If a,b,c are the `p^("th") , q^("th") and r^("th") ` terms of G.P then the angle between the vector `vecu = (log a) hati + (log b)hatj + (log c) hatk and vecv -( q -r) hati + ( r -p) hati + ( r-p) hatj + ( p-q) hatk`, is

Options

A. `pi/3`
B. `pi/6`
C. `pi`
D. ` pi/2`

Explanation

To solve the problem, we need to find the angle between the vectors \( \vec{u} \) and \( \vec{v} \) given the terms of a geometric progression (G.P).

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
Let the first term of the G.P. be \( A \) and the common ratio be \( R \). The terms can be expressed as:
- \( a = A R^{p-1} \) (the \( p \)-th term)
- \( b = A R^{q-1} \) (the \( q \)-th term)
- \( c = A R^{r-1} \) (the \( r \)-th term)

2. **Express the vectors**:
The vector \( \vec{u} \) is given by:
\[ \vec{u} = \log a \hat{i} + \log b \hat{j} + \log c \hat{k} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ \vec{u} = \log(A R^{p-1}) \hat{i} + \log(A R^{q-1}) \hat{j} + \log(A R^{r-1}) \hat{k} \]
Using the logarithmic property \( \log(xy) = \log x + \log y \):
\[ \vec{u} = \left( \log A + (p-1) \log R \right) \hat{i} + \left( \log A + (q-1) \log R \right) \hat{j} + \left( \log A + (r-1) \log R \right) \hat{k} \]

3. **Simplify \( \vec{u} \)**:
\[ \vec{u} = \log A \hat{i} + \log A \hat{j} + \log A \hat{k} + (p-1) \log R \hat{i} + (q-1) \log R \hat{j} + (r-1) \log R \hat{k} \]
\[ \vec{u} = \log A \hat{i} + \log A \hat{j} + \log A \hat{k} + \log R \left( (p-1) \hat{i} + (q-1) \hat{j} + (r-1) \hat{k} \right) \]

4. **Define vector \( \vec{v} \)**:
The vector \( \vec{v} \) is given as:
\[ \vec{v} = (q - r) \hat{i} + (r - p) \hat{j} + (p - q) \hat{k} \]

5. **Calculate the dot product \( \vec{u} \cdot \vec{v} \)**:
\[ \vec{u} \cdot \vec{v} = \left( \log A + (p-1) \log R \right)(q - r) + \left( \log A + (q-1) \log R \right)(r - p) + \left( \log A + (r-1) \log R \right)(p - q) \]

6. **Simplify the dot product**:
Expanding the terms:
\[ = \log A (q - r) + (p-1) \log R (q - r) + \log A (r - p) + (q-1) \log R (r - p) + \log A (p - q) + (r-1) \log R (p - q) \]
Combine like terms:
\[ = \log A \left( (q - r) + (r - p) + (p - q) \right) + \log R \left( (p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q) \right) \]
The first part simplifies to zero:
\[ (q - r) + (r - p) + (p - q) = 0 \]
The second part also simplifies to zero due to the cyclic nature of the terms.

7. **Conclusion**:
Since \( \vec{u} \cdot \vec{v} = 0 \), the vectors \( \vec{u} \) and \( \vec{v} \) are orthogonal. Therefore, the angle \( \theta \) between them is:
\[ \theta = 90^\circ \]