Question

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) i +\left(\log b^{2}\right) j +\left(\log c^{2}\right) k $ and $(q-r) i +(r-p) j +(p-q) k$ is

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{2}$
• C. $\sin ^{-1} \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}}}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (B)

Let first term of a GP be $u$ and common ratio $z$.
$\therefore T_{p}=u z^{p-1}=a$
$\Rightarrow \log u+(p-1) \log z=\log a$......(i)
$T_{q}=u z^{q-1}=b$
$\Rightarrow \log u+(q-1) \log z=\log b$.......(ii)
and $T_{r}=u z^{r-1}=c$
$\Rightarrow \log u+(r-1) \log z=\log c $......(iii)
Let $\theta$ be the angle between
$\left(\log a^{2}\right) i +\left(\log b^{2}\right) j +\left(\log c^{2}\right) k$
and $(q-r) i +(r-p) j +(p-q) k$ is
$\cos \theta=\frac{\left[\left(\log a^{2}\right)\left(q-r\right)+\left(\log b^{2}\right)\left(r-p\right)+\left(\log c^{2}\right)\left(p-q\right)\right]}{\left[\sqrt{\left(log a^{2}\right)^{2} +\left(\log b^{2}\right)^{2} +\left(\log c^{2}\right)^{2} } \sqrt{\left(q-r\right)^{2} +\left(r-p\right)^{2} +\left(p-q\right)^{2}}\right]}$......(iv)
From Eqs. (i), (ii) and (iii)
$q-r=\log b-\log c, $
$r-p=\log c-\log a $
$p-q=\log a-\log b $
$\therefore $ From Eq. (iv), taking numerator term
$=2 \log a(\log b-\log c)+ 2 \log b(\log c-\log a) $
$+2 \log c(\log a-\log b)$
$=0$
$\therefore $ From Eq. (i), we get
$\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$