Question

If $p^{th},q^{th},r^{th}$ terms of G.P. Are the possitive numbers a, b, c respectively then angle between the vectors
$\left(\log a^2\right)\hat{i}+\left(\log b^2\right)\hat{j}+\left(\log c^2\right)\hat{k}$ and
$(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k}$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. ${\sin }^{-1}\left(\frac{1}{\sqrt{a^2+b^2+c^2}}\right)$
• D. ${\cos }^{-1}\left(\frac{pqr}{\sqrt{p^2+q^2+r^2}}\right)$

Answer 1

The correct option is B $\frac{\pi }{2}$

According to the question,

a,b,c are the pth, qth and rth terms of G.P.

Let A be the first term and R be the common ratio.

Then,

$a=A{R}^{p-1}$

$b=A{R}^{q-1}$

$c=A{R}^{r-1}$

Taking log on both sides, we have,

$\log a=\log A+(p-1)\log R$......(1)

$\log b=\log A+(q-1)\log R$......(2)

$\log c=\log A+(r-1)\log R$......(3)

$\left(1\right)-\left(2\right)\Rightarrow \log a-\log b=\log R(p-q)$

$\left(2\right)-\left(3\right)\Rightarrow \log b-\log c=\log R(q-r)$

$\left(3\right)-\left(1\right)\Rightarrow \log c-\log a=\log R(r-p)$

Let $\theta$ be the desired angle between the 2 vectors,

Then,

$\cos \theta =\frac{(\log a^3\hat{i}+\log b^3\hat{j}+\log c^3\hat{k})\times [(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k}]}{\sqrt{\left[\right(\log a^3{)}^2+(\log b^3{)}^2+(\log c^3{)}^2]\times [(q-r{)}^2+(r-p{)}^2+(p-q{)}^2]}}$

Substituting the values of $(p-q),(q-r),(r-p)$ from above, we have,

$\cos \theta =\frac{\log a\left(\frac{\log b-\log c}{\log R}\right)+\log b\left(\frac{\log c-\log a}{\log R}\right)+\log c\left(\frac{\log a-\log b}{\log R}\right)}{\sqrt{\left[\right(\log a{)}^2+(\log b{)}^2+(\log c{)}^2]\{{\left(\frac{\log b-\log c}{\log R}\right)}^2+{\left(\frac{\log c-\log a}{\log R}\right)}^2+{\left(\frac{\log a-\log b}{\log R}\right)}^2\}}}$

$\Rightarrow \cos \theta =\frac{0}{\sqrt{\left[\right(\log a{)}^2+(\log b{)}^2+(\log c{)}^2]\{{\left(\log b-\log c\right)}^2+{\left(\log c-\log a\right)}^2+{\left(\log a-\log b\right)}^2\}}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta =90°$

Thus, the angle between the 2 given vectors is $90°$