If `T_(p), T_(q) and T_(r)` of a G.P. are `+`ive numbers a, b, c respectively, the vectors
`alpha =ilog a+jlogb+klogc`,
`beta =i(q-r)+j(r-p)+k(p-q)`
are perpendicular . True and False ?

Answer

To determine whether the vectors \(\alpha\) and \(\beta\) are perpendicular, we need to check if their dot product is zero. Let's break down the solution step by step.

### Step 1: Define the vectors
Given:
- \( T_p = a \)
- \( T_q = b \)
- \( T_r = c \)

The vectors are defined as:
\[ \alpha = i \log a + j \log b + k \log c \]
\[ \beta = i(q - r) + j(r - p) + k(p - q) \]

### Step 2: Calculate the components of \(\beta\)
From the definitions of \(T_p\), \(T_q\), and \(T_r\), we can express \(q - r\), \(r - p\), and \(p - q\) in terms of logarithms:
- \(q - r = \log b - \log c = \log \left(\frac{b}{c}\right)\)
- \(r - p = \log c - \log a = \log \left(\frac{c}{a}\right)\)
- \(p - q = \log a - \log b = \log \left(\frac{a}{b}\right)\)

Thus, we can rewrite \(\beta\) as:
\[ \beta = i \log \left(\frac{b}{c}\right) + j \log \left(\frac{c}{a}\right) + k \log \left(\frac{a}{b}\right) \]

### Step 3: Compute the dot product \(\alpha \cdot \beta\)
The dot product of two vectors \(\alpha\) and \(\beta\) is given by:
\[ \alpha \cdot \beta = \log a \cdot \log \left(\frac{b}{c}\right) + \log b \cdot \log \left(\frac{c}{a}\right) + \log c \cdot \log \left(\frac{a}{b}\right) \]

### Step 4: Simplify the dot product
Substituting the logarithmic expressions:
\[ \alpha \cdot \beta = \log a \cdot (\log b - \log c) + \log b \cdot (\log c - \log a) + \log c \cdot (\log a - \log b) \]

Expanding this:
\[ = \log a \log b - \log a \log c + \log b \log c - \log b \log a + \log c \log a - \log c \log b \]

### Step 5: Combine like terms
Notice that:
- \(\log a \log b\) and \(-\log b \log a\) cancel each other.
- \(\log b \log c\) and \(-\log c \log b\) cancel each other.
- \(\log c \log a\) and \(-\log a \log c\) cancel each other.

Thus, we find:
\[ \alpha \cdot \beta = 0 \]

### Conclusion
Since the dot product \(\alpha \cdot \beta = 0\), the vectors \(\alpha\) and \(\beta\) are indeed perpendicular.

### Final Answer
**True**: The vectors \(\alpha\) and \(\beta\) are perpendicular.

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