\sqrt{a^2+b^2+c^2}}$
$\frac{\pi}{4}$
Solution
Let first term of a GP be $u$ and common ratio $z$.
$\therefore \quad T_p=u z^{p-1}=a$
$\begin{aligned} & \Rightarrow \quad \begin{array}{l}\log u+(p-1) \log z=\log a \\ T_q=u z^{q-1}=b\end{array} \\ & \Rightarrow \log u+(q-1) \log z=\log b \\ & \text { and } \quad T_r=u z^{r-1}=c\end{aligned}$
$\Rightarrow \log u+(r-1) \log z=\log c$
Let $\theta$ be the angle between
$\left(\log a^2\right) \mathbf{i}+\left(\log b^2\right) \mathbf{j}+\left(\log c^2\right) \mathbf{k}$
and
$(q-r) \mathbf{i}+(r-p) \mathbf{j}+(p-q) \mathbf{k}$ is
$\cos \theta=\frac{\left[\begin{array}{r}\left(\log a^2\right)(q-r)+\left(\log b^2\right)(r-p) \\ +\left(\log c^2\right)(p-q)\end{array}\right]}{\left[\begin{array}{r}\sqrt{\left(\log a^2\right)^2+\left(\log b^2\right)^2+\left(\log c^2\right)^2} \\ \sqrt{(q-r)^2+(r-p)^2+(p-q)^2}\end{array}\right]}$
From Eqs. (i), (ii) and (iii)
$\begin{gathered}q-r=\log b-\log c, \quad r-p=\log c-\log a \\ p-q=\log a-\log b\end{gathered}$
$\therefore$ From Eq. (iv), taking numerator term
$\begin{aligned} &=2 \log a(\log b-\log c)+2 \log b(\log c-\log a) \\ &+2 \log c(\log a-\log b) \\ &=0\end{aligned}$
$\therefore$ From Eq. (i), we get
$\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$