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, $a=A{x}^{p-1}\Rightarrow loga=logA+\left(p-1\right)logx$
$Similarly,logb=logA+\left(q-1\right)logx$
$andlogc=logA+\left(r-1\right)logx$
$If\vec{a}=log{a}^2\hat{i}+log{b}^2\hat{j}+log{c}^2\hat{k}$
$and\vec{\beta }=\left(q-r\right)\hat{i}+\left(r-p\right)\hat{j}+\left(p-q\right)\hat{k}then$
$\overrightarrow{\alpha .}\vec{\beta }=2\left[loga\left(q-r\right)+logb\left(r-p\right)+logc\left(p-q\right)\right]$
$=2[\left(q-r\right)\left\{logA+\left(p-1\right)logx\right\}+\left(r-p\right)\left\{logA+\left(q-1\right)logx\right\}$
$+\left(p-q\right)\left\{logA+\left(r-1\right)logx\right\}]$
$=2[\left(q-r+r-p+p-q\right)logA+(qp-pr-p+r+qr-pq$
$-r+p+pr-qr-p+q)logx]=0$
Hence, the angle between $\vec{\alpha }and\vec{\beta }is\frac{\pi }{2}\cdot$