If `u = q-r,r-p,p-q and v = loga^(2), logb^(2), logc^(2)` and a,b, c and `T_(p), T_(q), T_(r)`of a G.P. then angle between vectors u and v is ........

Answer

To find the angle between the vectors \( \mathbf{u} \) and \( \mathbf{v} \), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \mathbf{u} = (q - r, r - p, p - q) \]
\[ \mathbf{v} = (\log a^2, \log b^2, \log c^2) \]

### Step 2: Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \)
The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by:
\[ \mathbf{u} \cdot \mathbf{v} = (q - r) \log a^2 + (r - p) \log b^2 + (p - q) \log c^2 \]

### Step 3: Substitute values for \( q - r \), \( r - p \), and \( p - q \)
Since \( a, b, c \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a geometric progression (G.P.), we can express them as:
- \( a = A r^{p-1} \)
- \( b = A r^{q-1} \)
- \( c = A r^{r-1} \)

Using logarithmic properties:
\[ \log a = \log A + (p-1) \log r \]
\[ \log b = \log A + (q-1) \log r \]
\[ \log c = \log A + (r-1) \log r \]

### Step 4: Find \( q - r \), \( r - p \), and \( p - q \)
Using the definitions of \( a, b, c \):
\[ q - r = \frac{\log b - \log c}{\log r} \]
\[ r - p = \frac{\log c - \log a}{\log r} \]
\[ p - q = \frac{\log a - \log b}{\log r} \]

### Step 5: Substitute these values back into the dot product
Substituting these values into the dot product:
\[ \mathbf{u} \cdot \mathbf{v} = 2 \left( \log a \cdot \frac{\log b - \log c}{\log r} + \log b \cdot \frac{\log c - \log a}{\log r} + \log c \cdot \frac{\log a - \log b}{\log r} \right) \]

### Step 6: Simplify the expression
When you simplify this expression, you will find that:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \]

### Step 7: Determine the angle between the vectors
The angle \( \theta \) between two vectors can be found using the formula:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \]
Since \( \mathbf{u} \cdot \mathbf{v} = 0 \), this implies:
\[ \cos \theta = 0 \]
Thus, the angle \( \theta \) is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer
The angle between the vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \frac{\pi}{2} \).

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