Question

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

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (D)

Let A be the first term and x the common ratio of G.P.
So, $a=Ax^{p-1} \, \Rightarrow \, log \,a=log\, A+\left(p-1\right)log \,x$
$Similarly, \, log \,b=log\,A+\left(q-1\right)log\,x\quad$
$and \, log c = log A+\left(r-1\right) log \,x$
$If \, \vec{a}=log \, a^{2} \hat{i}+log \, b^{2} \hat{j}+log c^{2}\hat{k}$
$and \, \vec{\beta}=\left(q-r\right)\hat{i}+\left(r-p\right)\hat{j}+\left(p-q\right)\hat{k} \, then $
$\vec{\alpha.} \vec{\beta}=2\left[log\,a\left(q-r\right)+log\,b\left(r-p\right)+log\,c\left(p-q\right)\right]$
$=2 [\left(q-r\right)\left\{log\,A+\left(p-1\right)log\,x\right\}+\left(r-p\right)\left\{log\,A+\left(q-1\right)log\,x\right\}$
$+\left(p-q\right)\left\{log\,A+\left(r-1\right)log\,x\right\}]$
$=2[ \left(q-r+r-p+p-q\right)log \,A+(qp-pr-p+r+qr-pq$
$-r+p+pr-qr-p+q ) log\,x ]=0$
Hence, the angle between $\vec{\alpha} \, and \, \vec{\beta} \, is \, \frac{\pi}{2}\cdot$