If the p^{th},q^{th},and r^{th} terms of a ge | vedclass

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(D) Let the first term of the geometric progression be $A$ and the common ratio be $R$. Then the terms are given by:$a = AR^{p-1}$,$b = AR^{q-1}$,$c =

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$\frac{\pi}{2}$

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(D) Let the first term of the geometric progression be $A$ and the common ratio be $R$. Then the terms are given by:$a = AR^{p-1}$,$b = AR^{q-1}$,$c = AR^{r-1}$.Taking the logarithm on both sides:$\log a = \log A + (p-1)\log R$$\log b = \log A + (q-1)\log R$$\log c = \log A + (r-1)\log R$Now,consider the dot product $\vec{u} \cdot \vec{v} = (\log a)(q-r) + (\log b)(r-p) + (\log c)(p-q)$.Substituting the expressions for $\log a$,$\log b$,and $\log c$:$\vec{u} \cdot \vec{v} = [\log A + (p-1)\log R](q-r) + [\log A + (q-1)\log R](r-p) + [\log A + (r-1)\log R](p-q)$$= \log A(q-r+r-p+p-q) + \log R[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$$= \log A(0) + \log R[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]$$= 0 + \log R(0) = 0$.Since the dot product $\vec{u} \cdot \vec{v} = 0$,the vectors are perpendicular to each other.Therefore,the angle between them is $\frac{\pi}{2}$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a geometric progression are $a$,$b$,and $c$ respectively,then find the angle between the vectors $\vec{u} = (\log a)\hat{i} + (\log b)\hat{j} + (\log c)\hat{k}$ and $\vec{v} = (q - r)\hat{i} + (r - p)\hat{j} + (p - q)\hat{k}$.

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