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
$\cos \theta =0\Rightarrow \theta =\frac{\pi }{2}$