Question

If the $pth,qth$ and $rth$ terms of $G.P.$ are positive numbers $a,b$ and $c$, respectively, then the angle between the vectors $i{l}_{n}a+j{l}_{n}b+k{l}_{n}c$ and $i(q-r)+j(r-p)+k(p-q)$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

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

Let $x$ be the first term and $y$ be the common ratio of the $G.P.$

then $pth$ term $=a=x{y}^{p-1}$

$qth$ term $=b=x{y}^{q-1}$,

$rth$ term $=c=x{y}^{r-1}$

Now find product of the given vectors

$\Rightarrow (i{l}_{n}a+j{l}_{n}b+k{l}_{n}c).(i(q-r)+j(r-p)+k(p-q))=l_{n}a(q-r)+l_{n}b(r-p)+l_{n}c(p-q)$

Putting values of $a,b,c$ to get

$=l_{n}x\{(q-r)+(r-p)+(p-q)\}+l_{n}y\{(q-r)(r-p)(p-1)q-1+(p-q)(r-1)\}=0$

Hence the vectors are perpendicular