If the harmonic conjugate of \( P(2,3,4) \) with respect to the line segment joining the points
\[
A(3,-2,2) \quad \text{and} \quad B(6,-17,-4)
\]
is \( Q(\alpha, \beta, \gamma) \), then the value of \( \alpha + \beta + \gamma \) is:
Show Hint
- \( \frac{2}{5} \)
- \( -\frac{3}{5} \)
- \( \frac{7}{5} \)
\( \frac{8}{5} \)
The Correct Option is B
Solution and Explanation
Step 1: Harmonic Conjugate Formula
The harmonic conjugate \( Q(\alpha, \beta, \gamma) \) of \( P(2,3,4) \) with respect to the line segment joining \( A(3,-2,2) \) and \( B(6,-17,-4) \) satisfies the section formula in harmonic division: \[ Q = \frac{mA - nB}{m - n}, \] where \( P \) divides \( AB \) internally in the ratio \( m:n \) and \( Q \) divides \( AB \) externally in the same ratio.
Step 2: Find the internal ratio of division
Using the section formula in 3D, we set: \[ P = \left( \frac{m(6) + n(3)}{m+n}, \frac{m(-17) + n(-2)}{m+n}, \frac{m(-4) + n(2)}{m+n} \right). \] Equating coordinates with \( P(2,3,4) \): \[ \frac{6m + 3n}{m+n} = 2, \quad \frac{-17m -2n}{m+n} = 3, \quad \frac{-4m + 2n}{m+n} = 4. \] Solving for \( m:n \): From the first equation: \[ 6m + 3n = 2(m+n). \] \[ 6m + 3n = 2m + 2n. \] \[ 4m + n = 0. \] \[ n = -4m. \] Substituting into the second equation: \[ -17m -2(-4m) = 3(m - 4m). \] \[ -17m + 8m = 3m - 12m. \] \[ -9m = -9m. \] So, \( m:n = 1:-4 \).
Step 3: Find \( Q(\alpha, \beta, \gamma) \)
Using the external section formula: \[ Q = \frac{mB - nA}{m - n}. \] \[ \alpha = \frac{1(6) - (-4)(3)}{1 + 4} = \frac{6 + 12}{5} = \frac{18}{5}. \] \[ \beta = \frac{1(-17) - (-4)(-2)}{1 + 4} = \frac{-17 - 8}{5} = \frac{-25}{5} = -5. \] \[ \gamma = \frac{1(-4) - (-4)(2)}{1 + 4} = \frac{-4 + 8}{5} = \frac{4}{5}. \]
Step 4: Compute \( \alpha + \beta + \gamma \)
\[ \alpha + \beta + \gamma = \frac{18}{5} + (-5) + \frac{4}{5}. \] \[ = \frac{18}{5} - \frac{25}{5} + \frac{4}{5}. \] \[ = \frac{18 - 25 + 4}{5} = \frac{-3}{5}. \]
Step 5: Conclusion
Thus, the correct answer is: \[ \mathbf{-\frac{3}{5}}. \]
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