If `u = q-r,r-p, p-q and v = (1)/(a),(1)/(b),(1)/(c)` and a, b, c are `T_(p), T_(q), T_(r)` of an HP then the angle between the 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:
\[ \mathbf{u} = (q - r, r - p, p - q) \]
\[ \mathbf{v} = \left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right) \]
where \( a, b, c \) are the \( T_p, T_q, T_r \) of a Harmonic Progression (HP).

### Step 2: Understand the properties of HP
Since \( a, b, c \) are in HP, their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) will be in Arithmetic Progression (AP). This means:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

### Step 3: Calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \)
The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by:
\[ \mathbf{u} \cdot \mathbf{v} = (q - r) \cdot \frac{1}{a} + (r - p) \cdot \frac{1}{b} + (p - q) \cdot \frac{1}{c} \]

### Step 4: Substitute the values
Using the relationships from the AP:
- From \( 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \), we can express \( \frac{1}{a} + \frac{1}{c} \) in terms of \( \frac{1}{b} \):
\[ \mathbf{u} \cdot \mathbf{v} = (q - r) \cdot \frac{1}{a} + (r - p) \cdot \frac{1}{b} + (p - q) \cdot \frac{1}{c} \]

### Step 5: Analyze the terms
Using the relationships derived from the AP:
1. \( q - r \) can be expressed in terms of \( \frac{1}{b} \).
2. \( r - p \) can also be expressed similarly.
3. \( p - q \) can be expressed in terms of \( \frac{1}{b} \).

### Step 6: Conclude the dot product
After substituting and simplifying, we find that:
\[ \mathbf{u} \cdot \mathbf{v} = 0 \]

### Step 7: Find the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \)
The magnitudes of the vectors are not needed to find the angle since the dot product is zero.

### Step 8: Determine the angle
Since the dot product \( \mathbf{u} \cdot \mathbf{v} = 0 \), this implies that the vectors are perpendicular:
\[ \theta = \cos^{-1}(0) = \frac{\pi}{2} \]

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

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