In a G.P. `T_(p) = a, T_(q) = b and T_(r) = c` where a, b, c are `+ive` then angle between the vectors `log a^(2)i + log b^(2)j + logc^(2)k` and `(q-r)i+(r-p)j+(p-q)k` is :

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `sin^(-1) ""(1)/(sqrt(suma^(2)))`
D. none

Answer

To find the angle between the vectors \( \log a^2 \mathbf{i} + \log b^2 \mathbf{j} + \log c^2 \mathbf{k} \) and \( (q-r) \mathbf{i} + (r-p) \mathbf{j} + (p-q) \mathbf{k} \), we can follow these steps:

### Step 1: Understand the terms in the G.P.
Given that \( T_p = a \), \( T_q = b \), and \( T_r = c \) are terms of a geometric progression (G.P.), we can express them in terms of the first term \( A \) and the common ratio \( R \):
- \( T_p = A R^{p-1} = a \)
- \( T_q = A R^{q-1} = b \)
- \( T_r = A R^{r-1} = c \)

### Step 2: Set up the equations
From the above expressions, we can derive:
1. \( A R^{p-1} = a \)
2. \( A R^{q-1} = b \)
3. \( A R^{r-1} = c \)

### Step 3: Take logarithms
Taking logarithms of each equation:
1. \( \log a = \log A + (p-1) \log R \)
2. \( \log b = \log A + (q-1) \log R \)
3. \( \log c = \log A + (r-1) \log R \)

### Step 4: Rearranging the equations
By rearranging these equations, we can express \( p-q \), \( q-r \), and \( r-p \):
- From \( \log a \) and \( \log b \):
\[ p - q = \frac{\log a - \log b}{\log R} = \frac{\log \frac{a}{b}}{\log R} \]
- From \( \log b \) and \( \log c \):
\[ q - r = \frac{\log b - \log c}{\log R} = \frac{\log \frac{b}{c}}{\log R} \]
- From \( \log c \) and \( \log a \):
\[ r - p = \frac{\log c - \log a}{\log R} = \frac{\log \frac{c}{a}}{\log R} \]

### Step 5: Define the vectors
Now we can define the vectors:
- \( \mathbf{u} = \log a^2 \mathbf{i} + \log b^2 \mathbf{j} + \log c^2 \mathbf{k} \)
- \( \mathbf{v} = (q-r) \mathbf{i} + (r-p) \mathbf{j} + (p-q) \mathbf{k} \)

### Step 6: Calculate the dot product
To find the angle between the vectors, we calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \):
\[ \mathbf{u} \cdot \mathbf{v} = (\log a^2)(q-r) + (\log b^2)(r-p) + (\log c^2)(p-q) \]

### Step 7: Substitute the values
Substituting the values of \( q-r \), \( r-p \), and \( p-q \):
\[ \mathbf{u} \cdot \mathbf{v} = (\log a^2) \left( \frac{\log \frac{b}{c}}{\log R} \right) + (\log b^2) \left( \frac{\log \frac{c}{a}}{\log R} \right) + (\log c^2) \left( \frac{\log \frac{a}{b}}{\log R} \right) \]

### Step 8: Simplify the expression
On simplifying, we find that the terms cancel out, leading to:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \]

### Step 9: Conclusion
Since the dot product is zero, the vectors are perpendicular. Therefore, the angle \( \theta \) between the vectors is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer
The angle between the vectors is \( \frac{\pi}{2} \) radians.