Question

Let the distinct numbers \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be the \(\mathrm{p}^{\text{th}}, \mathrm{q}^{\text{th}}\) and \(\mathrm{r}^{\text{th}}\) terms of a geometric progression of positive terms. Then the angle between the vectors \(\overrightarrow{\mathrm{u}}=\left(\log_{\mathrm{e}} \mathrm{a}^{2}\right) \hat{i}+\left(\log_{\mathrm{e}} \mathrm{b}^{2}\right) \hat{j}+\left(\log_{\mathrm{e}} \mathrm{c}^{2}\right) \hat{k}\) and \(\overrightarrow{\mathrm{v}}=(\mathrm{q}-\mathrm{r}) \hat{i}+(\mathrm{r}-\mathrm{p}) \hat{j}+(\mathrm{p}-\mathrm{q}) \hat{k}\) is :

• A. \(0\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)

Hint:

Write each of a, b, c as A R^(n-1), take logarithms and find the dot product u.v; it simplifies to zero.

Answer 1

The Correct Option is D

Let the GP have first term A and common ratio R, so \(a=AR^{p-1}\), \(b=AR^{q-1}\), \(c=AR^{r-1}\).
Taking logarithms: \(\log_e a = \log_e A + (p-1)\log_e R\), and similarly for b and c.
\(\vec u\cdot\vec v = 2\left[(q-r)\log_e a + (r-p)\log_e b + (p-q)\log_e c\right]\).
The terms with \(\log_e A\) give \(\log_e A\,[(q-r)+(r-p)+(p-q)] = 0\).
The terms with \(\log_e R\) give \(\log_e R\,[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]\).
Now \(p(q-r)+q(r-p)+r(p-q)=0\) and \((q-r)+(r-p)+(p-q)=0\), so the bracket is \(0-0=0\).
Hence \(\vec u\cdot\vec v = 0\), so the vectors are perpendicular and the angle is \(\frac{\pi}{2}\).
Answer: (4) \(\frac{\pi}{2}\)