Question

If Q \( (\alpha, \beta, \gamma) \) is the harmonic conjugate of the point P(0,-7,1) with respect to the line segment joining the points (2,-5,3) and (-1,-8,0), then \( \alpha - \beta + \gamma = \)

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

If P and Q are harmonic conjugates with respect to line segment AB, then P and Q divide AB internally and externally in the same ratio \(m:n\). Internal division: \( \left(\frac{mx_B+nx_A}{m+n}, \dots\right) \). External division: \( \left(\frac{mx_B-nx_A}{m-n}, \dots\right) \). First, find the ratio in which P divides AB. Then use that ratio for Q's external division.

Answer 1

The Correct Option is A

Let A=(2,-5,3) and B=(-1,-8,0).
Let P=(0,-7,1).
Let Q=\((\alpha, \beta, \gamma)\).
If Q is the harmonic conjugate of P with respect to segment AB, it means that P and Q divide the segment AB internally and externally in the same ratio.
Let P divide AB in ratio \(m:n\).
\( P_x = \frac{m x_B + n x_A}{m+n} \implies 0 = \frac{m(-1) + n(2)}{m+n} \implies -m+2n=0 \implies m=2n \).
So P divides AB internally in ratio \(2n:n = 2:1\).
Let's check: \( P_y = \frac{2(-8)+1(-5)}{2+1} = \frac{-16-5}{3} = \frac{-21}{3} = -7 \).
(Matches y-coordinate of P) \( P_z = \frac{2(0)+1(3)}{2+1} = \frac{0+3}{3} = \frac{3}{3} = 1 \).
(Matches z-coordinate of P) So P divides AB internally in the ratio 2:1.
Then Q must divide AB externally in the ratio 2:1.
The formula for external division of segment AB (from A to B) by Q in ratio \(m:n\) is: \( Q = \left( \frac{mx_B - nx_A}{m-n}, \frac{my_B - ny_A}{m-n}, \frac{mz_B - nz_A}{m-n} \right) \).
Here \(m=2, n=1\).
A=(2,-5,3), B=(-1,-8,0).
\[ \alpha = \frac{2(-1) - 1(2)}{2-1} = \frac{-2-2}{1} = -4 \] \[ \beta = \frac{2(-8) - 1(-5)}{2-1} = \frac{-16+5}{1} = -11 \] \[ \gamma = \frac{2(0) - 1(3)}{2-1} = \frac{0-3}{1} = -3 \] So \( Q = (-4, -11, -3) \).
Thus \( \alpha = -4, \beta = -11, \gamma = -3 \).
We need to find \( \alpha - \beta + \gamma \).
\[ \alpha - \beta + \gamma = (-4) - (-11) + (-3) = -4 + 11 - 3 = 7 - 3 = 4 \] This matches option (1).