Question

If $p^{th},q^{th},r^{th}$ terms of a $G.P$. are the positive numbers $a,b,c$, then angle between the vectors $\log a^3\hat{i}+\log b^3\hat{j}+\log c^3\hat{k}$ and $(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k}$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. ${\sin }^{-1}\left(\frac{1}{\sqrt{a^2+b^2+c^2}}\right)$

Answer 1

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

Taking $a_1$ as the first term and $d$ the common difference of a G.P.

$a=a_1\times d^{p-1}$....(i)

$b=a_1\times d^{q-1}$.....(ii)

$c=a_1\times d^{r-1}$.....(iii)

From (i) and (ii)

$\frac{a}{b}=d^{(p-q)}$...(iv)

Similarly we get $\frac{b}{c}=d^{(q-r)}$.....(v),

$\frac{c}{a}=d^{(r-p)}$....(vi)

We need angle $\theta$ between vectors $3\log a\hat{i}+3\log b\hat{j}+3\log c\hat{k}$ and $(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k}$

.$\cos \theta =\frac{3(q-r)\log a+3(r-p)\log b+3(p-q)\log c}{3\sqrt{(\log a{)}^2+(\log b{)}^2+(\log c{)}^2}\times \sqrt{(p-q{)}^2+(r-p{)}^2+(q-r{)}^2}}$

Numerator part $=3\left[\right(q-r)\log a+(r-p)\log b+(p-q\left)\log c\right]=3[\left(\frac{\log b-\log c}{\log d}\right)\log a+\left(\frac{\log c-\log a}{\log d}\right)\log b+\left(\frac{\log a-\log b}{\log d}\right)\log c]=0$

(from (v), $\log \left(\frac{b}{c}\right)=(q-r)\log d$; similarly we can find $p-q$ and $r-p$)

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