Vector Algebra Question 50
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 $ (q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k} $
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: D
Solution:
- [d] Let A be the first and x the common ration of G.P.
So, $ a=A{x^{p-1}}\Rightarrow \log a=\log A+(p-1)\log ,x $
Similarly, $ \log ,b=\log A+(q-1)log,x $
and $ \log ,c=\log A+(r-1)log,x $
if $ \overset{\to }{\mathop{\alpha }},=\log ,a^{2}\hat{i}+\log b^{2}\hat{j}+\log ,c^{2}\hat{k} $
and $ \overset{\to }{\mathop{\beta }},=(q-r)\hat{i}+(r-p)\hat{j}+(p-q)\hat{k} $ then $ \overset{\to }{\mathop{\alpha }},.\overset{\to }{\mathop{\beta }},=2[log,a(q-r)+log,b(r-p)+log,c(p-q)] $
$ =2[(q-r){log,A+(p-1)log,x} $
$ +(r-p){log,A+(q-1)log,x} $
$ +(p-q){log,A+(r-1)log,x}] $
$ =2[(q-r+r-p+p-q)log,A $
$ +(qp-pr-p+r+qr-pq $
$ -r+p+pr-qr-p+q)\log x]=0 $
Hence, the angle between $ \overset{\to }{\mathop{\alpha }}, $ and $ \overset{\to }{\mathop{\beta }}, $ is $ \frac{\pi }{2}. $
BETA
